# Definite integrals and series involving exponential, rational, and factorial parameters

Robert Reynolds

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Robert Reynolds. Definite integrals and series involving exponential, rational, and factorial parame-
ters. 2024. hal-04596863v3

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# Definite integrals and series involving exponential, rational, and factorial parameters

Robert Reynolds

<figure><img src="image_1.png" /></figure>

## To cite this version:

Robert Reynolds. Definite integrals and series involving exponential, rational, and factorial parame-
ters. 2024. hal-04596863v4

HAL Id: hal-04596863

https://hal.science/hal-04596863v4

Preprint submitted on 18 Sep 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_2.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
# Definite integrals and series involving exponential, rational, and factorial parameters

Robert Reynolds

ABSTRACT. In this compendium, we derive definite integrals, finite and infinite series involving trigonometric functions with exponential, rational, and factorial parameters. These derivations are achieved by applying contour integration to trigonometric formulae and definite integrals.

## 1. Introduction

Series that involve exponential, rational, and factorial terms can be intriguing and complex. They appear frequently in mathematical analysis and applied mathematics [DLMF, (6.16)]. These series are widely used in various applications across different fields. In differential equations, these series often emerge as solutions, providing methods to solve both ordinary and partial differential equations [DLMF, (3.7)]. In mathematical physics, such series are crucial for approximations and expansions, helping to model complex physical phenomena with high precision [DLMF, (18.38)]. Furthermore, in combinatorics, generating functions frequently utilize these series to encode sequences and solve enumeration problems, offering a robust toolset for tackling a broad range of combinatorial structures [DLMF, (18.12)]. In this work we apply the contour integral method in [1] to the equations quoted, to derive theorems and examples that follow. Definite integrals involving finite and infinite series are also derived, where in special cases expressions for the

2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.
Key words and phrases. Special functions, finite series, infinite series, contour integration.
Definite integrals and series involving exponential, rational,
and factorial parameters

Robert Reynolds

ABSTRACT. In this compendium, we derive definite integrals, finite and infi-
nite series involving trigonometric functions with exponential, rational, and
factorial parameters. These derivations are achieved by applying contour in-
tegration to trigonometric formulae and definite integrals.

1. Introduction

Series that involve exponential, rational, and factorial terms can be intriguing and complex. They appear frequently in mathematical analysis and applied mathematics [DLMF, (6.16)]. These series are widely used in various applications across different fields. In differential equations, these series often emerge as solutions, providing methods to solve both ordinary and partial differential equations [DLMF, (3.7)]. In mathematical physics, such series are crucial for approximations and expansions, helping to model complex physical phenomena with high precision [DLMF, (18.38)]. Furthermore, in combinatorics, generating functions frequently utilize these series to encode sequences and solve enumeration problems, offering a robust toolset for tackling a broad range of combinatorial structures [DLMF, (18.12)]. In this work we apply the contour integral method in [1] to the equations quoted, to derive theorems and examples that follow. Definite integrals involving finite and infinite series are also derived, where in special cases expressions for the

2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.
Key words and phrases. Special functions, finite series, infinite series, contour integration.
log-gamma function are derived. The results in this work represent an extension of work found in current literature. The formulae derived involve new series and integrals involving the Hurwitz-Lerch zeta function, Laplace and Mellin transforms, Theta functions, trigonometric functions and other Special functions. A few errata formulae are also derived to add to existing literature. The preparation of this list is to provide researchers an easily accessed document with mathematical formulae to assist in their research where applicable.

## 2. Series involving rational functions

THEOREM 2.1. From eq. (14.8.1) in [2].

$$ (2.1) \quad \sum_{j=1}^{n} \frac{e^{-ij\alpha} \left( (-1+a)^k - (1+a)^k e^{2i(m+j\alpha)} \right) j^p (-n)_j}{j!} \\ = \sum_{j=1}^{p} (-1)^n 2^{-j+n} e^{-\frac{1}{2}i(j+n)(\pi+\alpha)} \left( (-1+a)^k - (1+a)^k e^{i(2m+(j+n)(\pi+\alpha))} \right) \\ (-n)_j \sin^{-j+n}\left(\frac{\alpha}{2}\right) S_p^{(j)} $$

THEOREM 2.2. From eq. (14.17.3) in [2]. The Theta functions.

$$ (2.2) \quad \sum_{j=0}^{\infty} (-1)^j e^{-2ijm} \left( (-1+a-2j)^k - e^{2i(1+2j)m} (1+a+2j)^k \right) q^{\left(\frac{1}{2}+j\right)^2} \\ = \sum_{j=0}^{\infty} (-1)^j e^{-2ijm} \left( (-1+a-2j)^k - e^{2i(1+2j)m} (1+a+2j)^k \right) q^{\frac{1}{4}+j+\frac{j^2}{4}} $$

where $|\mathrm{Re}(q)|<1$.

THEOREM 2.3. From eq. (16.1.3) in [2].

$$ (2.3) \quad \sum_{j=1}^{\infty} \frac{2(-1)^j e^{-i\sqrt{b^2+j^2}m} \left( \left( a - \sqrt{b^2+j^2} \right)^k - e^{2i\sqrt{b^2+j^2}m} \left( a + \sqrt{b^2+j^2} \right)^k \right) \alpha^2}{\sqrt{b^2+j^2}(j^2-\alpha^2)} \\ = \frac{e^{-ibm} \left( (a-b)^k - (a+b)^k e^{2ibm} \right)}{b} $$
log-gamma function are derived. The results in this work represent an extension of work found in current literature. The formulae derived involve new series and integrals involving the Hurwitz-Lerch zeta function, Laplace and Mellin transforms, Theta functions, trigonometric functions and other Special functions. A few errata formulae are also derived to add to existing literature. The preparation of this list is to provide researchers an easily accessed document with mathematical formulae to assist in their research where applicable.

## 2. Series involving rational functions

THEOREM 2.1. From eq. (14.8.1) in [2].

$$ (2.1) \quad \sum_{j=1}^{n} \frac{e^{-ij\alpha} \left( (-1+a)^k - (1+a)^k e^{2i(m+j\alpha)} \right) j^p (-n)_j}{j!} \\ = \sum_{j=1}^{p} (-1)^n 2^{-j+n} e^{-\frac{1}{2}i(j+n)(\pi+\alpha)} \left( (-1+a)^k - (1+a)^k e^{i(2m+(j+n)(\pi+\alpha))} \right) \\ (-n)_j \sin^{-j+n}\left(\frac{\alpha}{2}\right) S_p^{(j)} $$

THEOREM 2.2. From eq. (14.17.3) in [2]. The Theta functions.

$$ (2.2) \quad \sum_{j=0}^{\infty} (-1)^j e^{-2ijm} \left( (-1+a-2j)^k - e^{2i(1+2j)m} (1+a+2j)^k \right) q^{\left(\frac{1}{2}+j\right)^2} \\ = \sum_{j=0}^{\infty} (-1)^j e^{-2ijm} \left( (-1+a-2j)^k - e^{2i(1+2j)m} (1+a+2j)^k \right) q^{\frac{1}{4}+j+\frac{j^2}{4}} $$

where $|\mathrm{Re}(q)|<1$.

THEOREM 2.3. From eq. (16.1.3) in [2].

$$ (2.3) \quad \sum_{j=1}^{\infty} \frac{2(-1)^j e^{-i\sqrt{b^2+j^2}m} \left( \left( a - \sqrt{b^2+j^2} \right)^k - e^{2i\sqrt{b^2+j^2}m} \left( a + \sqrt{b^2+j^2} \right)^k \right) \alpha^2}{\sqrt{b^2+j^2}(j^2-\alpha^2)} \\ = \frac{e^{-ibm} \left( (a-b)^k - (a+b)^k e^{2ibm} \right)}{b} $$
$$ + \frac{e^{-im\sqrt{b^2+\alpha^2}}\pi\alpha \left( - (a - \sqrt{b^2+\alpha^2})^k + e^{2im\sqrt{b^2+\alpha^2}}(a + \sqrt{b^2+\alpha^2})^k \right) \csc(\pi\alpha)}{\sqrt{b^2+\alpha^2}} $$

where $Re(a) > 3\pi$ in order for the series to converge.

**THEOREM 2.4.** *From eq. (17.11.3) in [2]. An application of Abel inverse relation and contour integration.*

$$ (2.4) \qquad \sum_{j=0}^{\infty} \frac{(e^{-ijx}(-1+a)^k + (1+a)^k e^{i(2m+jx)}) (\frac{1}{2}j(\pi-2x)+z)^{-1+j}}{j!} \\ = \frac{e^{-iz}((-1+a)^k + (1+a)^k e^{2i(m+z)})}{z} $$

where $4/5 < Re(x) < 12/5$.

**THEOREM 2.5.** *From eq. (17.11.3) in [2] and pp. 97 in [3].*

$$ (2.5) \qquad \sum_{p=0}^{\infty} \frac{e^{-ipx} \left( \frac{1}{2}p(\pi - 2x) + z \right)^{-1+p} \left( \Phi(-e^{2i(m+z)}, -k, a) + e^{2i(m+px)} \Phi(-e^{2i(m+z)}, -k, 1+a) \right)}{p!} \\ = \frac{a^k e^{-iz}}{z} $$

**EXAMPLE 2.6.**

$$ (2.6) \qquad \prod_{p=0}^{\infty} a^{\frac{2-p}{\Gamma(1+p)}e^{i\frac{px-2z}{(p(\pi-2x)+2z)-1+p}} \left( \frac{\Gamma\left(\frac{a}{2}\right)}{\sqrt{2\Gamma\left(\frac{1+a}{2}\right)}} \right)^{\frac{ie^{-iz}\left(\frac{1}{2}p(\pi-2x)+z\right)^{-1+p}\sin(px-z)}{\Gamma(1+p)}}} = a^{\frac{e^{-iz}}{2z}} $$

**EXAMPLE 2.7.**

$$ (2.7) \qquad \sum_{k=0}^{\infty} \frac{(kx + z)^{k-1} e^{ik(\frac{\pi}{2} - x)}}{k!} = \frac{e^{iz}}{z} $$

where $|Re(x)|< 1$.

**THEOREM 2.8.** *Alternate form. Abel-Lerch series.*
$$ + \frac{e^{-im\sqrt{b^2+\alpha^2}}\pi\alpha \left( - (a - \sqrt{b^2+\alpha^2})^k + e^{2im\sqrt{b^2+\alpha^2}}(a + \sqrt{b^2+\alpha^2})^k \right) \csc(\pi\alpha)}{\sqrt{b^2+\alpha^2}} $$

where $Re(a) > 3\pi$ in order for the series to converge.

**THEOREM 2.4.** *From eq. (17.11.3) in [2]. An application of Abel inverse relation and contour integration.*

$$ (2.4) \qquad \sum_{j=0}^{\infty} \frac{e^{-ijx}(-1+a)^k + (1+a)^k e^{i(2m+jx)}}{j!} \left(\frac{1}{2}j(\pi-2x)+z\right)^{-1+j} \\ = \frac{e^{-iz}((-1+a)^k + (1+a)^k e^{2i(m+z)})}{z} $$

where $4/5 < Re(x) < 12/5$.

**THEOREM 2.5.** *From eq. (17.11.3) in [2] and pp. 97 in [3].*

$$ (2.5) \qquad \sum_{p=0}^{\infty} \frac{e^{-ipx} \left( \frac{1}{2}p(\pi - 2x) + z \right)^{-1+p} \left( \Phi(-e^{2i(m+z)}), -k, a \right) + e^{2i(m+px)} \Phi(-e^{2i(m+z)}), -k, 1+a \right)}{p!} \\ = \frac{a^k e^{-iz}}{z} $$

**EXAMPLE 2.6.**

$$ (2.6) \qquad \prod_{p=0}^{\infty} a^{\frac{2-p}{\Gamma(1+p)} e^{i \frac{px-2z}{(p(\pi-2x)+2z)-1+p}} \left( \frac{\Gamma\left(\frac{a}{2}\right)}{\sqrt{2\Gamma\left(\frac{1+a}{2}\right)}} \right)^{\frac{e^{-iz} \left( \frac{1}{2}p(\pi-2x)+z \right)^{-1+p} \sin(px-z)}{\Gamma(1+p)}}} = a^{\frac{e^{-iz}}{2z}} $$

**EXAMPLE 2.7.**

$$ (2.7) \qquad \sum_{k=0}^{\infty} \frac{(kx+z)^{k-1} e^{ik(\frac{\pi}{2}-x)}}{k!} = \frac{e^{iz}}{z} $$

where $|Re(x)|< 1$.

**THEOREM 2.8.** *Alternate form. Abel-Lerch series.*
$$
(2.8) \qquad \sum_{k=0}^{\infty} \frac{(k+u)^{-1+k} (e^{-x} x)^k \left( \Phi(-e^{2im-2ux}, s, a) + e^{2im+ik\pi+2kx} \Phi(-e^{2im-2ux}, s, 1+a) \right)}{k!} \\
\qquad \qquad = \frac{a^{-s} e^{ux}}{u}
$$

where $|\mathrm{Re}(x)|<1, |\mathrm{Re}(u)|<1, |\mathrm{Re}(m)|<1$.

THEOREM 2.9. *The Polygamma function definition.*

$$
(2.9) \quad (-1)^k \Gamma(k+1) \sum_{j=0}^{n} j \left( \frac{1}{(m-j)^{k+1}} - \frac{1}{(m+j)^{k+1}} \right) \\
\qquad = \int_0^1 \frac{t^{-m+n} \left( -\log^{-1+k}\left(\frac{1}{t}\right) (k+m\log\left(\frac{1}{t}\right)) + t^{2m}\log^{-1+k}(t)(k+m\log(t)) \right)}{-1+t} dt \\
\qquad \quad - (2\pi i)^k (k \mathrm{Li}_{1-k}(e^{2im\pi}) + 2im\pi \mathrm{Li}_{-k}(e^{2im\pi}))
$$

where $\mathrm{Re}(k) > \frac{3}{2}$.

EXAMPLE 2.10.

$$
\begin{align*}
(2.10) \quad & \sum_{p=0}^{\infty} \sum_{j=0}^{n} \frac{1}{(j^2 - m^2)^{p+1} (-1 + k - p)! (k - p)!} \\
& \qquad \times (j((j-m)^p - (-1)^p(j+m)^p) - m((j-m)^p + (-1)^p(j+m)^p)) \\
& \qquad \left(\binom{-1}{p}\right) \log^{-1+k-p}(a)((k-p)! + m(-1+k-p)! \log(a)) \\
&= \sum_{j=0}^{n} \frac{a^{-m}}{\Gamma(1+k)} ((-1)^k (-j+m)^{-1-k} (a^j j k \Gamma(k, (j-m) \log(a)) + a^m m ((j-m) \log(a))^k) \\
& \quad + a^{-j} (-j-m)^{-1-k} (jk \Gamma(k, -((j+m) \log(a))) - a^{j+m} m ( -((j+m) \log(a)))^k)
\end{align*}
$$

where $1 < \mathrm{Re}(k) < 5/2, \mathrm{Re}(a) < -3\pi, \mathrm{Im}(m) \neq 0$.

THEOREM 2.11. *From (2.2.4.12) in [4]. Extended Brychkov definite integral.*

$$
(2.11) \quad \sum_{j=0}^{n} \sum_{p=0}^{j} \frac{m^{j-p} (b(j-n) + \alpha)^{-s} \binom{j}{p} \binom{n}{j} \log^{-p}(a)}{(k-p)!}
$$
$$
(2.8) \qquad \sum_{k=0}^{\infty} \frac{(k+u)^{-1+k} (e^{-x}x)^k \left( \Phi(-e^{2im-2ux}, s, a) + e^{2im+ik\pi+2kx} \Phi(-e^{2im-2ux}, s, 1+a) \right)}{k!} \\
\qquad \qquad \qquad \qquad = \frac{a^{-s}e^{ux}}{u}
$$

where $|\mathrm{Re}(x)|<1, |\mathrm{Re}(u)|<1, |\mathrm{Re}(m)|<1$.

THEOREM 2.9. *The Polygamma function definition.*

$$
(2.9) \quad (-1)^k \Gamma(k+1) \sum_{j=0}^{n} j \left( \frac{1}{(m-j)^{k+1}} - \frac{1}{(m+j)^{k+1}} \right) \\
\qquad = \int_0^1 \frac{t^{-m+n} \left( -\log^{-1+k}\left(\frac{1}{t}\right) (k+m\log\left(\frac{1}{t}\right)) + t^{2m}\log^{-1+k}(t)(k+m\log(t)) \right)}{-1+t} dt \\
\qquad \quad - (2\pi i)^k (kLi_{1-k}(e^{2im\pi}) + 2im\pi Li_{-k}(e^{2im\pi}))
$$

where $\mathrm{Re}(k) > \frac{3}{2}$.

EXAMPLE 2.10.

$$
\begin{align*}
(2.10) \quad & \sum_{p=0}^{\infty} \sum_{j=0}^{n} \frac{1}{(j^2 - m^2)^{p+1} (-1 + k - p)! (k - p)!} \\
& \qquad \times (j((j-m)^p - (-1)^p(j+m)^p) - m((j-m)^p + (-1)^p(j+m)^p)) \\
& \qquad \qquad \binom{-1}{p} \log^{-1+k-p}(a)((k-p)! + m(-1+k-p)! \log(a)) \\
&= \sum_{j=0}^{n} \frac{a^{-m}}{\Gamma(1+k)} ((-1)^k (-j+m)^{-1-k} (a^j j k \Gamma(k, (j-m)\log(a)) + a^m m ((j-m)\log(a))^k) \\
& \quad + a^{-j} (-j-m)^{-1-k} (jk\Gamma(k, -((j+m)\log(a))) - a^{j+m} m ( -((j+m)\log(a)))^k)
\end{align*}
$$

where $1 < \mathrm{Re}(k) < 5/2, \mathrm{Re}(a) < -3\pi, \mathrm{Im}(m) \neq 0$.

THEOREM 2.11. *From (2.2.4.12) in [4]. Extended Brychkov definite integral.*

$$
(2.11) \quad \sum_{j=0}^{n} \sum_{p=0}^{j} \frac{m^{j-p} (b(j-n) + \alpha)^{-s} \binom{j}{p} \binom{n}{j} \log^{-p}(a)}{(k-p)!}
$$
$$= \sum_{j=0}^{n} \sum_{l=0}^{n-j} \frac{m^{-j-l+n} (-bj + \alpha)^{-s} \binom{n}{j} \binom{-j+n}{l} \log^{-l}(a)}{(k-l)!}$$

THEOREM 2.12. From eq. (16.1.4) in [2].

$$ (2.12) \qquad \sum_{j=1}^{\infty} \frac{(-1)^j e^{-i\sqrt{b^2+j^2}m} \left( (-1)^{-k} \left(\sqrt{b^2+j^2}-a\right)^k - e^{2i\sqrt{b^2+j^2}m} \left(a+\sqrt{b^2+j^2}\right)^k \right)}{j^2\sqrt{b^2+j^2}} \\ = \frac{e^{-ibm}}{12b^3} \left( \begin{aligned} & -(-1)^{-k}(-a+b)^{-1+k} \left(3(a+b(-1+k)+i(a-b)bm)+b^2(-a+b)\pi^2\right) \\ & \qquad + (a+b)^{-1+k}e^{2ibm} \left(-3(a+b-bk)+3ib(a+b)m+b^2(a+b)\pi^2\right) \end{aligned} \right) $$

### 3. Series in terms of definite integrals

THEOREM 3.1. From eq. (1.12.28) [5]. An application of the definite integral involving Bessel functions is in the study on deep-sea water waves caused by a local disturbance on or beneath the surface [6].

$$ (3.1) \qquad \int_0^\infty e^{-imx} Y_0(x\alpha) \left( \begin{aligned} & \left((-ix + \log(a))\right)^k + e^{2imx}(ix + \log(a))^k \\ & \qquad - \sum_{j=0}^\infty \sum_{p=0}^\infty 2(-1)^j m^{-1-2j-p} \alpha^{2j} \binom{-\frac{1}{2}}{j} \binom{-1-2j}{p} (1+k-p)_p \log^{k-p}(a) \end{aligned} \right) dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty 2(-1)^j m^{-1-2j-p} \alpha^{2j} \binom{-\frac{1}{2}}{j} \binom{-1-2j}{p} (1+k-p)_p \log^{k-p}(a) $$

where $Re(k) \le -3/2, Re(a) < -1, Re(m) < -2, |Re(\alpha)| < Re(m).$

THEOREM 3.2. From eq. (2.4.14) in [5].

$$ (3.2) \qquad \int_0^\infty \frac{ie^{-imx-nx} \left( \begin{aligned} & \left((-ix + \log(a))\right)^k - e^{2imx}(ix + \log(a))^k \\ & \qquad + \sum_{j=1}^{n-1} (a^{ij-m}(ij-m)^{-1-k}\Gamma(1+k,-((-ij+m)\log(a))) \\ & \qquad + a^{-ij-m}(-ij-m)^{-1-k}\Gamma(1+k,-((ij+m)\log(a)))) \\ & \qquad - \pi \left( 2^{1+k} e^{2m\pi} \pi^k \Phi \left( e^{2m\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) + \log^k(a) \right) \end{aligned} \right) dx \\ = a^{-m}(-m)^{-1-k}\Gamma(1+k,-m\log(a)) $$
$$= \sum_{j=0}^{n} \sum_{l=0}^{n-j} \frac{m^{-j-l+n} (-bj + \alpha)^{-s} \binom{n}{j} \binom{-j+n}{l} \log^{-l}(a)}{(k-l)!}$$

THEOREM 2.12. From eq. (16.1.4) in [2].

$$(2.12) \qquad \sum_{j=1}^{\infty} \frac{(-1)^j e^{-i\sqrt{b^2+j^2}m} \left( (-1)^{-k} \left(\sqrt{b^2+j^2}-a\right)^k - e^{2i\sqrt{b^2+j^2}m} \left(a+\sqrt{b^2+j^2}\right)^k \right)}{j^2\sqrt{b^2+j^2}} \\ = \frac{e^{-ibm}}{12b^3} \left( \begin{aligned} & -(-1)^{-k}(-a+b)^{-1+k} \left(3(a+b(-1+k)+i(a-b)bm)+b^2(-a+b)\pi^2\right) \\ & \qquad + (a+b)^{-1+k}e^{2ibm} \left(-3(a+b-bk)+3ib(a+b)m+b^2(a+b)\pi^2\right) \end{aligned} \right)$$

### 3. Series in terms of definite integrals

THEOREM 3.1. From eq. (1.12.28) [5]. An application of the definite integral involving Bessel functions is in the study on deep-sea water waves caused by a local disturbance on or beneath the surface [7].

$$(3.1) \qquad \int_0^\infty e^{-imx} Y_0(x\alpha) \left( \begin{aligned} & \left((-ix + \log(a))\right)^k + e^{2imx}(ix + \log(a))^k \\ & \qquad - \sum_{j=0}^\infty \sum_{p=0}^\infty 2(-1)^j m^{-1-2j-p} \alpha^{2j} \binom{-\frac{1}{2}}{j} \binom{-1-2j}{p} (1+k-p)_p \log^{k-p}(a) \end{aligned} \right) dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty 2(-1)^j m^{-1-2j-p} \alpha^{2j} \binom{-\frac{1}{2}}{j} \binom{-1-2j}{p} (1+k-p)_p \log^{k-p}(a)$$

where $Re(k) \le -3/2, Re(a) < -1, Re(m) < -2, |Re(\alpha)| < Re(m).$

THEOREM 3.2. From eq. (2.4.14) in [5].

$$(3.2) \qquad \int_0^\infty \frac{ie^{-imx-nx} \left( \begin{aligned} & \left((-ix + \log(a))\right)^k - e^{2imx}(ix + \log(a))^k \\ & \qquad + \sum_{j=1}^{n-1} (a^{ij-m}(ij-m)^{-1-k}\Gamma(1+k,-((-ij+m)\log(a))) \\ & \qquad + a^{-ij-m}(-ij-m)^{-1-k}\Gamma(1+k,-((ij+m)\log(a)))) \\ & \qquad - \pi \left( 2^{1+k} e^{2m\pi} \pi^k \Phi \left( e^{2m\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) + \log^k(a) \right) \end{aligned} \right) dx \\ = a^{-m}(-m)^{-1-k}\Gamma(1+k,-m\log(a))$$
where $Im(a) \le 0$.

EXAMPLE 3.3.

$$
\begin{align*}
& (3.3) \quad \int_0^\infty \frac{e^{x-nx} \cosh(mx) ((-ix + \log(a))^k - (ix + \log(a))^k)}{-1 + e^x} dx \\
&= -\frac{1}{2}i (a^{-im}(-im)^{-1-k}\Gamma(1+k, -im \log(a)) + a^{im}(im)^{-1-k}\Gamma(1+k, im \log(a))) \\
& \quad -\sum_{j=1}^{n-1} \frac{1}{2}ia^{-ij-im} (a^{2ij}(ij-im)^{-1-k}\Gamma(1+k, i(j-m) \log(a)) \\
& \quad \quad + (-ij-im)^{-1-k}\Gamma(1+k, -((ij+im) \log(a))) \\
& \quad \quad + a^{2im} ((-ij+im)^{-1-k}\Gamma(1+k, (-ij+im) \log(a)) \\
& \quad \quad \quad + a^{2ij}(ij+im)^{-1-k}\Gamma(1+k, (ij+im) \log(a))) \\
& \quad -\frac{1}{2}i \left( -e^{-2im\pi}(2\pi)^{1+k} \left( \Phi \left( e^{-2im\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) \right. \right. \\
& \quad \quad \quad \left. \left. + e^{4im\pi}\Phi \left( e^{2im\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) \right) - 2\pi \log^k(a) \right)
\end{align*}
$$

EXAMPLE 3.4. Mellin transform.

$$
\begin{align*}
& (3.4) \quad \int_0^\infty e^{-nx} x^{-1+s} \cosh(mx) \left(1 + \coth\left(\frac{x}{2}\right)\right) dx \\
&= -m^{-s} \Gamma(s) + \sum_{j=1}^{n-1} \frac{1}{2} \left( - \left( \left((-i(j-m))^s + (i(j-m))^s\right) (j-m)^{-s} \right) \right. \\
& \qquad \left. - \left( (j+m)^2 \right)^{-s} \left( \left((-i(j+m))^s + (i(j+m))^s\right) \right) \right) \Gamma(s) \sec\left(\frac{\pi s}{2}\right) \\
& \qquad + 2^{-1+s} \pi^s \left( \mathrm{Li}_{1-s}\left(e^{-2im\pi}\right) + \mathrm{Li}_{1-s}\left(e^{2im\pi}\right) \right) \sec\left(\frac{\pi s}{2}\right)
\end{align*}
$$

EXAMPLE 3.5.

$$
\begin{align*}
& (3.5) \quad \int_0^\infty \frac{e^{x-nx} \log(x^2 + \log^2(a)) \sinh(mx)}{-1 + e^x} dx \\
&= \frac{1}{2m} (a^{-im}\Gamma(0, -im \log(a)) + a^{im}\Gamma(0, im \log(a)) - \log(-im) \\
& \phantom{= \frac{1}{2m} (a^{-im}\Gamma(0, -im \log(a)) + a^{im}\Gamma(0, im \log(a))} - \log(im) + \log(-im \log(a)) + \log(im \log(a)))
\end{align*}
$$
where $Im(a) \le 0$.

EXAMPLE 3.3.

$$
\begin{align*}
& (3.3) \quad \int_0^\infty \frac{e^{x-nx} \cosh(mx) ((-ix + \log(a))^k - (ix + \log(a))^k)}{-1 + e^x} dx \\
&= -\frac{1}{2}i (a^{-im}(-im)^{-1-k}\Gamma(1+k, -im \log(a)) + a^{im}(im)^{-1-k}\Gamma(1+k, im \log(a))) \\
& \quad -\sum_{j=1}^{n-1} \frac{1}{2}ia^{-ij-im} (a^{2ij}(ij-im)^{-1-k}\Gamma(1+k, i(j-m) \log(a)) \\
& \quad \quad + (-ij-im)^{-1-k}\Gamma(1+k, -((ij+im) \log(a))) \\
& \quad \quad + a^{2im} ((-ij+im)^{-1-k}\Gamma(1+k, (-ij+im) \log(a)) \\
& \quad \quad \quad + a^{2ij}(ij+im)^{-1-k}\Gamma(1+k, (ij+im) \log(a))) \\
& \quad \quad -\frac{1}{2}i \left( -e^{-2im\pi}(2\pi)^{1+k} \left( \Phi \left( e^{-2im\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) \right. \right. \\
& \quad \quad \quad \left. \left. + e^{4im\pi}\Phi \left( e^{2im\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) \right) - 2\pi \log^k(a) \right)
\end{align*}
$$

EXAMPLE 3.4. Mellin transform.

$$
\begin{align*}
& (3.4) \quad \int_0^\infty e^{-nx} x^{-1+s} \cosh(mx) \left(1 + \coth\left(\frac{x}{2}\right)\right) dx \\
&= -m^{-s} \Gamma(s) + \sum_{j=1}^{n-1} \frac{1}{2} \left( - \left( \left((-i(j-m))^s + (i(j-m))^s\right) (j-m)^{-s} \right) \right. \\
& \qquad \left. - \left( (j+m)^2 \right)^{-s} \left( \left((-i(j+m))^s + (i(j+m))^s\right) \right) \right) \Gamma(s) \sec\left(\frac{\pi s}{2}\right) \\
& \qquad + 2^{-1+s} \pi^s \left( \operatorname{Li}_{1-s}(e^{-2im\pi}) + \operatorname{Li}_{1-s}(e^{2im\pi}) \right) \sec\left(\frac{\pi s}{2}\right)
\end{align*}
$$

EXAMPLE 3.5.

$$
\begin{align*}
& (3.5) \quad \int_0^\infty \frac{e^{x-nx} \log(x^2 + \log^2(a)) \sinh(mx)}{-1 + e^x} dx \\
&= \frac{1}{2m} (a^{-im}\Gamma(0, -im \log(a)) + a^{im}\Gamma(0, im \log(a)) - \log(-im) \\
& \phantom{= \frac{1}{2m} (a^{-im}\Gamma(0, -im \log(a)) + a^{im}\Gamma(0, im \log(a))} - \log(im) + \log(-im \log(a)) + \log(im \log(a)))
\end{align*}
$$
$$
\begin{aligned}
& -\sum_{j=1}^{n-1} \frac{1}{2(j-m)(j+m)} \left( a^{-i(j-m)}(j+m)E_1(-i(j-m)\log(a)) \right. \\
& \qquad \left. +a^{i(j-m)}(j+m)E_1(i(j-m)\log(a)) + a^{-i(j+m)}(-j+m)E_1(-i(j+m)\log(a)) \right. \\
& \qquad \left. -a^{i(j+m)}(j-m)E_1(i(j+m)\log(a)) - (j+m)\log(-i(j-m)) - (j+m)\log(i(j-m)) \right. \\
& \qquad \left. +(j-m)\log(-i(j+m)) + (j-m)\log(i(j+m)) + (j+m)\log(-i(j-m)\log(a)) \right. \\
& \qquad \left. +(j+m)\log(i(j-m)\log(a)) + (-j+m)\log(-i(j+m)\log(a)) + (-j+m)\log(i(j+m)\log(a)) \right) \\
& \qquad + \pi \left( -\cot(m\pi)\log(2\pi) + ie^{-2im\pi} \left( \Phi' \left( e^{-2im\pi}, 0, 1 + \frac{\log(a)}{2\pi} \right) \right. \right. \\
& \qquad \left. \left. -e^{4im\pi}\Phi' \left( e^{2im\pi}, 0, 1 + \frac{\log(a)}{2\pi} \right) \right) \right)
\end{aligned}
$$

THEOREM 3.6. From Eq's. (2.9.42) and (2.9.43) in [5].

$$
\begin{aligned}
& (3.6) \quad \int_0^\infty e^{ix(-m+\pi x)} \coth(\pi x) \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) dx \\
& = i \left( -2e^m \Phi(-e^m, -k, 1 + \log(a)) + \log^k(a) \right) \\
& \quad - \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{i \left(-\frac{1}{4}\right)^j i^j m^{-l+2p} \pi^{j-2p} \binom{j}{p} \binom{2p}{l} \log^{-l}(a)}{j!} \\
& \quad \times \left( \log^k(a) - 2e^m \Phi(-e^m, -k+l, 1+\log(a)) \log^l(a) \right) (1+k-l)_l
\end{aligned}
$$

where $Re(a) \le -3\pi, Re(m) \le -3/2$.

EXAMPLE 3.7.

$$
\begin{aligned}
& (3.7) \quad \int_0^\infty \frac{e^{i\pi x^2} \coth(\pi x)(x \cos(mx) - \log(a) \sin(mx))}{x^2 + \log^2(a)} dx \\
& = \frac{1}{2} \left( -2e^m \Phi(-e^m, 1, 1+\log(a)) + \frac{1}{\log(a)} \right) \\
& \quad + \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{(-1)^j i^j 2^{-1-2j} m^{-l+2p} \pi^{j-2p} \binom{j}{p} \binom{2p}{l} \log^{-1-l}(a)}{j!} \\
& \quad \times \left( -1 + 2e^m \Phi(-e^m, 1+l, 1+\log(a)) \log^{1+l}(a) \right) (-l)_l
\end{aligned}
$$

EXAMPLE 3.8.
$$
\begin{aligned}
& -\sum_{j=1}^{n-1} \frac{1}{2(j-m)(j+m)} \left( a^{-i(j-m)}(j+m)E_1(-i(j-m)\log(a)) \right. \\
& \qquad \left. +a^{i(j-m)}(j+m)E_1(i(j-m)\log(a)) + a^{-i(j+m)}(-j+m)E_1(-i(j+m)\log(a)) \right. \\
& \qquad \left. -a^{i(j+m)}(j-m)E_1(i(j+m)\log(a)) - (j+m)\log(-i(j-m)) - (j+m)\log(i(j-m)) \right. \\
& \qquad \left. +(j-m)\log(-i(j+m)) + (j-m)\log(i(j+m)) + (j+m)\log(-i(j-m)\log(a)) \right. \\
& \qquad \left. +(j+m)\log(i(j-m)\log(a)) + (-j+m)\log(-i(j+m)\log(a)) + (-j+m)\log(i(j+m)\log(a)) \right) \\
& \qquad + \pi \left( -\cot(m\pi)\log(2\pi) + ie^{-2im\pi} \left( \Phi' \left( e^{-2im\pi}, 0, 1 + \frac{\log(a)}{2\pi} \right) \right. \right. \\
& \qquad \left. \left. -e^{4im\pi}\Phi' \left( e^{2im\pi}, 0, 1 + \frac{\log(a)}{2\pi} \right) \right) \right)
\end{aligned}
$$

THEOREM 3.6. From Eq's. (2.9.42) and (2.9.43) in [5].

$$
\begin{aligned}
& (3.6) \quad \int_0^\infty e^{ix(-m+\pi x)} \coth(\pi x) \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) dx \\
& \qquad = i \left( -2e^m \Phi(-e^m, -k, 1 + \log(a)) + \log^k(a) \right) \\
& \qquad - \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{i \left(-\frac{1}{4}\right)^j i^j m^{-l+2p} \pi^{j-2p} \binom{j}{p} \binom{2p}{l} \log^{-l}(a)}{j!} \\
& \qquad \times \left( \log^k(a) - 2e^m \Phi(-e^m, -k+l, 1+\log(a)) \log^l(a) \right) (1+k-l)_l
\end{aligned}
$$

where $Re(a) \le -3\pi, Re(m) \le -3/2$.

EXAMPLE 3.7.

$$
\begin{aligned}
& (3.7) \quad \int_0^\infty \frac{e^{i\pi x^2} \coth(\pi x) (x \cos(mx) - \log(a) \sin(mx))}{x^2 + \log^2(a)} dx \\
& \qquad = \frac{1}{2} \left( -2e^m \Phi(-e^m, 1, 1 + \log(a)) + \frac{1}{\log(a)} \right) \\
& \qquad + \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{(-1)^j i^{2-1-2j} m^{-l+2p} \pi^{j-2p} \binom{j}{p} \binom{2p}{l} \log^{-1-l}(a)}{j!} \\
& \qquad \times \left( -1 + 2e^m \Phi(-e^m, 1+l, 1+\log(a)) \log^{1+l}(a) \right) (-l)_l
\end{aligned}
$$

EXAMPLE 3.8.
$$
(3.8) \quad \int_0^\infty \frac{e^{i\pi x^2} \coth(\pi x) \left(x \cos\left(\frac{\pi x}{2}\right) + \pi \sin\left(\frac{\pi x}{2}\right)\right)}{\pi^2 + x^2} dx \\
= \frac{1}{2} \left( \frac{1}{\pi} - 2e^{-\frac{\pi}{2}} \Phi(-e^{-\frac{\pi}{2}}, 1, 1 + \pi) \right) \\
- \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{(-1)^{j-l+2p} i j 2^{-1-2j+l-2p} \pi^{j-2l} \binom{j}{p} \binom{2p}{l} \left(\frac{1}{\pi} - 2e^{-\frac{\pi}{2}} \pi^l \Phi(-e^{-\frac{\pi}{2}}, 1+l, 1+\pi)\right) (-l)_l}{j!}
$$

EXAMPLE 3.9.

$$
(3.9) \quad \int_0^\infty e^{\frac{1}{2}ix(\pi+2\pi x)} \left( \frac{1}{3-ix} - \frac{e^{-i\pi x}}{3+ix} \right) \coth(\pi x) dx \\
= \frac{1}{3}i \left( -1 + 3e^{\pi/2} - 6e^\pi + 6e^{\frac{3\pi}{2}} \log \left( 1 + e^{-\frac{\pi}{2}} \right) \right) \\
- \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{i(-1)^{j-l+2p} ij 2^{-2j+l-2p} 3^{-l} \pi^{j-l} \binom{j}{p} \binom{2p}{l} \left( \frac{1}{3} - 2 \right.}{j!} \left. 3^l e^{-\frac{\pi}{2}} \Phi(-e^{-\frac{\pi}{2}}, 1+l, 4) \right) (-l)_l
$$

THEOREM 3.10. From Eq's. (2.9.44) and (2.9.45) in [5].

$$
(3.10) \quad \int_0^\infty e^{\frac{ix(-m\pi+b^2x)}{\pi}} \operatorname{csch}(bx) \left( (-ix + \log(a))^k - e^{2imx}(ix + \log(a))^k \right) dx \\
= \frac{i\pi \left( 2\left(\frac{1}{b}\right)^k e^{\frac{m\pi}{b}} \pi^k \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \log^k(a) \right)}{b} \\
- \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i(-1)^j ij 2^{1-2jb-1-2j-k+p} e^{\frac{m\pi}{2b}} m^{2j-p} \pi^{1+j+k-p} \binom{2j}{p} (1+k-p)_p}{j!} \\
\times \Phi\left(e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right)
$$

where Re(a) ≤ − 1.

EXAMPLE 3.11.

$$
(3.11) \quad \int_0^\infty e^{\frac{ib^2x^2}{\pi}} \cos(mx)\operatorname{csch}(bx) \left( (-ix + \log(a))^k - (ix + \log(a))^k \right) dx \\
= \frac{i\pi \left( \left(\frac{1}{b}\right)^k e^{-\frac{m\pi}{b}} \pi^k \Phi\left(e^{-\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \left(\frac{1}{b}\right)^k e^{\frac{m\pi}{b}} \pi^k \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \log^k(a) \right)}{b} \\
- \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i \left(-\frac{1}{4}\right)^j i^j b^{1-2j-k+p} e^{-\frac{m\pi}{2b}} (-m^2)^{-p} \pi^{1+j+k-p} \binom{2j}{p}}{j!}
$$
$$
(3.8) \quad \int_0^\infty \frac{e^{i\pi x^2} \coth(\pi x) \left(x \cos\left(\frac{\pi x}{2}\right) + \pi \sin\left(\frac{\pi x}{2}\right)\right)}{\pi^2 + x^2} dx \\
= \frac{1}{2} \left( \frac{1}{\pi} - 2e^{-\frac{\pi}{2}} \Phi(-e^{-\frac{\pi}{2}}, 1, 1 + \pi) \right) \\
- \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{(-1)^{j-l+2p} i j 2^{-1-2j+l-2p} \pi^{j-2l} \binom{j}{p} \binom{2p}{l} \left(\frac{1}{\pi} - 2e^{-\frac{\pi}{2}} \pi^l \Phi(-e^{-\frac{\pi}{2}}, 1+l, 1+\pi)\right) (-l)_l}{j!}
$$

EXAMPLE 3.9.

$$
(3.9) \quad \int_0^\infty e^{\frac{1}{2}ix(\pi+2\pi x)} \left( \frac{1}{3-ix} - \frac{e^{-i\pi x}}{3+ix} \right) \coth(\pi x) dx \\
= \frac{1}{3}i \left( -1 + 3e^{\pi/2} - 6e^\pi + 6e^{\frac{3\pi}{2}} \log \left( 1 + e^{-\frac{\pi}{2}} \right) \right) \\
- \sum_{j=0}^\infty \sum_{p=0}^j \sum_{l=0}^{2p} \frac{i(-1)^{j-l+2p} ij 2^{-2j+l-2p} 3^{-l} \pi^{j-l} \binom{j}{p} \binom{2p}{l} \left( \frac{1}{3} - 2 \right.}{j!} \left. 3^l e^{-\frac{\pi}{2}} \Phi(-e^{-\frac{\pi}{2}}, 1+l, 4) \right) (-l)_l
$$

THEOREM 3.10. From Eq's. (2.9.44) and (2.9.45) in [5].

$$
(3.10) \quad \int_0^\infty e^{\frac{ix(-m\pi+b^2x)}{\pi}} \operatorname{csch}(bx) \left( (-ix + \log(a))^k - e^{2imx}(ix + \log(a))^k \right) dx \\
= \frac{i\pi \left( 2\left(\frac{1}{b}\right)^k e^{\frac{m\pi}{b}} \pi^k \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \log^k(a) \right)}{b} \\
- \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i(-1)^j ij 2^{1-2jb-1-2j-k+p} e^{\frac{m\pi}{2b}} m^{2j-p} \pi^{1+j+k-p} \binom{2j}{p} (1+k-p)_p}{j!} \\
\times \Phi\left(e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right)
$$

where Re(a) ≤ − 1.

EXAMPLE 3.11.

$$
(3.11) \quad \int_0^\infty e^{\frac{ib^2x^2}{\pi}} \cos(mx)\operatorname{csch}(bx) \left( (-ix + \log(a))^k - (ix + \log(a))^k \right) dx \\
= \frac{i\pi \left( \left(\frac{1}{b}\right)^k e^{-\frac{m\pi}{b}} \pi^k \Phi\left(e^{-\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \left(\frac{1}{b}\right)^k e^{\frac{m\pi}{b}} \pi^k \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) + \log^k(a) \right)}{b} \\
- \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i \left(-\frac{1}{4}\right)^j i^j b^{1-2j-k+p} e^{-\frac{m\pi}{2b}} (-m^2)^{-p} \pi^{1+j+k-p} \binom{2j}{p}}{j!}
$$
$$ \times \left( (-m)^{2j} m^p \Phi \left( e^{-\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b \log(a)}{\pi} \right) + e^{\frac{m\pi}{b}} (-m)^p m^{2j} \Phi \left( e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b \log(a)}{\pi} \right) \right) (1+k-p)_p $$

where $Re(a) \le -1$.

### EXAMPLE 3.12.

$$ \begin{aligned} \int_0^\infty \frac{e^{\frac{4ix^2}{\pi}} x \cos\left(\frac{x}{2}\right) \operatorname{csch}(x) \operatorname{sech}(x)}{-\pi^2 + x^2} dx = & -\frac{i}{2} + \left(\frac{1}{5} - \frac{2i}{5}\right) e^{-\frac{\pi}{4}} {}_2F_1(1, 1+2i; 2+2i; e^{-\frac{\pi}{4}}) \\ & + \left(\frac{1}{5} - \frac{2i}{5}\right) e^{\pi/4} {}_2F_1(1, 1+2i; 2+2i; e^{\pi/4}) \\ & - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i^j 4^{-3j+p} p e^{-\frac{\pi}{8}} (-\pi)^{j-p} \binom{2j}{p}}{j!} \\ & \times \left( (-1)^{2j} \Phi\left(e^{-\frac{\pi}{4}}, 1+p, \frac{1}{2}+2i\right) + (-1)^p e^{\pi/4} \Phi\left(e^{\pi/4}, 1+p, \frac{1}{2}+2i\right) \right) (-p)_p \end{aligned} $$

### EXAMPLE 3.13.

$$ \begin{aligned} & \int_0^\infty e^{\frac{ib^2x^2}{\pi}} \operatorname{csch}(bx) \left( \left(-ix + \log(a)\right)^k + \left(ix + \log(a)\right)^k \right) \sin(mx) dx \\ = & \left(\frac{1}{b}\right)^{1+k} e^{-\frac{m\pi}{b}} \pi^{1+k} \left( \Phi\left(e^{-\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) - e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) \right) \\ & + \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{(-\frac{1}{4})^j i^j b^{1-2j-k+p} p e^{-\frac{m\pi}{2b}}}{j!} \\ & \times (-m^2)^{-p} \pi^{1+j+k-p} \binom{2j}{p} \left( \begin{aligned} & -(-m)^{2j} m^p \Phi\left(e^{-\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right) \\ & + e^{\frac{m\pi}{b}} (-m)^p m^{2j} \Phi\left(e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right) \end{aligned} \right) (1+k-p)_p \end{aligned} $$

where $Re(a) \le -1$.

### EXAMPLE 3.14.

$$ (3.14) \quad \int_0^\infty \frac{e^{\frac{4ix^2}{\pi}} (2+2i\pi) \operatorname{csch}(2x) \sin\left(\frac{3x}{2}\right)}{1+2i\pi - \pi^2 + x^2} dx $$
$$ \times \left( (-m)^{2j} m^p \Phi \left( e^{-\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b \log(a)}{\pi} \right) + e^{\frac{m\pi}{b}} (-m)^p m^{2j} \Phi \left( e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b \log(a)}{\pi} \right) \right) (1+k-p)_p $$

where $Re(a) \le -1$.

### EXAMPLE 3.12.

$$ \begin{aligned} \int_0^\infty \frac{e^{\frac{4ix^2}{\pi}} x \cos\left(\frac{x}{2}\right) \operatorname{csch}(x) \operatorname{sech}(x)}{-\pi^2 + x^2} dx = & -\frac{i}{2} + \left(\frac{1}{5} - \frac{2i}{5}\right) e^{-\frac{\pi}{4}} {}_2F_1(1, 1+2i; 2+2i; e^{-\frac{\pi}{4}}) \\ & + \left(\frac{1}{5} - \frac{2i}{5}\right) e^{\pi/4} {}_2F_1(1, 1+2i; 2+2i; e^{\pi/4}) \\ & - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i^j 4^{-3j+p} p e^{-\frac{\pi}{8}} (-\pi)^{j-p} \binom{2j}{p}}{j!} \\ & \times \left( (-1)^{2j} \Phi\left(e^{-\frac{\pi}{4}}, 1+p, \frac{1}{2}+2i\right) + (-1)^p e^{\pi/4} \Phi\left(e^{\pi/4}, 1+p, \frac{1}{2}+2i\right) \right) (-p)_p \end{aligned} $$

### EXAMPLE 3.13.

$$ \begin{aligned} & \int_0^\infty e^{\frac{ib^2x^2}{\pi}} \operatorname{csch}(bx) \left( \left(-ix + \log(a)\right)^k + \left(ix + \log(a)\right)^k \right) \sin(mx) dx \\ = & \left(\frac{1}{b}\right)^{1+k} e^{-\frac{m\pi}{b}} \pi^{1+k} \left( \Phi\left(e^{-\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) - e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{m\pi}{b}}, -k, 1 + \frac{b\log(a)}{\pi}\right) \right) \\ & + \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{(-\frac{1}{4})^j i^j b^{1-2j-k+p} p e^{-\frac{m\pi}{2b}}}{j!} \\ & \times (-m^2)^{-p} \pi^{1+j+k-p} \binom{2j}{p} \left( \begin{aligned} & -(-m)^{2j} m^p \Phi\left(e^{-\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right) \\ & + e^{\frac{m\pi}{b}} (-m)^p m^{2j} \Phi\left(e^{\frac{m\pi}{b}}, -k+p, \frac{1}{2} + \frac{b\log(a)}{\pi}\right) \end{aligned} \right) (1+k-p)_p \end{aligned} $$

where $Re(a) \le -1$.

### EXAMPLE 3.14.

$$ (3.14) \quad \int_0^\infty \frac{e^{\frac{4ix^2}{\pi}} (2+2i\pi) \operatorname{csch}(2x) \sin\left(\frac{3x}{2}\right)}{1+2i\pi - \pi^2 + x^2} dx $$
$$ = e^{-\frac{3\pi}{4}} \left( \Phi \left( e^{-\frac{3\pi}{4}}, 1, (1+2i) + \frac{2}{\pi} \right) - e^{\frac{3\pi}{2}} \Phi \left( e^{\frac{3\pi}{4}}, 1, (1+2i) + \frac{2}{\pi} \right) \right) \\ + \sum_{j=0}^{\infty} \sum_{p=0}^{2j} \frac{i^j 3^{2j-p} 4^{-3j+p} e^{-\frac{3\pi}{8}} (-\pi)^{j-p} \binom{2j}{p}}{j!} \\ \times \left( (-1)^{2j} \Phi \left( e^{-\frac{3\pi}{4}}, 1+p, \left( \frac{1}{2} + 2i \right) + \frac{2}{\pi} \right) \right. \\ \left. + (-1)^p e^{\frac{3\pi}{4}} \Phi \left( e^{\frac{3\pi}{4}}, 1+p, \left( \frac{1}{2} + 2i \right) + \frac{2}{\pi} \right) \right) (-p)_p $$

THEOREM 3.15. From Eq. (2.10.63) in [5]. Struve function $H_v(x\alpha)$.

$$ (3.15) \quad \int_0^\infty e^{-imx} x^{-v} \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) H_v(x\alpha) dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{i(-1)^j 2^{1-v} m^{2j-p} \sqrt{\pi\alpha^{-1-2j+v}} \binom{2j}{p} \binom{-\frac{1}{2}+v}{j} \log^{k-p}(a)(1+k-p)_p}{\Gamma\left(\frac{1}{2}+v\right)} $$

where $\mathrm{Re}(k) \le 0, |\mathrm{Re}(a)| < 3\pi$.

EXAMPLE 3.16.

$$ (3.16) \quad \int_0^\infty \frac{x^{-v} (x \cos(mx) - ia \sin(mx)) H_v(x\alpha)}{a^2 - x^2} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j 2^{-v} (ia)^{-1-p} m^{2j-p} \sqrt{\pi\alpha^{-1-2j+v}} \binom{2j}{p} \binom{-\frac{1}{2}+v}{j} (-p)_p}{\Gamma\left(\frac{1}{2}+v\right)} $$

EXAMPLE 3.17.

$$ (3.17) \quad \int_0^\infty \frac{(x \cos(\frac{\pi x}{4}) + \pi \sin(\frac{\pi x}{4})) H_1(6x)}{x(\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{j-p} 4^{-3j+p} 9^{-j} \pi^{-1+2j-2p} \binom{\frac{1}{2}}{j} \binom{2j}{p} (-p)_p $$

THEOREM 3.18. From Eq. (2.14.7) in [5]. Series and integrals involving higher functions.

$$ (3.18) \quad \int_0^\infty ie^{-imx} x_1 F_2 \left( \alpha; \frac{3}{2}; \beta; -\frac{x^2}{4} \right) \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) dx $$
$$ = e^{-\frac{3\pi}{4}} \left( \Phi \left( e^{-\frac{3\pi}{4}}, 1, (1+2i) + \frac{2}{\pi} \right) - e^{\frac{3\pi}{2}} \Phi \left( e^{\frac{3\pi}{4}}, 1, (1+2i) + \frac{2}{\pi} \right) \right) \\ + \sum_{j=0}^{\infty} \sum_{p=0}^{2j} \frac{i^j 3^{2j-p} 4^{-3j+p} e^{-\frac{3\pi}{8}} (-\pi)^{j-p} \binom{2j}{p}}{j!} \\ \times \left( (-1)^{2j} \Phi \left( e^{-\frac{3\pi}{4}}, 1+p, \left( \frac{1}{2} + 2i \right) + \frac{2}{\pi} \right) \right. \\ \left. + (-1)^p e^{\frac{3\pi}{4}} \Phi \left( e^{\frac{3\pi}{4}}, 1+p, \left( \frac{1}{2} + 2i \right) + \frac{2}{\pi} \right) \right) (-p)_p $$

THEOREM 3.15. From Eq. (2.10.63) in [5]. Struve function $H_v(x\alpha)$.

$$ (3.15) \quad \int_0^\infty e^{-imx} x^{-v} \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) H_v(x\alpha) dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{i(-1)^j 2^{1-v} m^{2j-p} \sqrt{\pi\alpha^{-1-2j+v}} \binom{2j}{p} \binom{-\frac{1}{2}+v}{j} \log^{k-p}(a)(1+k-p)_p}{\Gamma\left(\frac{1}{2}+v\right)} $$

where $Re(k) \le 0, |Re(a)| < 3\pi$.

EXAMPLE 3.16.

$$ (3.16) \quad \int_0^\infty \frac{x^{-v} (x \cos(mx) - ia \sin(mx)) H_v(x\alpha)}{a^2 - x^2} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j 2^{-v} (ia)^{-1-p} m^{2j-p} \sqrt{\pi\alpha^{-1-2j+v}} \binom{2j}{p} \binom{-\frac{1}{2}+v}{j} (-p)_p}{\Gamma\left(\frac{1}{2}+v\right)} $$

EXAMPLE 3.17.

$$ (3.17) \quad \int_0^\infty \frac{(x \cos(\frac{\pi x}{4}) + \pi \sin(\frac{\pi x}{4})) H_1(6x)}{x(\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{j-p} 4^{-3j+p} 9^{-j} \pi^{-1+2j-2p} \binom{\frac{1}{2}}{j} \binom{2j}{p} (-p)_p $$

THEOREM 3.18. From Eq. (2.14.7) in [5]. Series and integrals involving higher functions.

$$ (3.18) \quad \int_0^\infty ie^{-imx} x_1 F_2 \left( \alpha; \frac{3}{2}; \beta; -\frac{x^2}{4} \right) \left( (-ix + \log(a))^k - e^{2imx} (ix + \log(a))^k \right) dx $$
$$= \sum_{j=0}^{\infty} \sum_{p=0}^{\infty} \frac{2(-1)^j m^{-2+2j-p+2\alpha} \pi \binom{-2+2j+2\alpha}{p} \binom{-1-\alpha+\beta}{j} (1+k-p)_p \Gamma(\beta) \log^{k-p}(a)}{\Gamma(\alpha)\Gamma(-\alpha+\beta)}$$

where $Re(a) > e^{3\pi}$ in order for the series to converge.

### EXAMPLE 3.19.

$$ (3.19) \quad \int_0^\infty \frac{x_1 F_2\left(\alpha; \frac{3}{2}, \beta; -\frac{x^2}{4}\right) (-x \cos(mx) + a\pi \sin(mx))}{a^2 \pi^2 + x^2} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j a^{-1-p} m^{-2+2j-p+2\alpha} \pi^{-p} \binom{-2+2j+2\alpha}{p} \binom{-1-\alpha+\beta}{j} \Gamma(\beta) (-p)_p}{\Gamma(\alpha)\Gamma(-\alpha+\beta)} $$

### EXAMPLE 3.20.

$$ (3.20) \quad \int_0^\infty \frac{(-x \cos(\frac{x}{2}) + 4\pi \sin(\frac{x}{2})) (3x \cos(x) + (-3+x^2) \sin(x))}{16\pi^2 x^4 + x^6} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^j 2^{-5-2j-p} \pi^{-p} \binom{1}{j} \binom{1+2j}{p} (-p)_p $$

### EXAMPLE 3.21.

$$ (3.21) \quad \int_0^\infty \frac{2 \times 2^{2/3} J_\frac{5}{3}(x) \Gamma(\frac{8}{3}) (-x \cos(\frac{x}{4}) + 5\pi \sin(\frac{x}{4}))}{x^{2/3} (25\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j 2^{-1-4j+2p} 5^{-1-p} \pi^{-\frac{1}{2}-p} \binom{\frac{1}{6}}{j} \binom{1+2j}{p} \Gamma(\frac{8}{3}) (-p)_p}{\Gamma(\frac{7}{6})} $$

### EXAMPLE 3.22.

$$ (3.22) \quad \int_0^\infty \frac{(-x \cos(\frac{x}{2}) - 11\pi \sin(\frac{x}{2})) (3x \cos(x) - 3\sin(x) + x^2 \sin(x))}{x^4 (121\pi^2 + x^2)} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{1+3j-p} 2^{-3-2j+p} 11^{-1-p} \pi^{-p} \binom{1}{j} \binom{1+2j}{p} (-p)_p $$

### EXAMPLE 3.23.

$$ (3.23) \quad \int_0^\infty \frac{J_3(x) (-x \cos(\frac{x}{3}) - 11\pi \sin(\frac{x}{3}))}{x^2 (121\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{1+3j-p} 3^{-2-2j+p} (11\pi)^{-1-p} \binom{\frac{3}{2}}{j} \binom{1+2j}{p} (-p)_p $$
$$= \sum_{j=0}^{\infty} \sum_{p=0}^{\infty} \frac{2(-1)^j m^{-2+2j-p+2\alpha} \pi \binom{-2+2j+2\alpha}{p} \binom{-1-\alpha+\beta}{j} (1+k-p)_p \Gamma(\beta) \log^{k-p}(a)}{\Gamma(\alpha)\Gamma(-\alpha+\beta)}$$

where $Re(a) > e^{3\pi}$ in order for the series to converge.

### EXAMPLE 3.19.

$$ (3.19) \quad \int_0^\infty \frac{x_1 F_2\left(\alpha; \frac{3}{2}, \beta; -\frac{x^2}{4}\right) (-x \cos(mx) + a\pi \sin(mx))}{a^2 \pi^2 + x^2} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j a^{-1-p} m^{-2+2j-p+2\alpha} \pi^{-p} \binom{-2+2j+2\alpha}{p} \binom{-1-\alpha+\beta}{j} \Gamma(\beta) (-p)_p}{\Gamma(\alpha) \Gamma(-\alpha+\beta)} $$

### EXAMPLE 3.20.

$$ (3.20) \quad \int_0^\infty \frac{(-x \cos(\frac{x}{2}) + 4\pi \sin(\frac{x}{2})) (3x \cos(x) + (-3+x^2) \sin(x))}{16\pi^2 x^4 + x^6} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^j 2^{-5-2j-p} \pi^{-p} \binom{1}{j} \binom{1+2j}{p} (-p)_p $$

### EXAMPLE 3.21.

$$ (3.21) \quad \int_0^\infty \frac{2 \times 2^{2/3} J_\frac{5}{3}(x) \Gamma(\frac{8}{3}) (-x \cos(\frac{x}{4}) + 5\pi \sin(\frac{x}{4}))}{x^{2/3} (25\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{(-1)^j 2^{-1-4j+2p} 5^{-1-p} \pi^{-\frac{1}{2}-p} \binom{\frac{1}{6}}{j} \binom{1+2j}{p} \Gamma(\frac{8}{3}) (-p)_p}{\Gamma(\frac{7}{6})} $$

### EXAMPLE 3.22.

$$ (3.22) \quad \int_0^\infty \frac{(-x \cos(\frac{x}{2}) - 11\pi \sin(\frac{x}{2})) (3x \cos(x) - 3\sin(x) + x^2 \sin(x))}{x^4 (121\pi^2 + x^2)} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{1+3j-p} 2^{-3-2j+p} 11^{-1-p} \pi^{-p} \binom{1}{j} \binom{1+2j}{p} (-p)_p $$

### EXAMPLE 3.23.

$$ (3.23) \quad \int_0^\infty \frac{J_3(x) (-x \cos(\frac{x}{3}) - 11\pi \sin(\frac{x}{3}))}{x^2 (121\pi^2 + x^2)} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty (-1)^{1+3j-p} 3^{-2-2j+p} (11\pi)^{-1-p} \binom{\frac{3}{2}}{j} \binom{1+2j}{p} (-p)_p $$
EXAMPLE 3.24.

$$ (3.24) \quad \int_0^\infty \frac{x_1 F_2\left(\frac{4}{3}; \frac{3}{2}; 6; -\frac{x^2}{4}\right) (-x \cos\left(\frac{x}{2}\right) + 13\pi \sin\left(\frac{x}{2}\right))}{169\pi^2 + x^2} dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{15(-1)^j 2^{\frac{7}{3}-2j+p} 13^{-1-p}\pi^{-p} \binom{\frac{11}{3}}{j} \binom{\frac{2}{3}+2j}{p} (-p)_p}{\Gamma\left(\frac{4}{3}\right) \Gamma\left(\frac{14}{3}\right)} $$

THEOREM 3.25. From Eq. (2.4.10.5) in [7]. A finite series involving the in-complete gamma function.

$$ (3.25) \quad \int_0^\infty e^{-mx} \cosh^n(bx)(a-x)^k dx \\ = - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{j} \left( \frac{\exp(a(bn-2bj-m))\Gamma(1+k, -(2bj+m-bn)a)}{2^n(bn-2bj-m)^{k+1}} \right. \\ \left. + \frac{\exp(a(2bj-m-bn))\Gamma(1+k, -(-2bj+m+bn)a)}{2^n(2bj-m-bn)^{k+1}} \right) \\ + \frac{2^{-1-n}(1+(-1)^n)e^{-am}(-m)^{-k}(\lfloor \frac{n}{2} \rfloor)\Gamma(1+k,-am)}{m} $$

where $Re(m) > nRe(b)$, $Re(a) < 0$.

EXAMPLE 3.26.

$$ (3.26) \quad \int_0^\infty \frac{e^{-mx} x \cosh^n(bx)}{a^2 + x^2} dx \\ = \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} 2^{-1-n} e^{-ia(m+b(2j+n))} \binom{n}{j} \left( e^{4iabj} \Gamma(0, ia(2bj-m-bn)) + e^{2iabn} \Gamma(0, -ia(2bj+m-bn)) \right. \\ \left. + e^{2ia(2bj+m)} \Gamma(0, ia(2bj+m-bn)) + e^{2ia(m+bn)} \Gamma(0, ia(-2bj+m+bn)) \right) \\ + 2^{-2-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} \left( \Gamma(0, -iam) + e^{2iam} \Gamma(0, iam) \right) $$

EXAMPLE 3.27.

$$ (3.27) \quad \int_0^\infty \frac{e^{-mx} \cosh^n(bx) \left(-\pi x - 2 \coth\left(\frac{\pi x}{2}\right) + \pi x \coth^2\left(\frac{\pi x}{2}\right)\right) \tanh\left(\frac{\pi x}{2}\right)}{4x} dx $$
EXAMPLE 3.24.

$$
(3.24) \quad \int_0^\infty \frac{x_1 F_2\left(\frac{4}{3}; \frac{3}{2}; 6; -\frac{x^2}{4}\right) (-x \cos\left(\frac{x}{2}\right) + 13\pi \sin\left(\frac{x}{2}\right))}{169\pi^2 + x^2} dx \\
= \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{15(-1)^j 2^{\frac{7}{3}-2j+p} 13^{-1-p}\pi^{-p} \binom{\frac{11}{3}}{j} \binom{\frac{2}{3}+2j}{p} (-p)_p}{\Gamma\left(\frac{4}{3}\right) \Gamma\left(\frac{14}{3}\right)}
$$

THEOREM 3.25. *From Eq. (2.4.10.5) in [8].* A finite series involving the in-
complete gamma function.

$$
(3.25) \quad \int_0^\infty e^{-mx} \cosh^n(bx)(a-x)^k dx \\
= - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{j} \left( \frac{\exp(a(bn-2bj-m))\Gamma(1+k,-(2bj+m-bn)a)}{2^n(bn-2bj-m)^{k+1}} \right. \\
\left. \qquad + \frac{\exp(a(2bj-m-bn))\Gamma(1+k,-(-2bj+m+bn)a)}{2^n(2bj-m-bn)^{k+1}} \right) \\
\qquad + \frac{2^{-1-n}(1+(-1)^n)e^{-am}(-m)^{-k}(\lfloor \frac{n}{2} \rfloor)\Gamma(1+k,-am)}{m}
$$

where $Re(m) > nRe(b), Re(a) < 0$.

EXAMPLE 3.26.

$$
(3.26) \quad \int_0^\infty \frac{e^{-mx} x \cosh^n(bx)}{a^2 + x^2} dx \\
= \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} 2^{-1-n} e^{-ia(m+b(2j+n))} \binom{n}{j} \left( \begin{aligned}[t]
& e^{4iabj} \Gamma(0, ia(2bj-m-bn)) + e^{2iabn} \Gamma(0, -ia(2bj+m-bn)) \\
& + e^{2ia(2bj+m)} \Gamma(0, ia(2bj+m-bn)) + e^{2ia(m+bn)} \Gamma(0, ia(-2bj+m+bn)) \\
& + 2^{-2-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} (\Gamma(0, -iam) + \Gamma(0, iam))
\end{aligned} \right)
$$

EXAMPLE 3.27.

$$
(3.27) \quad \int_0^\infty \frac{e^{-mx} \cosh^n(bx) \left(-\pi x - 2 \coth\left(\frac{\pi x}{2}\right) + \pi x \coth^2\left(\frac{\pi x}{2}\right)\right) \tanh\left(\frac{\pi x}{2}\right)}{4x} dx
$$
$$
= \sum_{a=1}^{\infty} \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^a 2^{-1-n} e^{-ia(m+b(2j+n))} \binom{n}{j} \\
\left( e^{4iabj} \Gamma(0, ia(2bj-m-bn)) + e^{2iabn} \Gamma(0, -ia(2bj+m-bn)) \right. \\
\left. + e^{2ia(2bj+m)} \Gamma(0, ia(2bj+m-bn)) + e^{2ia(m+bn)} \Gamma(0, ia(-2bj+m+bn)) \right) \\
+ \sum_{a=1}^{\infty} (-1)^a 2^{-2-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} \left( \Gamma(0, -iam) + e^{2iam} \Gamma(0, iam) \right)
$$

EXAMPLE 3.28.

$$
\begin{align*}
& (3.28) \quad \int_0^\infty \frac{e^{-mx} x \cosh^n(bx)}{(a^2 + x^2)^2} dx \\
&= - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \frac{2^{-2-n} \binom{n}{j}}{a} \left( -\frac{4}{a} + i e^{-ia(m+b(2j+n))} \right) \\
& \quad \left( e^{4iabj} (2bj - m - bn) E_1(ia(2bj - m - bn)) \right. \\
& \quad \left. + e^{2iabn} (-2bj - m + bn) E_1(-ia(2bj + m - bn)) \right. \\
& \quad \left. + e^{2ia(2bj+m)} (2bj + m - bn) E_1(ia(2bj + m - bn)) \right. \\
& \quad \left. + e^{2ia(m+bn)} (-2bj + m + bn) E_1(ia(-2bj + m + bn)) \right) \\
& \quad + \frac{2^{-3-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{n}{2} \rfloor} (2e^{iam} + iam (\Gamma(0, -iam) - e^{2iam}\Gamma(0, iam)))}{a^2}
\end{align*}
$$

EXAMPLE 3.29.

$$
\begin{align*}
& (3.29) \quad \int_0^\infty \frac{e^{-mx} (a-x)^k \cosh(bx) (-1+(-1)^p \cosh^p(bx))}{1+\cosh(bx)} dx \\
&= \sum_{n=1}^p \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^n \binom{n}{j} \left( 2^{-n} e^{a(2bj-m-bn)} (2bj-m-bn)^{-1-k} \Gamma(1+k, a(2bj-m-bn)) \right. \\
& \qquad \left. + 2^{-n} e^{a(-2bj-m+bn)} (-2bj-m+bn)^{-1-k} \Gamma(1+k, a(-2bj-m+bn)) \right) \\
& \qquad + \sum_{n=1}^p \frac{(-1)^n 2^{-1-n} (1+(-1)^n) e^{-am} (-m)^{-k} \binom{n}{\lfloor \frac{n}{2} \rfloor} \Gamma(1+k, -am)}{m}
\end{align*}
$$

EXAMPLE 3.30.

$$
(3.30) \quad \int_0^\infty e^{-mx} \cosh^n(bx) \log(a-x) dx
$$
$$
= \sum_{a=1}^{\infty} \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^a 2^{-1-n} e^{-ia(m+b(2j+n))} \binom{n}{j} \\
\qquad (e^{4iabj} \Gamma(0, ia(2bj-m-bn)) + e^{2iabn} \Gamma(0, -ia(2bj+m-bn)) \\
\qquad \qquad + e^{2ia(2bj+m)} \Gamma(0, ia(2bj+m-bn)) + e^{2ia(m+bn)} \Gamma(0, ia(-2bj+m+bn))) \\
\qquad + \sum_{a=1}^{\infty} (-1)^a 2^{-2-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} (\Gamma(0, -iam) + e^{2iam} \Gamma(0, iam))
$$

EXAMPLE 3.28.

$$
\begin{equation} \tag{3.28}
\begin{split}
& \int_0^\infty \frac{e^{-mx} x \cosh^n(bx)}{(a^2 + x^2)^2} dx \\
& = - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \frac{2^{-2-n} \binom{n}{j}}{a} \left( -\frac{4}{a} + i e^{-ia(m+b(2j+n))} \right) \\
& \qquad \left( e^{4iabj} (2bj - m - bn) E_1(ia(2bj - m - bn)) \right. \\
& \qquad \qquad + e^{2iabn} (-2bj - m + bn) E_1(-ia(2bj + m - bn)) \\
& \qquad \qquad + e^{2ia(2bj+m)} (2bj + m - bn) E_1(ia(2bj + m - bn)) \\
& \qquad \qquad \left. + e^{2ia(m+bn)} (-2bj + m + bn) E_1(ia(-2bj + m + bn)) \right) \\
& \qquad + \frac{2^{-3-n} (1+(-1)^n) e^{-iam} \binom{n}{\lfloor \frac{n}{2} \rfloor} (2e^{iam} + iam (\Gamma(0,-iam) - e^{2iam}\Gamma(0,iam)))}{a^2}
\end{split}
\end{equation}
$$

EXAMPLE 3.29.

$$
\begin{equation} \tag{3.29}
\begin{split}
& \int_0^\infty \frac{e^{-mx}(a-x)^k \cosh(bx) (-1+(-1)^p \cosh^p(bx))}{1+\cosh(bx)} dx \\
& = \sum_{n=1}^p \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^n \binom{n}{j} \left( 2^{-n} e^{a(2bj-m-bn)} (2bj-m-bn)^{-1-k} \Gamma(1+k, a(2bj-m-bn)) \right. \\
& \qquad \left. + 2^{-n} e^{a(-2bj-m+bn)} (-2bj-m+bn)^{-1-k} \Gamma(1+k, a(-2bj-m+bn)) \right) \\
& \qquad + \sum_{n=1}^p \frac{(-1)^n 2^{-1-n} (1+(-1)^n) e^{-am} (-m)^{-k} \binom{n}{\lfloor \frac{n}{2} \rfloor} \Gamma(1+k, -am)}{m}
\end{split}
\end{equation}
$$

EXAMPLE 3.30.

$$
(3.30) \quad \int_0^\infty e^{-mx} \cosh^n(bx) \log(a-x) dx
$$
$$
\begin{aligned}
&= - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{j} \left( - \frac{2^{-n} \log(2bj - m - bn)}{2bj - m - bn} - \frac{2^{-n} \log(-2bj - m + bn)}{-2bj - m + bn} \right. \\
&\qquad \left. + \frac{2^{-n} e^{a(2bj-m-bn)} \left( e^{-a(2bj-m-bn)} \log(a(2bj-m-bn)) + G_{1,2}^{2,0} \left( a(2bj-m-bn) \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{2bj - m - bn} \right. \\
&\qquad \left. + \frac{2^{-n} e^{a(-2bj-m+bn)} \left( e^{-a(-2bj-m+bn)} \log(a(-2bj-m+bn)) + G_{1,2}^{2,0} \left( a(-2bj-m+bn) \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{-2bj - m + bn} \right) \\
&\qquad - \frac{2^{-1-n} (1+(-1)^n) \binom{\lfloor \frac{n}{2} \rfloor}{\lfloor \frac{n}{2} \rfloor} \log(-m)}{m} \\
&\qquad + \frac{2^{-1-n} (1+(-1)^n) e^{-am} \binom{\lfloor \frac{n}{2} \rfloor}{\lfloor \frac{n}{2} \rfloor} \left( e^{am} \log(-am) + G_{1,2}^{2,0} \left( -am \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{m}
\end{aligned}
$$

EXAMPLE 3.31. Entry (2.4.10.10(ii)) in [7].

$$
\begin{aligned}
& (3.31) \quad \int_0^\infty e^{-mx} x^{-1+s} \cosh^n(bx) dx \\
& = \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^s 2^{-n} \left( (2bj - m - bn)^{-s} + (-2bj - m + bn)^{-s} \right) \binom{n}{j} \Gamma(s) \\
& \qquad + 2^{-1-n} (1+(-1)^n) m^{-s} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} \Gamma(s)
\end{aligned}
$$

EXAMPLE 3.32. Entry (2.4.10.7(ii)) in [7]. Errata. The quoted equation is in error.

$$
(3.32) \quad \int_0^\infty e^{-mx} x^{-1+s} \cosh(bx) dx = \frac{\Gamma(s)}{2} (-1)^s ((-b-m)^{-s} + (b-m)^{-s})
$$
$$
\begin{aligned}
&= - \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{j} \left( - \frac{2^{-n} \log(2bj - m - bn)}{2bj - m - bn} - \frac{2^{-n} \log(-2bj - m + bn)}{-2bj - m + bn} \right. \\
&\qquad \left. + \frac{2^{-n} e^{a(2bj-m-bn)} \left( e^{-a(2bj-m-bn)} \log(a(2bj-m-bn)) + G_{1,2}^{2,0} \left( a(2bj-m-bn) \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{2bj - m - bn} \right. \\
&\qquad \left. + \frac{2^{-n} e^{a(-2bj-m+bn)} \left( e^{-a(-2bj-m+bn)} \log(a(-2bj-m+bn)) + G_{1,2}^{2,0} \left( a(-2bj-m+bn) \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{-2bj - m + bn} \right) \\
&\qquad - \frac{2^{-1-n} (1+(-1)^n) \binom{\lfloor \frac{n}{2} \rfloor}{\lfloor \frac{n}{2} \rfloor} \log(-m)}{m} \\
&\qquad + \frac{2^{-1-n} (1+(-1)^n) e^{-am} \binom{\lfloor \frac{n}{2} \rfloor}{\lfloor \frac{n}{2} \rfloor} \left( e^{am} \log(-am) + G_{1,2}^{2,0} \left( -am \middle| \begin{array}{c} 1 \\ 0,0 \end{array} \right) \right)}{m}
\end{aligned}
$$

EXAMPLE 3.31. Entry (2.4.10.10(ii)) in [8].

$$
\begin{aligned}
& (3.31) \quad \int_0^\infty e^{-mx} x^{-1+s} \cosh^n(bx) dx \\
& = \sum_{j=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^s 2^{-n} \left( (2bj - m - bn)^{-s} + (-2bj - m + bn)^{-s} \right) \binom{n}{j} \Gamma(s) \\
& \qquad + 2^{-1-n} (1+(-1)^n) m^{-s} \binom{n}{\lfloor \frac{1+n}{2} \rfloor} \Gamma(s)
\end{aligned}
$$

EXAMPLE 3.32. Entry (2.4.10.7(ii)) in [8]. Errata. The quoted equation is in error.

$$
(3.32) \quad \int_0^\infty e^{-mx} x^{-1+s} \cosh(bx) dx = \frac{\Gamma(s)}{2} (-1)^s ((-b-m)^{-s} + (b-m)^{-s})
$$
EXAMPLE 3.33. Entry (2.4.10.7(i)) in [7]. Errata. The quoted equation is in error.

$$ (3.33) \qquad \int_0^\infty e^{-mx} x^{-1+s} \sinh(bx) dx = \frac{\Gamma(s)}{2} (-1)^s ((b-m)^{-s} - (-b-m)^{-s}) $$

EXAMPLE 3.34. Entry (2.4.10.8) in [7].

$$ (3.34) \qquad \int_0^\infty \frac{e^{-mx} \sinh(bx)}{x} dx = -\frac{1}{2} \log \left( \frac{m-b}{m+b} \right) $$

EXAMPLE 3.35. Entry (2.4.10.8) in [7].

$$ (3.35) \qquad \int_0^\infty \frac{e^{-mx} \sinh^2(bx)}{x} dx = -\frac{1}{2} \log \left( \sqrt{1 - \frac{4b^2}{m^2}} \right) $$

EXAMPLE 3.36. Shifted Mellin transform or Laplace transform.

$$ (3.36) \qquad \int_0^\infty e^{-mx} (a-x)^{-1+s} \cosh(bx) dx = - \frac{e^{a(-b-m)} \Gamma(s, a(-b-m))}{2(-b-m)^s} - \frac{e^{a(b-m)} \Gamma(s, a(b-m))}{2(b-m)^s} $$

EXAMPLE 3.37.

$$ (3.37) \qquad \int_0^\infty \frac{e^{-7\pi x} \cosh\left(\frac{\pi x}{2}\right)}{\sqrt{\sqrt{2}-x}} dx = \frac{ie^{-\frac{15\pi}{\sqrt{2}}} \Gamma\left(\frac{1}{2}, -\frac{15\pi}{\sqrt{2}}\right)}{\sqrt{30\pi}} + \frac{ie^{-\frac{13\pi}{\sqrt{2}}} \Gamma\left(\frac{1}{2}, -\frac{13\pi}{\sqrt{2}}\right)}{\sqrt{26\pi}} $$

EXAMPLE 3.38. Laplace transform involving the arctangent function.

$$ (3.38) \qquad \int_0^\infty e^{-mx} \cosh(bx) \tanh^{-1}\left(\frac{x}{a}\right) dx \\ = \frac{1}{4} e^{-a(b+m)} \left( \frac{e^{2ab} E_1(a(b-m)) - e^{2am} E_1(a(-b+m)) + e^{a(b+m)} (\log(a(b-m)) - \log(a(-b+m)))}{b-m} \right. \\ \left. + \frac{-E_1(-a(b+m)) + e^{a(b+m)} (e^{a(b+m)} E_1(a(b+m)) - \log(-a(b+m)) + \log(a(b+m)))}{b+m} \right) \\ + \frac{im\pi}{2(-b^2 + m^2)} $$

THEOREM 3.39. From Eq. (2.5.6.6) in [7]. Extended Catalan definite integral in terms of a finite series, pp.110 in [9] and pp.90 in [10].
EXAMPLE 3.33. Entry (2.4.10.7(i)) in [8]. Errata. The quoted equation is in error.

$$ (3.33) \quad \int_0^\infty e^{-mx} x^{-1+s} \sinh(bx) dx = \frac{\Gamma(s)}{2} (-1)^s ((b-m)^{-s} - (-b-m)^{-s}) $$

EXAMPLE 3.34. Entry (2.4.10.8) in [8].

$$ (3.34) \quad \int_0^\infty \frac{e^{-mx} \sinh(bx)}{x} dx = -\frac{1}{2} \log \left( \frac{m-b}{m+b} \right) $$

EXAMPLE 3.35. Entry (2.4.10.8) in [8].

$$ (3.35) \quad \int_0^\infty \frac{e^{-mx} \sinh^2(bx)}{x} dx = -\frac{1}{2} \log \left( \sqrt{1 - \frac{4b^2}{m^2}} \right) $$

EXAMPLE 3.36. Shifted Mellin transform or Laplace transform.

$$ (3.36) \quad \int_0^\infty e^{-mx} (a-x)^{-1+s} \cosh(bx) dx = - \frac{e^{a(-b-m)} \Gamma(s, a(-b-m))}{2(-b-m)^s} - \frac{e^{a(b-m)} \Gamma(s, a(b-m))}{2(b-m)^s} $$

EXAMPLE 3.37.

$$ (3.37) \quad \int_0^\infty \frac{e^{-7\pi x} \cosh\left(\frac{\pi x}{2}\right)}{\sqrt{\sqrt{2}-x}} dx = \frac{ie^{-\frac{15\pi}{\sqrt{2}}} \Gamma\left(\frac{1}{2}, -\frac{15\pi}{\sqrt{2}}\right)}{\sqrt{30\pi}} + \frac{ie^{-\frac{13\pi}{\sqrt{2}}} \Gamma\left(\frac{1}{2}, -\frac{13\pi}{\sqrt{2}}\right)}{\sqrt{26\pi}} $$

EXAMPLE 3.38. Laplace transform involving the arctangent function.

$$ \begin{aligned} & (3.38) \quad \int_0^\infty e^{-mx} \cosh(bx) \tanh^{-1}\left(\frac{x}{a}\right) dx \\ & = \frac{1}{4} e^{-a(b+m)} \left( \frac{e^{2ab} E_1(a(b-m)) - e^{2am} E_1(a(-b+m)) + e^{a(b+m)} (\log(a(b-m)) - \log(a(-b+m)))}{b-m} \right. \\ & \qquad \left. + \frac{-E_1(-a(b+m)) + e^{a(b+m)} (e^{a(b+m)} E_1(a(b+m)) - \log(-a(b+m)) + \log(a(b+m)))}{b+m} \right) \\ & \qquad + \frac{im\pi}{2(-b^2 + m^2)} \end{aligned} $$

THEOREM 3.39. From Eq. (2.5.6.6) in [8]. Extended Catalan definite integral in terms of a finite series, pp.110 in [10] and pp.90 in [11].
$$ (3.39) \quad \int_0^\infty \frac{e^{-imx} ((a-x)^k + e^{2imx}(a+x)^k)}{(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{i^{-k} 2^{j-2n} e^{-mz} m^{j-p} \pi (ia-z)^{k-p} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (1+k-p)_p}{j! n! (-j+n)!} $$

where $Re(m) > 0, Re(z) > 0$

### EXAMPLE 3.40.

$$ (3.40) \quad \int_0^\infty \frac{a \cos(mx) - x \sin(mx)}{(a^2 + x^2)(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{(-1)^{-p} 2^{-1+j-2n} e^{-mz} m^{j-p} \pi z^{-1+j-2n} (a+z)^{-1-p} \binom{j}{p} (-j+2n)! (-p)_p}{j! n! (-j+n)!} $$

### EXAMPLE 3.41. Mellin transform.

$$ (3.41) \quad \int_0^\infty x^{-1+s} \frac{(e^{imx} - e^{i(\pi s - mx)})}{(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{i^{1-s} 2^{j-2n} e^{-mz} m^{j-p} \pi (-z)^{-1-p+s} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (-p+s)_p}{j! n! (-j+n)!} $$

### EXAMPLE 3.42.

$$ (3.42) \quad m^2|m|^{2n-s} \Gamma(-2-2n+s) {}_1F_2\left(1+n; \frac{3}{2}+n-\frac{s}{2}, 2+n-\frac{s}{2}; \frac{m^2 z^2}{4}\right) \\ \qquad \qquad \qquad \left( (1+\operatorname{sgn}(m))\sin(n\pi) + (-1+\operatorname{sgn}(m))\sin(\pi(n-s)) \right) \\ \qquad \qquad \qquad + \frac{z^{-2-2n+s}}{\Gamma(1+n)} \left( mz \cos\left(\frac{\pi s}{2}\right) \Gamma\left(\frac{1}{2}+n-\frac{s}{2}\right) \Gamma\left(\frac{1+s}{2}\right) {}_1F_2\left(\frac{1}{2}+\frac{s}{2}; \frac{3}{2}, \frac{1}{2}-n+\frac{s}{2}; \frac{m^2 z^2}{4}\right) \right. \\ \qquad \qquad \qquad \left. - \Gamma\left(1+n-\frac{s}{2}\right) \Gamma\left(\frac{s}{2}\right) {}_1F_2\left(\frac{s}{2}; \frac{1}{2}, -n+\frac{s}{2}; \frac{m^2 z^2}{4}\right) \sin\left(\frac{\pi s}{2}\right) \right) \\ \qquad \qquad \qquad = \sum_{j=0}^n \sum_{p=0}^j \frac{2^{j-2n} e^{-i\pi s - m\pi} m^{j-p} \pi (-z)^{1-p+s} z^{-3+j-2n} \binom{j}{p} (-j+2n)! (-p+s)_p}{j! n! (-j+n)!} $$

where $Re(m) > 0, Im(m) = 0$.

### EXAMPLE 3.43.
$$ (3.39) \quad \int_0^\infty \frac{e^{-imx} ((a-x)^k + e^{2imx}(a+x)^k)}{(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{i^{-k} 2^{j-2n} e^{-mz} m^{j-p} \pi (ia-z)^{k-p} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (1+k-p)_p}{j! n! (-j+n)!} $$

where $Re(m) > 0, Re(z) > 0$

### EXAMPLE 3.40.

$$ (3.40) \quad \int_0^\infty \frac{a \cos(mx) - x \sin(mx)}{(a^2 + x^2)(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{(-1)^p 2^{-1+j-2n} e^{-mz} m^{j-p} \pi z^{-1+j-2n} (a+z)^{-1-p} \binom{j}{p} (-j+2n)! (-p)_p}{j! n! (-j+n)!} $$

### EXAMPLE 3.41. Mellin transform.

$$ (3.41) \quad \int_0^\infty x^{-1+s} \frac{(e^{imx} - e^{i(\pi s - mx)})}{(x^2 + z^2)^{n+1}} dx \\ = \sum_{j=0}^n \sum_{p=0}^j \frac{i^{1-s} 2^{j-2n} e^{-mz} m^{j-p} \pi (-z)^{-1-p+s} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (-p+s)_p}{j! n! (-j+n)!} $$

### EXAMPLE 3.42.

$$ (3.42) \quad m^2|m|^{2n-s} \Gamma(-2-2n+s) {}_1F_2\left(1+n; \frac{3}{2}+n-\frac{s}{2}, 2+n-\frac{s}{2}; \frac{m^2 z^2}{4}\right) \\ \qquad \qquad \qquad \left( (1+\operatorname{sgn}(m))\sin(n\pi) + (-1+\operatorname{sgn}(m))\sin(\pi(n-s)) \right) \\ \qquad \qquad \qquad + \frac{z^{-2-2n+s}}{\Gamma(1+n)} \left( mz \cos\left(\frac{\pi s}{2}\right) \Gamma\left(\frac{1}{2}+n-\frac{s}{2}\right) \Gamma\left(\frac{1+s}{2}\right) {}_1F_2\left(\frac{1}{2}+\frac{s}{2}; \frac{3}{2}, \frac{1}{2}-n+\frac{s}{2}; \frac{m^2 z^2}{4}\right) \right. \\ \qquad \qquad \qquad \left. - \Gamma\left(1+n-\frac{s}{2}\right) \Gamma\left(\frac{s}{2}\right) {}_1F_2\left(\frac{s}{2}; \frac{1}{2}, -n+\frac{s}{2}; \frac{m^2 z^2}{4}\right) \sin\left(\frac{\pi s}{2}\right) \right) \\ \qquad \qquad \qquad = \sum_{j=0}^n \sum_{p=0}^j \frac{2^{j-2n} e^{-i\pi s - m\pi} m^{j-p} \pi (-z)^{1-p+s} z^{-3+j-2n} \binom{j}{p} (-j+2n)! (-p+s)_p}{j! n! (-j+n)!} $$

where $Re(m) > 0, Im(m) = 0$.

### EXAMPLE 3.43.
$$
(3.43) \quad \int_0^\infty \frac{((a-x)^k - (a+x)^k) \sin(mx)}{(x^2+z^2)^{n+1}} dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^j \frac{i(-1)^{-p} i^{-k} 2^{-1+j-2n} e^{-mz}}{j! n! (-j+n)!} \\
\qquad \times ((-1)^p - (-1)^j e^{2mz}) m^{j-p} \pi (ia-z)^{k-p} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (1+k-p)_p
$$

THEOREM 3.44. From (2.6.5.2) in [7]. Errata, the quoted equation is in error.

$$
(3.44) \quad \int_0^b x^{-1+m} (b^u - x^u)^n \log^k(ax) \log^z \left(\frac{x}{b}\right) dx \\
\qquad = b^{m+nu} z! \sum_{j=0}^n \sum_{p=0}^{\infty} \frac{(-1)^{j+z} \binom{n}{j} (-1-z)^p (1+k-p)_p \log^{k-p}(ab)}{(m+ju)^{1+p+z}}
$$

where Re(a) < 0, Re(m) > 2π in order for the series to converge.

EXAMPLE 3.45.

$$
(3.45) \quad \int_0^1 x^{-1+m} (1-x^u)^n \log^z(x) \log^k(ax) dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^{\infty} (-1)^{j+z} (m+ju)^{-1-p-z} \binom{n}{j} \binom{-1-z}{p} z! \log^{k-p}(a) (1+k-p)_p
$$

EXAMPLE 3.46.

$$
(3.46) \quad \int_0^1 x^{-1+s} (1-x^2)^n \log^z(x) \log^k(ax) dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^{\infty} (-1)^{j+z} (2j+s)^{-1-p-z} \binom{n}{j} \binom{-1-z}{p} z! \log^{k-p}(a) (1+k-p)_p
$$

THEOREM 3.47. From Eq. (1.11.11.1) in [11]. The exponential integral func-
tion.

$$
(3.47) \quad \int_0^\infty e^{-imx} Ei(-x\alpha) ((-ix + \log(a))^k + e^{2imx} (ix + \log(a))^k) dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{2(-1)^j m^{2j-p} \alpha^{-1-2j} \binom{2j}{p} (1+k-p)_p \log^{k-p}(a)}{1+2j}
$$

where Re(a) ≥ 2π, Re(α) ≥ 2π in order for the series to converge.
$$
(3.43) \quad \int_0^\infty \frac{((a-x)^k - (a+x)^k) \sin(mx)}{(x^2+z^2)^{n+1}} dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^j \frac{i(-1)^{-p} i^{-k} 2^{-1+j-2n} e^{-mz}}{j! n! (-j+n)!} \\
\qquad \times ((-1)^p - (-1)^j e^{2mz}) m^{j-p} \pi (ia-z)^{k-p} z^{-1+j-2n} \binom{j}{p} (-j+2n)! (1+k-p)_p
$$

THEOREM 3.44. From (2.6.5.2) in [8]. Errata, the quoted equation is in error.

$$
(3.44) \quad \int_0^b x^{-1+m} (b^u - x^u)^n \log^k(ax) \log^z \left(\frac{x}{b}\right) dx \\
\qquad = b^{m+nu} z! \sum_{j=0}^n \sum_{p=0}^\infty \frac{(-1)^{j+z} \binom{n}{j} (-1-z)^p (1+k-p)_p \log^{k-p}(ab)}{(m+ju)^{1+p+z}}
$$

where Re(a) < 0, Re(m) > 2π in order for the series to converge.

EXAMPLE 3.45.

$$
(3.45) \quad \int_0^1 x^{-1+m} (1-x^u)^n \log^z(x) \log^k(ax) dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^\infty (-1)^{j+z} (m+ju)^{-1-p-z} \binom{n}{j} \binom{-1-z}{p} z! \log^{k-p}(a) (1+k-p)_p
$$

EXAMPLE 3.46.

$$
(3.46) \quad \int_0^1 x^{-1+s} (1-x^2)^n \log^z(x) \log^k(ax) dx \\
\qquad = \sum_{j=0}^n \sum_{p=0}^{\infty} (-1)^{j+z} (2j+s)^{-1-p-z} \binom{n}{j} \binom{-1-z}{p} z! \log^{k-p}(a) (1+k-p)_p
$$

THEOREM 3.47. From Eq. (1.11.11.1) in [12]. The exponential integral func-
tion.

$$
(3.47) \quad \int_0^\infty e^{-imx} Ei(-x\alpha) ((-ix + \log(a))^k + e^{2imx} (ix + \log(a))^k) dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{2(-1)^j m^{2j-p} \alpha^{-1-2j} \binom{2j}{p} (1+k-p)_p \log^{k-p}(a)}{1+2j}
$$

where Re(a) ≥ 2π, Re(α) ≥ 2π in order for the series to converge.
EXAMPLE 3.48. Fourier sine transform.

$$
(3.48) \quad \int_0^\infty \mathrm{Ei}(-x\alpha) ((-ix + \log(a))^k - (ix + \log(a))^k) \sin(mx) \, dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i(-1)^j (-m^2)^{-p} ((-m)^p m^{2j} - (-m)^{2j} m^p) \alpha^{-1-2j} \binom{2j}{p} \log^{k-p}(a) (1+k-p)_p}{1+2j}
$$

THEOREM 3.49. *From pp.507, Eq. (362(16)) in [12]. Modular Lerch integral.*

$$
(3.49) \quad \int_0^\infty e^{x(2im-p^2x)} x^{1+k} \Phi \left( e^{2imx}, -k, 1 - \frac{ia\pi}{x} \right) dx \\
= \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{2j+1} \frac{i(-1)^j i^{-k} 2^{-1-l} a^{k-l} m^{1+2j-l} n^{1+2j} p^{-3-2j} \pi^{\frac{1}{2}+k-l} \binom{1+2j}{l} (1+k-l)_l}{j!} - \frac{i^{-k} a^k \pi^k}{4p^2}
$$

where $\mathrm{Re}(m) > 0, \mathrm{Re}(a) > 2\pi$ in order for the series to converge.

EXAMPLE 3.50.

$$
(3.50) \quad \int_0^\infty e^{x(2i-4\pi^2x)} \Phi \left( e^{2ix}, 1, 1 - \frac{2i\pi^2}{x} \right) dx \\
= -\frac{i}{32\pi^4} - \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{1+2j} \frac{(-1)^j 2^{-5-2j-2l} n^{1+2j} \pi^{-\frac{9}{2}-2j-2l} \binom{1+2j}{l} (-l)_l}{j!}
$$

EXAMPLE 3.51.

$$
(3.51) \quad \int_0^\infty \frac{e^{x(3i-4\pi^2x)} \Phi\left(e^{3ix}, \frac{3}{2}, 1-\frac{5i\pi}{x}\right)}{\sqrt{x}} dx \\
= -\frac{(-1)^{3/4}}{80\sqrt{5}\pi^{7/2}} + \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{1+2j} \frac{i(-1)^{\frac{3}{4}+j} 2^{-5-4j} 3^{1+2j-l} 5^{-\frac{3}{2}-l} n^{1+2j} \pi^{-4-2j-l} \binom{1+2j}{l} (-\frac{1}{2}-l)_l}{j!}
$$

THEOREM 3.52. *From Eq. (4.134) in [13].*

$$
(3.52) \quad \int_0^\infty e^{-mx^2} \cosh(x) (a^2 - x^2)^k dx = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-1-2j} m^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} a^{2(k-p)} (1+k-p)_p}{j!}
$$

where $\mathrm{Re}(a) > \pi, \mathrm{Re}(m) > \pi$ in order for the series to converge.
EXAMPLE 3.48. Fourier sine transform.

$$
(3.48) \quad \int_0^\infty \mathrm{Ei}(-x\alpha) ((-ix + \log(a))^k - (ix + \log(a))^k) \sin(mx) \, dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^{2j} \frac{i(-1)^j (-m^2)^{-p} ((-m)^p m^{2j} - (-m)^{2j} m^p) \alpha^{-1-2j} \binom{2j}{p} \log^{k-p}(a) (1+k-p)_p}{1+2j}
$$

THEOREM 3.49. *From pp.507, Eq. (362(16)) in [13]. Modular Lerch integral.*

$$
(3.49) \quad \int_0^\infty e^{x(2im-p^2x)} x^{1+k} \Phi \left( e^{2imx}, -k, 1 - \frac{ia\pi}{x} \right) dx \\
= \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{2j+1} \frac{i(-1)^j i^{-k} 2^{-1-l} a^{k-l} m^{1+2j-l} n^{1+2j} p^{-3-2j} \pi^{\frac{1}{2}+k-l} \binom{1+2j}{l} (1+k-l)_l}{j!} - \frac{i^{-k} a^k \pi^k}{4p^2}
$$

where Re(m) > 0, Re(a) > 2π in order for the series to converge.

EXAMPLE 3.50.

$$
(3.50) \quad \int_0^\infty e^{x(2i-4\pi^2x)} \Phi \left( e^{2ix}, 1, 1 - \frac{2i\pi^2}{x} \right) dx \\
= -\frac{i}{32\pi^4} - \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{1+2j} \frac{(-1)^j 2^{-5-2j-2l} n^{1+2j} \pi^{-\frac{9}{2}-2j-2l} \binom{1+2j}{l} (-l)_l}{j!}
$$

EXAMPLE 3.51.

$$
(3.51) \quad \int_0^\infty \frac{e^{x(3i-4\pi^2x)} \Phi\left(e^{3ix}, \frac{3}{2}, 1-\frac{5i\pi}{x}\right)}{\sqrt{x}} dx \\
= -\frac{(-1)^{3/4}}{80\sqrt{5}\pi^{7/2}} + \sum_{n=1}^\infty \sum_{j=0}^\infty \sum_{l=0}^{1+2j} \frac{i(-1)^{\frac{3}{4}+j} 2^{-5-4j} 3^{1+2j-l} 5^{-\frac{3}{2}-l} n^{1+2j} \pi^{-4-2j-l} \binom{1+2j}{l} (-\frac{1}{2}-l)_l}{j!}
$$

THEOREM 3.52. *From Eq. (4.134) in [14].*

$$
(3.52) \quad \int_0^\infty e^{-mx^2} \cosh(x) (a^2 - x^2)^k dx = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-1-2j} m^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} a^{2(k-p)} (1+k-p)_p}{j!}
$$

where Re(a) > π, Re(m) > π in order for the series to converge.
EXAMPLE 3.53.

$$
(3.53) \qquad \int_0^\infty \frac{e^{-mx^2} \cosh(x)(1 - \pi x \cot(\pi x))}{x^2} dx = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-2j} m^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} (-p)_p \zeta(2(1+p))}{j!}
$$

where $Re(m) > 5\pi$ in order for the series to converge.

EXAMPLE 3.54. The *n*-th harmonic number.

$$
(3.54) \quad \int_0^\infty \frac{e^{-mx^2} \cosh(x) \log(a^2 - x^2)}{\sqrt{(a-x)(a+x)}} dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-1-2j} a^{-1-2p} m^{-\frac{1}{2}-j-p} \pi \binom{-\frac{1}{2}-j}{p} \left(H_{-\frac{1}{2}-p} + \log(4) - 2\log(a)\right)}{j! \Gamma(\frac{1}{2}-p)}
$$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

EXAMPLE 3.55.

$$
(3.55) \quad \int_0^\infty \frac{\cosh(x) \log(a^2 - x^2) \sin(mx^2)}{\sqrt{a^2 - x^2}} dx \\
= - \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{\sqrt[4]{-1} \times 4^{-1-j} \left((-i)^{j+p} - i^{1+j+p}\right) a^{-1-2p} m^{-\frac{1}{2}-j-p} \pi \binom{-\frac{1}{2}-j}{p} \left(H_{-\frac{1}{2}-p} + \log(4) - 2\log(a)\right)}{j! \Gamma(\frac{1}{2}-p)}
$$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

EXAMPLE 3.56. Bierens de Haan form.

$$
(3.56) \quad \int_0^\infty ((a-x)(a+x))^k \cos(mx^2) \cosh(x) dx \\
= \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-2-2j} a^{2(k-p)} \left((-im)^{\frac{1}{2}+j+p} + (im)^{\frac{1}{2}+j+p}\right) (m^2)^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} (1+k-p)_p}{j!}
$$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

4. Definite integrals over the real line

In this section we apply the contour integral method in [1] to the equations
cited to derive definite integrals over the real line with examples that follow. These
EXAMPLE 3.53.

$$ (3.53) \qquad \int_0^\infty \frac{e^{-mx^2} \cosh(x)(1 - \pi x \cot(\pi x))}{x^2} dx = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-2j} m^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} (-p)_p \zeta(2(1+p))}{j!} $$

where $Re(m) > 5\pi$ in order for the series to converge.

EXAMPLE 3.54. The n-th harmonic number.

$$ (3.54) \qquad \int_0^\infty \frac{e^{-mx^2} \cosh(x) \log(a^2 - x^2)}{\sqrt{(a-x)(a+x)}} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-1-2j} a^{-1-2p} m^{-\frac{1}{2}-j-p} \pi \binom{-\frac{1}{2}-j}{p} \left(H_{-\frac{1}{2}-p} + \log(4) - 2\log(a)\right)}{j! \Gamma(\frac{1}{2}-p)} $$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

EXAMPLE 3.55.

$$ (3.55) \qquad \int_0^\infty \frac{\cosh(x) \log(a^2 - x^2) \sin(mx^2)}{\sqrt{a^2 - x^2}} dx \\ = - \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{\sqrt[4]{-1} \times 4^{-1-j} \left((-i)^{j+p} - i^{1+j+p}\right) a^{-1-2p} m^{-\frac{1}{2}-j-p} \pi \binom{-\frac{1}{2}-j}{p} \left(H_{-\frac{1}{2}-p} + \log(4) - 2\log(a)\right)}{j! \Gamma(\frac{1}{2}-p)} $$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

EXAMPLE 3.56. Bierens de Haan form.

$$ (3.56) \qquad \int_0^\infty ((a-x)(a+x))^k \cos(mx^2) \cosh(x) dx \\ = \sum_{j=0}^\infty \sum_{p=0}^\infty \frac{2^{-2-2j} a^{2(k-p)} \left((-im)^{\frac{1}{2}+j+p} + (im)^{\frac{1}{2}+j+p}\right) (m^2)^{-\frac{1}{2}-j-p} \sqrt{\pi} \binom{-\frac{1}{2}-j}{p} (1+k-p)_p}{j!} $$

where $Re(a) > \pi, Re(m) > \pi$, in order for the series to converge.

4. Definite integrals over the real line

In this section we apply the contour integral method in [1] to the equations cited to derive definite integrals over the real line with examples that follow. These
integrals are highly oscillatory and have closed form solutions in terms of Special functions.

**THEOREM 4.1.** *From Eqs. (17.23.20(9)) and (17.23.21(9)) in [13].*

$$ (4.1) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} \mathrm{csch}(x\beta)(ix + \log(a))^k dx \\ = -\frac{i\pi \left( -2e^{\frac{\pi(m+i\alpha)}{\beta}} \pi^k \left(\frac{1}{\beta}\right)^k \Phi\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}, -k, 1 + \frac{\beta \log(a)}{\pi}\right) + \log^k(a) \right)}{\beta} $$

where $Re(\alpha) < Re(\beta)$.

**EXAMPLE 4.2.** The polylogarithm function.

$$ (4.2) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} (ix)^k \mathrm{csch}(x\beta) dx = -2i \left(\frac{\pi}{\beta}\right)^{1+k} \mathrm{Li}_{-k}\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}\right) $$

**EXAMPLE 4.3.** The digamma function.

$$ (4.3) \quad \int_{-\infty}^{\infty} e^{-x\alpha} (ix)^k \mathrm{csch}(x\beta) \log(ix) \sin(mx) dx \\ = i2^{-2-k} e^{-\frac{1}{2}ik\pi} \left(\frac{1}{\beta}\right)^{1+k} \Gamma(1+k) \left( \left( \zeta\left(1+k, -\frac{im+\alpha-\beta}{2\beta}\right) - \zeta\left(1+k, \frac{im-\alpha+\beta}{2\beta}\right) \right) \right. \\ \left. \left(i\pi + \log(4) - 2\log\left(\frac{1}{\beta}\right) - 2\psi^{(0)}(1+k)\right) - 2\zeta'\left(1+k, -\frac{im+\alpha-\beta}{2\beta}\right) \right. \\ \left. + 2\zeta'\left(1+k, \frac{im-\alpha+\beta}{2\beta}\right) + e^{ik\pi} \left( \left( \zeta\left(1+k, \frac{-im+\alpha+\beta}{2\beta}\right) - \zeta\left(1+k, \frac{im+\alpha+\beta}{2\beta}\right) \right) \right. \right. \\ \left. \left. \left(-i\pi + \log(4) - 2\log\left(\frac{1}{\beta}\right) - 2\psi^{(0)}(1+k)\right) \right. \right. \\ \left. \left. - 2\zeta'\left(1+k, \frac{-im+\alpha+\beta}{2\beta}\right) + 2\zeta'\left(1+k, \frac{im+\alpha+\beta}{2\beta}\right) \right) \right) $$

where $Re(k) > -1/2$.

**EXAMPLE 4.4.**

$$ (4.4) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} (a+ix)^k \mathrm{csch}(x\beta) dx $$
integrals are highly oscillatory and have closed form solutions in terms of Special functions.

**THEOREM 4.1.** From Eqs. (17.23.20(9)) and (17.23.21(9)) in [14].

$$ (4.1) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} \mathrm{csch}(x\beta)(ix + \log(a))^k dx \\ = -\frac{i\pi \left( -2e^{\frac{\pi(m+i\alpha)}{\beta}} \pi^k \left(\frac{1}{\beta}\right)^k \Phi\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}, -k, 1 + \frac{\beta \log(a)}{\pi}\right) + \log^k(a) \right)}{\beta} $$

where $Re(\alpha) < Re(\beta)$.

**EXAMPLE 4.2.** The polylogarithm function.

$$ (4.2) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} (ix)^k \mathrm{csch}(x\beta) dx = -2i \left(\frac{\pi}{\beta}\right)^{1+k} \mathrm{Li}_{-k}\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}\right) $$

**EXAMPLE 4.3.** The digamma function.

$$ (4.3) \quad \int_{-\infty}^{\infty} e^{-x\alpha} (ix)^k \mathrm{csch}(x\beta) \log(ix) \sin(mx) dx \\ = i2^{-2-k} e^{-\frac{1}{2}ik\pi} \left(\frac{1}{\beta}\right)^{1+k} \Gamma(1+k) \left( \left( \zeta\left(1+k, -\frac{im+\alpha-\beta}{2\beta}\right) - \zeta\left(1+k, \frac{im-\alpha+\beta}{2\beta}\right) \right) \right. \\ \left. \left(i\pi + \log(4) - 2\log\left(\frac{1}{\beta}\right) - 2\psi^{(0)}(1+k)\right) - 2\zeta'\left(1+k, -\frac{im+\alpha-\beta}{2\beta}\right) \right. \\ \left. + 2\zeta'\left(1+k, \frac{im-\alpha+\beta}{2\beta}\right) + e^{ik\pi} \left( \left( \zeta\left(1+k, \frac{-im+\alpha+\beta}{2\beta}\right) - \zeta\left(1+k, \frac{im+\alpha+\beta}{2\beta}\right) \right) \right. \right. \\ \left. \left. \left(-i\pi + \log(4) - 2\log\left(\frac{1}{\beta}\right) - 2\psi^{(0)}(1+k)\right) \right. \right. \\ \left. \left. - 2\zeta'\left(1+k, \frac{-im+\alpha+\beta}{2\beta}\right) + 2\zeta'\left(1+k, \frac{im+\alpha+\beta}{2\beta}\right) \right) \right) $$

where $Re(k) > -1/2$.

**EXAMPLE 4.4.**

$$ (4.4) \quad \int_{-\infty}^{\infty} e^{ix(m+i\alpha)} (a+ix)^k \mathrm{csch}(x\beta) dx $$
$$= - \frac{i\pi \left( a^k - 2e^{\frac{\pi(m+i\alpha)}{\beta}} \pi^k \left(\frac{1}{\beta}\right)^k \Phi\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}, -k, 1+\frac{a\beta}{\pi}\right) \right)}{\beta}$$

where $Re(k) < 0, -1 < Im(k) < 1, 0 < Re(a) > Im(a) < 1, -1 < Im(m) < 1$.

**THEOREM 4.5.** *From Eq. (3.2.20) in [5].*

$$ (4.5) \quad \int_{-\infty}^{\infty} \frac{e^{-x(m+iy)} \log^k (ae^{-x})}{(1+e^{-x})^n} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{i(-1)^{-1-j+n+p} i^{k-p} e^{i\pi(m+iy)} (2\pi)^{1+k-p} (1-m-iy)^{j-p} \binom{j}{p} (1+k-p)_p}{(-1+n)!} \\ \times \Phi\left(e^{2i\pi(m+iy)}, -k+p, -\frac{i(i\pi+\log(a))}{2\pi}\right) S_{-1+n}^{(j)} $$

where $Re(m) > 0, Im(a) > 0$.

**EXAMPLE 4.6.**

$$ (4.6) \quad \int_{-\infty}^{\infty} \frac{e^{-\frac{x}{2}} (1 + e^{-x})^{-n} \log(\log(ae^{-x}))}{\sqrt{\log(ae^{-x})}} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{1}{\Gamma(n)\Gamma(\frac{1}{2}-p)} \left( (-1)^{\frac{3}{4}-j+n+p} i^{-p} 2^{-1-j-p} \pi^{1-p} \binom{j}{p} S_{-1+n}^{(j)} \left( \left( \zeta\left(\frac{1}{2}+p, \frac{\pi-i\log(a)}{4\pi}\right) - \zeta\left(\frac{1}{2}+p, \frac{i\log(a)}{4\pi}\right) \right) (i\pi - 2H_{-\frac{1}{2}-p} + 2\log(\pi)) \right. \right. \\ \left. \left. -2\zeta'\left(\frac{1}{2}+p, \frac{\pi-i\log(a)}{4\pi}\right) + 2\zeta'\left(\frac{1}{2}+p, \frac{3}{4}-\frac{i\log(a)}{4\pi}\right) \right) \right) $$

**EXAMPLE 4.7.**

$$ (4.7) \quad \int_{-\infty}^{\infty} \frac{\mathrm{csch}(x) \log(-x) \sinh\left(\frac{x}{2}\right)}{\sqrt{-x}} dx \\ = \frac{1}{2}(-1)^{3/4} \sqrt{\pi} \left( -i \left( \zeta\left(\frac{1}{2}, \frac{1}{4}\right) - \zeta\left(\frac{1}{2}, \frac{3}{4}\right) \right) (\pi - 2i \log(4\pi)) \right. \\ \left. + 2 \left( \zeta'\left(\frac{1}{2}, \frac{1}{4}\right) - \zeta'\left(\frac{1}{2}, \frac{3}{4}\right) \right) \right) $$

**EXAMPLE 4.8.**
$$ = - \frac{i\pi \left( a^k - 2e^{\frac{\pi(m+i\alpha)}{\beta}} \pi^k \left(\frac{1}{\beta}\right)^k \Phi\left(-e^{\frac{\pi(m+i\alpha)}{\beta}}, -k, 1+\frac{a\beta}{\pi}\right) \right)}{\beta} $$

where $Re(k) < 0, -1 < Im(k) < 1, 0 < Re(a) > Im(a) < 1, -1 < Im(m) < 1$.

**THEOREM 4.5.** *From Eq. (3.2.20) in [5].*

$$ (4.5) \quad \int_{-\infty}^{\infty} \frac{e^{-x(m+iy)} \log^k (ae^{-x})}{(1+e^{-x})^n} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{i(-1)^{-1-j+n+p} i^{k-p} e^{i\pi(m+iy)} (2\pi)^{1+k-p} (1-m-iy)^{j-p} \binom{j}{p} (1+k-p)_p}{(-1+n)!} \\ \times \Phi\left(e^{2i\pi(m+iy)}, -k+p, -\frac{i(i\pi+\log(a))}{2\pi}\right) S_{-1+n}^{(j)} $$

where $Re(m) > 0, Im(a) > 0$.

**EXAMPLE 4.6.**

$$ (4.6) \quad \int_{-\infty}^{\infty} \frac{e^{-\frac{x}{2}} (1 + e^{-x})^{-n} \log(\log(ae^{-x}))}{\sqrt{\log(ae^{-x})}} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{1}{\Gamma(n)\Gamma(\frac{1}{2}-p)} \left( (-1)^{\frac{3}{4}-j+n+p} i^{-p} 2^{-1-j-p} \pi^{1-p} \binom{j}{p} S_{-1+n}^{(j)} \left( \left( \zeta\left(\frac{1}{2}+p, \frac{\pi-i\log(a)}{4\pi}\right) - \zeta\left(\frac{1}{2}+p, \frac{i\log(a)}{4\pi}\right) \right) (i\pi - 2H_{-\frac{1}{2}-p} + 2\log(\pi)) \right. \right. \\ \left. \left. -2\zeta'\left(\frac{1}{2}+p, \frac{\pi-i\log(a)}{4\pi}\right) + 2\zeta'\left(\frac{1}{2}+p, \frac{3}{4}-\frac{i\log(a)}{4\pi}\right) \right) \right) $$

**EXAMPLE 4.7.**

$$ (4.7) \quad \int_{-\infty}^{\infty} \frac{\operatorname{csch}(x) \log(-x) \sinh\left(\frac{x}{2}\right)}{\sqrt{-x}} dx \\ = \frac{1}{2}(-1)^{3/4} \sqrt{\pi} \left( -i \left( \zeta\left(\frac{1}{2}, \frac{1}{4}\right) - \zeta\left(\frac{1}{2}, \frac{3}{4}\right) \right) (\pi - 2i \log(4\pi)) \right. \\ \left. + 2 \left( \zeta'\left(\frac{1}{2}, \frac{1}{4}\right) - \zeta'\left(\frac{1}{2}, \frac{3}{4}\right) \right) \right) $$

**EXAMPLE 4.8.**
$$
(4.8) \quad \int_{-\infty}^{\infty} e^{-2mx} (1+e^{-x})^{-n} \log^k (ae^{-x}) \, dx \\
\qquad = - \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{i(-1)^{-1-j+n+p} i^{k-p} e^{2im\pi} (1-2m)^{j-p} (2\pi)^{1+k-p} \binom{j}{p}}{(-1+n)!} \\
\qquad \qquad \times \Phi \left( e^{4im\pi}, -k+p, -\frac{i(i\pi+\log(a))}{2\pi} \right) (1+k-p)_p S_{-1+n}^{(j)}
$$

THEOREM 4.9. *From (2.4.10.11) in [7].*

$$
(4.9) \quad \int_0^\infty e^{-mx} x^s \log^k (ae^{-x}) \sinh^n(x\alpha) dx \\
\qquad = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{h=0}^n (-1)^h 2^{-n} (-n\alpha)^j (m + 2h\alpha)^{-1-j-l-s} \\
\qquad \qquad \binom{n}{h} \binom{-1-s}{j} \binom{-1-j-s}{l} (1+k-l)_l s! \log^{k-l}(a)
$$

where $Re(a) < -\pi$, $Re(m) > 2\pi$ in order for the series to converge.

EXAMPLE 4.10.

$$
(4.10) \quad \int_0^\infty e^{-mx} x^{-2+s} (-1 + \pi x \cot(\pi x)) \sinh^n(x\alpha) dx \\
\qquad = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{h=0}^n (-1)^h 2^{-n} \left(-1 + e^{i(1-l)\pi}\right) (-n\alpha)^j (m + 2h\alpha)^{-1-j-l-s} \\
\qquad \qquad \binom{n}{h} \binom{-1-s}{j} \binom{-1-j-s}{l} s! (-l)_l \zeta(2+l)
$$

where $Re(m) > 6\pi$ in order for the series to converge.

THEOREM 4.11. *From Eq. (4.11.27) in [5].*

$$
(4.11) \quad \int_0^\infty e^{-st} L_n^\alpha(t) \log^k (ae^{-t}) \, dt \\
\qquad = \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^\infty (-1)^j s^{-1-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \binom{-1+m+\alpha}{m} (1+k-l)_l \log^{k-l}(a)
$$

where $Re(a) > e^\pi$, $Re(m) > 0$ in order for the series to converge.

EXAMPLE 4.12.
$$
(4.8) \quad \int_{-\infty}^{\infty} e^{-2mx} (1+e^{-x})^{-n} \log^k (ae^{-x}) \, dx \\
\qquad = - \sum_{j=0}^{n-1} \sum_{p=0}^{j} \frac{i(-1)^{-1-j+n+p} i^{k-p} e^{2im\pi} (1-2m)^{j-p} (2\pi)^{1+k-p} \binom{j}{p}}{(-1+n)!} \\
\qquad \qquad \times \Phi \left( e^{4im\pi}, -k+p, -\frac{i(i\pi+\log(a))}{2\pi} \right) (1+k-p)_p S_{-1+n}^{(j)}
$$

THEOREM 4.9. *From (2.4.10.11) in [8].*

$$
(4.9) \quad \int_0^\infty e^{-mx} x^s \log^k (ae^{-x}) \sinh^n(x\alpha) dx \\
\qquad = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{h=0}^n (-1)^h 2^{-n} (-n\alpha)^j (m + 2h\alpha)^{-1-j-l-s} \\
\qquad \qquad \binom{n}{h} \binom{-1-s}{j} \binom{-1-j-s}{l} (1+k-l)_l s! \log^{k-l}(a)
$$

where $Re(a) < -\pi$, $Re(m) > 2\pi$ in order for the series to converge.

EXAMPLE 4.10.

$$
(4.10) \quad \int_0^\infty e^{-mx} x^{-2+s} (-1 + \pi x \cot(\pi x)) \sinh^n(x\alpha) dx \\
\qquad = \sum_{l=0}^\infty \sum_{j=0}^\infty \sum_{h=0}^n (-1)^h 2^{-n} \left(-1 + e^{i(1-l)\pi}\right) (-n\alpha)^j (m + 2h\alpha)^{-1-j-l-s} \\
\qquad \qquad \binom{n}{h} \binom{-1-s}{j} \binom{-1-j-s}{l} s! (-l)_l \zeta(2+l)
$$

where $Re(m) > 6\pi$ in order for the series to converge.

THEOREM 4.11. *From Eq. (4.11.27) in [5].*

$$
(4.11) \quad \int_0^\infty e^{-st} L_n^\alpha(t) \log^k (ae^{-t}) \, dt \\
\qquad = \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^\infty (-1)^j s^{-1-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \binom{-1+m+\alpha}{m} (1+k-l)_l \log^{k-l}(a)
$$

where $Re(a) > e^\pi$, $Re(m) > 0$ in order for the series to converge.

EXAMPLE 4.12.
$$
(4.12) \quad \int_0^\infty \frac{e^{-st} (-2 + t \coth(\frac{t}{2})) L_n^\alpha(t)}{t^2} dt \\
= \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^{\infty} (-1)^j (-i)^{-l} i^{-l} 2^{-1-2l} ((-2i\pi)^l + (2i\pi)^l) \pi^{-2-2l} s^{-1-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \binom{-1+m+\alpha}{m} (-l)_l \zeta(2+l)
$$

EXAMPLE 4.13.

$$
(4.13) \quad \int_0^\infty \left(1 - \frac{t}{2a\pi}\right)^{-1+z} L_n^\alpha(t) \sin(st) dt \\
= \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^\infty \frac{2^{-1-l} (1 + (-1)^{j+l}) a^{-l} (-\pi)^{-l} (is)^{-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \binom{-1+m+\alpha}{m} (-l+z)_l}{s}
$$

THEOREM 4.14. *From Eq. (2.1.1.2) in [14].*

$$
(4.14) \quad \int_0^\infty e^{-mx} (x+z)^n \log^k (ae^{-x}) \, dx = \sum_{j=0}^n \sum_{l=0}^\infty \frac{m^{-1+j-l-n} z^j (1+k-l)_l \binom{-1+j-n}{l} n! \log^{k-l}(a)}{j!}
$$
$$
\text{where } \operatorname{Re}(a) < 0, \operatorname{Im}(m) < 0
$$

EXAMPLE 4.15.

$$
(4.15) \quad \int_0^\infty \frac{e^{-mx}(x+z)^n(-1+\pi x \coth(\pi x))}{x^2} dx \\
= \sum_{j=0}^n \sum_{l=0}^\infty \frac{i^l (1+(-1)^l) m^{-1+j-l-n} z^j \binom{-1+j-n}{l} n! (-l)_l \zeta(2+l)}{j!}
$$

THEOREM 4.16. *From Eq. (3.911.4) in [13].*

$$
(4.16) \quad \int_0^\infty \frac{e^{-i(m-in)x} ((-ix + \log(a))^k - e^{2imx}(ix + \log(a))^k)}{1 - e^{-x}} dx \\
= -ia^{-m}(-m)^{-1-k}\Gamma(1+k, -m\log(a)) \\
- i \sum_{j=1}^{n-1} (a^{-ij-m}(-ij-m)^{-1-k}\Gamma(1+k, -((ij+m)\log(a))) \\
+ a^{ij-m}(ij-m)^{-1-k}\Gamma(1+k, -((-ij+m)\log(a))))
$$
$$
(4.12) \quad \int_0^\infty \frac{e^{-st} (-2 + t \coth(\frac{t}{2})) L_n^\alpha(t)}{t^2} dt \\
= \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^{\infty} (-1)^j (-i)^{-l} i^{-l} 2^{-1-2l} ((-2i\pi)^l + (2i\pi)^l) \pi^{-2-2l} s^{-1-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \\
\qquad \qquad \qquad \qquad \qquad \qquad \binom{-1+m+\alpha}{m} (-l)_l \zeta(2+l)
$$

EXAMPLE 4.13.

$$
(4.13) \quad \int_0^\infty \left(1 - \frac{t}{2a\pi}\right)^{-1+z} L_n^\alpha(t) \sin(st) dt \\
\qquad = \sum_{m=0}^n \sum_{j=0}^{n-m} \sum_{l=0}^\infty \frac{2^{-1-l} (1 + (-1)^{j+l}) a^{-l} (-\pi)^{-l} (is)^{-j-l} \binom{-1-j}{l} \binom{-m+n}{j} \binom{-1+m+\alpha}{m} (-l+z)_l}{s}
$$

THEOREM 4.14. From Eq. (2.1.1.2) in [15].

$$
(4.14) \quad \int_0^\infty e^{-mx} (x+z)^n \log^k (ae^{-x}) \, dx = \sum_{j=0}^n \sum_{l=0}^\infty \frac{m^{-1+j-l-n} z^j (1+k-l)_l \binom{-1+j-n}{l} n! \log^{k-l}(a)}{j!}
$$
$$
\text{where } \operatorname{Re}(a) < 0, \operatorname{Im}(m) < 0
$$

EXAMPLE 4.15.

$$
(4.15) \quad \int_0^\infty \frac{e^{-mx}(x+z)^n(-1+\pi x \coth(\pi x))}{x^2} dx \\
\qquad = \sum_{j=0}^n \sum_{l=0}^\infty \frac{i^l (1+(-1)^l) m^{-1+j-l-n} z^j \binom{-1+j-n}{l} n! (-l)_l \zeta(2+l)}{j!}
$$

THEOREM 4.16. From Eq. (3.911.4) in [14].

$$
(4.16) \quad \int_0^\infty \frac{e^{-i(m-in)x} ((-ix + \log(a))^k - e^{2imx}(ix + \log(a))^k)}{1 - e^{-x}} dx \\
= -ia^{-m}(-m)^{-1-k}\Gamma(1+k, -m\log(a)) \\
- i \sum_{j=1}^{n-1} (a^{-ij-m}(-ij-m)^{-1-k}\Gamma(1+k, -((ij+m)\log(a))) \\
+ a^{ij-m}(ij-m)^{-1-k}\Gamma(1+k, -((-ij+m)\log(a))))
$$
$$+ i\pi \left( 2^{1+k} e^{2m\pi} \pi^k \Phi \left( e^{2m\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) + \log^k(a) \right)$$

where $Im(a) \le 0$.

### EXAMPLE 4.17.

$$\begin{aligned} & (4.17) \quad \int_0^\infty \frac{e^{x-nx}(x \cos(mx) - a \sin(mx))}{(-1+e^x)(a^2+x^2)} dx \\ &= -\frac{1}{2}e^{-am}\Gamma(0,-am) - \sum_{j=1}^{n-1} \frac{1}{2}e^{a(-ij-m)} (e^{2iaj}\Gamma(0,ia(j+im)) + \Gamma(0,-a(ij+m))) \\ & \qquad + \frac{\pi + ae^{2m\pi}\Phi(e^{2m\pi}, 1, 1 + \frac{a}{2\pi})}{2a} \end{aligned}$$

### EXAMPLE 4.18.

$$\begin{aligned} & (4.18) \quad \int_0^\infty \frac{e^{x-nx} {}_4F_3(1,1-ix,1+ix,2-x \cot(mx);2-ix,2+ix,1-x \cot(mx);-1)}{(-1+e^x)(1+x^2)} \\ & \qquad \times (x \cos(mx) - \sin(mx)) dx \\ &= \sum_{a=1}^\infty \frac{1}{2}(-1)^a e^{-am}\Gamma(0,-am) \\ & \quad + \sum_{j=1}^{n-1} \sum_{a=1}^\infty \frac{1}{2}(-1)^a e^{a(-ij-m)} (e^{2iaj}\Gamma(0,ia(j+im)) + \Gamma(0,-a(ij+m))) \\ & \qquad + \sum_{a=1}^\infty \frac{(-1)^a (\pi + ae^{2m\pi}\Phi(e^{2m\pi}, 1, 1 + \frac{a}{2\pi}))}{2a} \end{aligned}$$

### EXAMPLE 4.19.

$$\begin{aligned} & (4.19) \quad \int_0^\infty \frac{e^{-\left(-\frac{1}{2}+n\right)x} (e^x \log(-ix + \log(a)) - \log(ix + \log(a)))}{-1+e^x} dx \\ &= 2 \left( a^{-\frac{1}{2}} \Gamma \left( 0, -\frac{1}{2}i \log(a) \right) + \log(\log(a)) \right) \\ & \quad + \sum_{j=1}^{n-1} \frac{a^{-\frac{1}{2}i(1+2j)}}{-1+4j^2} \left( -2a^{2ij}(1+2j)E_1 \left( \frac{1}{2}i(-1+2j)\log(a) \right) \right. \\ & \qquad \left. + (-2+4j)E_1 \left( -\frac{1}{2}i(1+2j)\log(a) \right) + 4a^{\frac{i}{2}+ij} \log \left( \frac{1}{\log(a)} \right) \right) \end{aligned}$$
$$+ i\pi \left( 2^{1+k} e^{2m\pi} \pi^k \Phi \left( e^{2m\pi}, -k, 1 + \frac{\log(a)}{2\pi} \right) + \log^k(a) \right)$$

where $Im(a) \le 0$.

### EXAMPLE 4.17.

$$\begin{aligned} & (4.17) \quad \int_0^\infty \frac{e^{x-nx}(x \cos(mx) - a \sin(mx))}{(-1+e^x)(a^2+x^2)} dx \\ &= -\frac{1}{2}e^{-am}\Gamma(0,-am) - \sum_{j=1}^{n-1} \frac{1}{2}e^{a(-ij-m)} \left( e^{2iaj}\Gamma(0,ia(j+im)) + \Gamma(0,-a(ij+m)) \right) \\ & \qquad + \frac{\pi+a e^{2m\pi}\Phi\left(e^{2m\pi}, 1, 1+\frac{a}{2\pi}\right)}{2a} \end{aligned}$$

### EXAMPLE 4.18.

$$\begin{aligned} & (4.18) \quad \int_0^\infty \frac{e^{x-nx} {}_4F_3(1,1-ix,1+ix,2-x \cot(mx);2-ix,2+ix,1-x \cot(mx);-1)}{(-1+e^x)(1+x^2)} \\ & \qquad \times (x \cos(mx) - \sin(mx)) dx \\ &= \sum_{a=1}^\infty \frac{1}{2}(-1)^a e^{-am}\Gamma(0,-am) \\ & \quad + \sum_{j=1}^{n-1} \sum_{a=1}^\infty \frac{1}{2}(-1)^a e^{a(-ij-m)} \left( e^{2iaj}\Gamma(0,ia(j+im)) + \Gamma(0,-a(ij+m)) \right) \\ & \qquad + \sum_{a=1}^\infty \frac{(-1)^a \left( \pi + a e^{2m\pi}\Phi\left(e^{2m\pi}, 1, 1+\frac{a}{2\pi}\right) \right)}{2a} \end{aligned}$$

### EXAMPLE 4.19.

$$\begin{aligned} & (4.19) \quad \int_0^\infty \frac{e^{-\left(-\frac{1}{2}+n\right)x} (e^x \log(-ix + \log(a)) - \log(ix + \log(a)))}{-1+e^x} dx \\ &= 2 \left( a^{-\frac{1}{2}} \Gamma \left( 0, -\frac{1}{2}i \log(a) \right) + \log(\log(a)) \right) \\ & \quad + \sum_{j=1}^{n-1} \frac{a^{-\frac{1}{2}i(1+2j)}}{-1+4j^2} \left( -2a^{2ij}(1+2j)E_1 \left( \frac{1}{2}i(-1+2j)\log(a) \right) \right. \\ & \qquad \left. + (-2+4j)E_1 \left( -\frac{1}{2}i(1+2j)\log(a) \right) + 4a^{\frac{i}{2}+ij} \log \left( \frac{1}{\log(a)} \right) \right) \end{aligned}$$
$$ -i\pi \log \left( \frac{\Gamma\left(\frac{\log(a)}{4\pi}\right)^2 \log(a)}{4\pi\Gamma\left(\frac{1}{4}\left(2+\frac{\log(a)}{\pi}\right)\right)^2} \right) $$

**THEOREM 4.20.** *From Eq. (3.935) in [13] where $\frac{1}{2} < \mathrm{Re}(p) < 1$.*

$$ (4.20) \qquad \int_0^\pi \frac{e^{e^{-ix}m} \left( (\log ae^{\cos(x)}) - i \sin(x) \right)^k + e^{2im\sin(x)} \left( \log ae^{\cos(x)} + i \sin(x) \right)^k \sin(2nx)}{\sin(x)} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{2j+1} \frac{2m^{1+2j-p}\pi \binom{1+2j}{p} (1+k-p)_p \log^{k-p}(a)}{(1+2j)!} $$

where $\mathrm{Re}(a) < 0$.

**EXAMPLE 4.21.**

$$ (4.21) \qquad \int_0^\pi \frac{(ia \cos(m \sin(x))) + \cos(x - m \sin(x))) \csc(x) \sin(2nx) (\cosh(m \cos(x))) + \sinh(m \cos(x)))}{1 - a^2 + 2ia \cos(x)} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{2j+1} \frac{(ia)^{-1-p} m^{1+2j-p} \pi \binom{1+2j}{p} (-p)_p}{(1+2j)!} $$

**THEOREM 4.22.** *From Eq. (3.915.1) in [13]. An extended Gröbner definite integral. In this derivation we derive the exponential of the exponential function with the cosine function raised to a complex number in terms of the incomplete gamma function.*

$$ (4.22) \qquad \int_0^\pi e^{m \cos(x)} \log^k (ae^{\cos(x)}) \sin(x) dx \\ = (-1)^{-k} a^{-m} m^{-1-k} (\Gamma(1+k, -m(1+\log(a)))) - \Gamma(1+k, m-m\log(a))) $$

where $\mathrm{Im}(a) < 0$.

**EXAMPLE 4.23.**

$$ (4.23) \qquad \int_0^\pi e^{m \cos(x)} \cos^{-1+s}(x) \sin(x) dx = \frac{\Gamma(s, m) - \Gamma(s, -m)}{(-m)^s} $$
$$ -i\pi \log \left( \frac{\Gamma\left(\frac{\log(a)}{4\pi}\right)^2 \log(a)}{4\pi\Gamma\left(\frac{1}{4}\left(2+\frac{\log(a)}{\pi}\right)\right)^2} \right) $$

**THEOREM 4.20.** *From Eq. (3.935) in [14] where $\frac{1}{2} < \text{Re}(p) < 1$.*

$$ (4.20) \qquad \int_0^\pi \frac{e^{e^{-ix}m} \left( (\log ae^{\cos(x)}) - i \sin(x) \right)^k + e^{2im \sin(x)} \left( \log ae^{\cos(x)} + i \sin(x) \right)^k \sin(2nx)}{\sin(x)} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{2j+1} \frac{2m^{1+2j-p} \pi \binom{1+2j}{p} (1+k-p)_p \log^{k-p}(a)}{(1+2j)!} $$

where $\text{Re}(a) < 0$.

**EXAMPLE 4.21.**

$$ (4.21) \qquad \int_0^\pi \frac{(ia \cos(m \sin(x))) + \cos(x - m \sin(x))) \csc(x) \sin(2nx) (\cosh(m \cos(x)) + \sinh(m \cos(x)))}{1 - a^2 + 2ia \cos(x)} dx \\ = \sum_{j=0}^{n-1} \sum_{p=0}^{2j+1} \frac{(ia)^{-1-p} m^{1+2j-p} \pi \binom{1+2j}{p} (-p)_p}{(1+2j)!} $$

**THEOREM 4.22.** *From Eq. (3.915.1) in [14]. An extended Gröbner definite integral. In this derivation we derive the exponential of the exponential function with the cosine function raised to a complex number in terms of the incomplete gamma function.*

$$ (4.22) \qquad \int_0^\pi e^{m \cos(x)} \log^k (ae^{\cos(x)}) \sin(x) dx \\ = (-1)^{-k} a^{-m} m^{-1-k} (\Gamma(1+k, -m(1+\log(a))) - \Gamma(1+k, m-m\log(a))) $$

where $\text{Im}(a) < 0$.

**EXAMPLE 4.23.**

$$ (4.23) \qquad \int_0^\pi e^{m \cos(x)} \cos^{-1+s}(x) \sin(x) dx = \frac{\Gamma(s, m) - \Gamma(s, -m)}{(-m)^s} $$
where $Im(a) < 0$.

EXAMPLE 4.24.

$$ (4.24) \quad \int_0^\pi \frac{e^{\cos(x)} \cos^{-1+s}(x) \sin(x)}{1 + e^{\cos(x)}} dx = \sum_{m=1}^\infty (-1)^m (-m)^{-s} (\Gamma(s, -m) - \Gamma(s, m)) $$

where $Im(a) \neq 0$.

THEOREM 4.25. *From Eq. (3.951.5) in [13].*

$$ (4.25) \quad \int_0^\infty \frac{ie^{-imx} \left( \left( -ix + \log(a) \right)^k - e^{2imx} (ix + \log(a))^k \right)}{-1 + e^{bx}} dx \\ = a^{-m} (-m)^{-1-k} \Gamma(1+k, -m \log(a)) \\ - \frac{\pi \left( 2^{1+k} \left( \frac{1}{b} \right)^k e^{\frac{2m\pi}{b}} \pi^k \Phi \left( e^{\frac{2m\pi}{b}}, -k, 1 + \frac{b \log(a)}{2\pi} \right) + \log^k(a) \right)}{b} $$

where $Re(b) > 0$.

EXAMPLE 4.26. Eq. (3.951.13) in [13]. Mellin transform representation with generalized parameters.

$$ (4.26) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx)}{-1 + e^{bx}} dx \\ = \frac{1}{4} \left( -2e^{-\frac{1}{2}i\pi s} m^{-s} \Gamma(s) + b^{-s} (2\pi)^s \left( \mathrm{Li}_{1-s} \left( e^{-\frac{2m\pi}{b}} \right) + \mathrm{Li}_{1-s} \left( e^{\frac{2m\pi}{b}} \right) \right) \sec\left(\frac{\pi s}{2}\right) \right) $$

where $Re(b) > 0$.

EXAMPLE 4.27. Eq. (3.951.12) in [13]. Mellin transform representation with generalized parameters.

$$ (4.27) \quad \int_0^\infty \frac{x^{-1+s} \sin(mx)}{-1 + e^{bx}} dx \\ = \frac{1}{4} \left( -2ie^{-\frac{1}{2}i\pi s} m^{-s} \Gamma(s) + b^{-s} (2\pi)^s \csc\left(\frac{\pi s}{2}\right) \left( \mathrm{Li}_{1-s}\left(e^{-\frac{2m\pi}{b}}\right) - \mathrm{Li}_{1-s}\left(e^{\frac{2m\pi}{b}}\right) \right) \right) $$

where $Re(b) > 0$.
where $Im(a) < 0$.

EXAMPLE 4.24.

$$ (4.24) \quad \int_0^\pi \frac{e^{\cos(x)} \cos^{-1+s}(x) \sin(x)}{1 + e^{\cos(x)}} dx = \sum_{m=1}^\infty (-1)^m (-m)^{-s} (\Gamma(s, -m) - \Gamma(s, m)) $$

where $Im(a) \neq 0$.

THEOREM 4.25. *From Eq. (3.951.5) in [14].*

$$ (4.25) \quad \int_0^\infty \frac{ie^{-imx} \left( \left( -ix + \log(a) \right)^k - e^{2imx} (ix + \log(a))^k \right)}{-1 + e^{bx}} dx \\ = a^{-m} (-m)^{-1-k} \Gamma(1+k, -m \log(a)) \\ - \frac{\pi \left( 2^{1+k} \left( \frac{1}{b} \right)^k e^{\frac{2m\pi}{b}} \pi^k \Phi \left( e^{\frac{2m\pi}{b}}, -k, 1 + \frac{b \log(a)}{2\pi} \right) + \log^k(a) \right)}{b} $$

where $Re(b) > 0$.

EXAMPLE 4.26. Eq. (3.951.13) in [14]. Mellin transform representation with generalized parameters.

$$ (4.26) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx)}{-1 + e^{bx}} dx \\ = \frac{1}{4} \left( -2e^{-\frac{1}{2}i\pi s} m^{-s} \Gamma(s) + b^{-s} (2\pi)^s \left( \mathrm{Li}_{1-s} \left( e^{-\frac{2m\pi}{b}} \right) + \mathrm{Li}_{1-s} \left( e^{\frac{2m\pi}{b}} \right) \right) \sec\left(\frac{\pi s}{2}\right) \right) $$

where $Re(b) > 0$.

EXAMPLE 4.27. Eq. (3.951.12) in [14]. Mellin transform representation with generalized parameters.

$$ (4.27) \quad \int_0^\infty \frac{x^{-1+s} \sin(mx)}{-1 + e^{bx}} dx \\ = \frac{1}{4} \left( -2ie^{-\frac{1}{2}i\pi s} m^{-s} \Gamma(s) + b^{-s} (2\pi)^s \csc\left(\frac{\pi s}{2}\right) \left( \mathrm{Li}_{1-s}\left(e^{-\frac{2m\pi}{b}}\right) - \mathrm{Li}_{1-s}\left(e^{\frac{2m\pi}{b}}\right) \right) \right) $$

where $Re(b) > 0$.
EXAMPLE 4.28. Mellin transform of the product of sine functions.

$$
(4.28) \quad \int_0^\infty \frac{x^{-1+s} \sin(mx) \sin(px)}{-1 + e^{bx}} dx \\
\qquad = -\frac{1}{8} \left( \begin{aligned} & \left( \left( -(m-p)^2 \right)^{-s} \left( (m-p)^s + (-m+p)^s \right) \right. \\ & \left. - \left( (-m+p)^2 \right)^{-s} \left( (-m-p)^s + (m+p)^s \right) \right) \Gamma(s) \\ & + b^{-s} (2\pi)^s \left( -\text{Li}_{1-s} \left( e^{\frac{2(m-p)\pi}{b}} \right) - \text{Li}_{1-s} \left( e^{\frac{2(-m+p)\pi}{b}} \right) + \text{Li}_{1-s} \left( e^{-\frac{2(m+p)\pi}{b}} \right) \right. \\ & \left. \left. + \text{Li}_{1-s} \left( e^{\frac{2(m+p)\pi}{b}} \right) \right) \right) \sec\left(\frac{\pi s}{2}\right) \end{aligned} \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.29. Eq. (3.951.20) in [13]. Generalized Lobachevsky integral.

$$
(4.29) \quad \int_0^\infty \frac{\sin(mx) \sin(px)}{(-1 + e^{bx})x} dx \\
= \frac{1}{8} \left( -\log\left(1 - e^{\frac{2(m-p)\pi}{b}}\right) - \log\left(1 - e^{\frac{2(-m+p)\pi}{b}}\right) + \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) + \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\
+ \log(-m-p) - \log(m-p) + 2\log(-(m-p)^2) - \log(-m+p) + \log(m+p) \\
\left. - 2\log(-(m+p)^2) \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.30. Eq. (3.951.21). Generalized Lobachevsky integral.

$$
(4.30) \quad \int_0^\infty \frac{\sin^2(px)}{(-1 + e^{bx})x} dx = \frac{1}{8} \log \left( \frac{b^2 \sinh^2\left(\frac{2p\pi}{b}\right)}{4p^2\pi^2} \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.31. Mellin Transform.

$$
(4.31) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx) \cos(2px)}{-1 + e^{bx}} dx \\
= \frac{1}{8} \left( \begin{aligned} & \left( \left( -\left(-(m-2p)^2\right)^{-s} \left( (m-2p)^s + (-m+2p)^s \right) \right. \\ & \left. - \left( (-m+2p)^2 \right)^{-s} \left( (-m-2p)^s + (m+2p)^s \right) \right) \Gamma(s) \end{aligned} \right)
$$
EXAMPLE 4.28. Mellin transform of the product of sine functions.

$$
(4.28) \quad \int_0^\infty \frac{x^{-1+s} \sin(mx) \sin(px)}{-1 + e^{bx}} dx \\
\qquad = -\frac{1}{8} \left( \begin{aligned}[t]
& \left( \left( -(m-p)^2 \right)^{-s} \left( (m-p)^s + (-m+p)^s \right) \right. \\
& \left. - \left( (-m+p)^2 \right)^{-s} \left( (-m-p)^s + (m+p)^s \right) \right) \Gamma(s) \\
& + b^{-s} (2\pi)^s \left( -\text{Li}_{1-s} \left( e^{\frac{2(m-p)\pi}{b}} \right) - \text{Li}_{1-s} \left( e^{\frac{2(-m+p)\pi}{b}} \right) + \text{Li}_{1-s} \left( e^{-\frac{2(m+p)\pi}{b}} \right) \right. \\
& \left. \left. + \text{Li}_{1-s} \left( e^{\frac{2(m+p)\pi}{b}} \right) \right) \right) \sec\left(\frac{\pi s}{2}\right)
\end{aligned} \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.29. Eq. (3.951.20) in [14]. Generalized Lobachevsky integral.

$$
(4.29) \quad \int_0^\infty \frac{\sin(mx) \sin(px)}{(-1 + e^{bx})x} dx \\
= \frac{1}{8} \left( -\log\left(1 - e^{\frac{2(m-p)\pi}{b}}\right) - \log\left(1 - e^{\frac{2(-m+p)\pi}{b}}\right) + \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) + \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\
+ \log(-m-p) - \log(m-p) + 2\log(-(m-p)^2) - \log(-m+p) + \log(m+p) \\
\left. - 2\log(-(m+p)^2) \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.30. Eq. (3.951.21). Generalized Lobachevsky integral.

$$
(4.30) \quad \int_0^\infty \frac{\sin^2(px)}{(-1 + e^{bx})x} dx = \frac{1}{8} \log \left( \frac{b^2 \sinh^2\left(\frac{2p\pi}{b}\right)}{4p^2\pi^2} \right)
$$

where $Re(b) > 0$.

EXAMPLE 4.31. Mellin Transform.

$$
(4.31) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx) \cos(2px)}{-1 + e^{bx}} dx \\
\qquad = \frac{1}{8} \left( \begin{aligned}
& \left( \left( -\left(-(m-2p)^2\right)^{-s} \left( (m-2p)^s + (-m+2p)^s \right) \right. \\
& \left. - \left( (-m+2p)^2 \right)^{-s} \left( (-m-2p)^s + (m+2p)^s \right) \right) \Gamma(s)
\end{aligned} \right)
$$
$$ +b^{-s}(2\pi)^s \left( \mathrm{Li}_{1-s}\left(e^{-\frac{2(m-2p)\pi}{b}}\right) + \mathrm{Li}_{1-s}\left(e^{\frac{2(m-2p)\pi}{b}}\right) + \mathrm{Li}_{1-s}\left(e^{-\frac{2(m+2p)\pi}{b}}\right) \right. \\ \left. \qquad + \mathrm{Li}_{1-s}\left(e^{\frac{2(m+2p)\pi}{b}}\right) \right) \sec\left(\frac{\pi s}{2}\right) $$

where $Re(b) > 0$.

EXAMPLE 4.32. Product of logarithm and sine functions.

$$ \begin{aligned} & (4.32) \quad \int_0^\infty \frac{\log(x) \sin(mx) \sin(px)}{(-1 + e^{bx})x} dx \\ &= \frac{1}{16} \left( -2 \log(2) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(b) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \right. \\ & \qquad \left. -2 \log(2) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(b) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(2) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\ & \qquad \left. -2 \log(b) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) + 2 \log(2) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) - 2 \log(b) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\ & \qquad \left. -2\gamma \log(-m-p) + \log^2(-m-p) + 2\gamma \log(m-p) - \log^2(m-p) - 4\gamma \log(-(m-p)^2) \right. \\ & \qquad \left. + 2 \log(m-p) \log(-(m-p)^2) - 2 \log^2(-(m-p)^2) + 2\gamma \log(-m+p) \right. \\ & \qquad \left. + 2 \log(-(m-p)^2) \log(-m+p) - \log^2(-m+p) - 2\gamma \log(m+p) + \log^2(m+p) \right. \\ & \qquad \left. + 4\gamma \log(-(m+p)^2) - 2 \log(-m-p) \log(-(m+p)^2) - 2 \log(m+p) \log(-(m+p)^2) \right. \\ & \qquad \left. + 2 \log^2(-(m+p)^2) - 2 \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \log(\pi) - 2 \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \log(\pi) \right. \\ & \qquad \left. + 2 \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \log(\pi) + 2 \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \log(\pi) - 2 \mathrm{Li}_1\left(e^{\frac{2(m-p)\pi}{b}}\right) \right. \\ & \qquad \left. - 2 \mathrm{Li}_1\left(e^{\frac{2(-m+p)\pi}{b}}\right) + 2 \mathrm{Li}_1\left(e^{-\frac{2(m+p)\pi}{b}}\right) + 2 \mathrm{Li}_1\left(e^{\frac{2(m+p)\pi}{b}}\right) \right) \end{aligned} $$

where $Re(b) > 0$.

EXAMPLE 4.33. Product of cosine functions.

$$ \begin{aligned} & (4.33) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx) \cos(2px)}{-1 + e^{bx}} dx \\ &= \frac{1}{8} \left( \left( \frac{(-1)^{1+s} ((m-2p)^s + (-m+2p)^s)}{(m-2p)^{2s}} + \frac{(-1)^{1-s} ((-m-2p)^s + (m+2p)^s)}{(m+2p)^{2s}} \right) \Gamma(s) \right. \\ & \qquad \left. + i^{-s} b^{-s} \Gamma(s) \left( \zeta\left(s, -\frac{i(m-2p)}{b}\right) + \zeta\left(s, \frac{i(m-2p)}{b}\right) + \zeta\left(s, -\frac{i(m+2p)}{b}\right) \right. \right. \\ & \qquad \left. \left. + \zeta\left(s, \frac{i(m+2p)}{b}\right) + i^{2s} \left( \zeta\left(s, 1+\frac{i(m-2p)}{b}\right) + \zeta\left(s, \frac{b-im+2ip}{b}\right) \right) \right) \right) \end{aligned} $$
$$+b^{-s}(2\pi)^s \left( \mathrm{Li}_{1-s}\left(e^{-\frac{2(m-2p)\pi}{b}}\right) + \mathrm{Li}_{1-s}\left(e^{\frac{2(m-2p)\pi}{b}}\right) + \mathrm{Li}_{1-s}\left(e^{-\frac{2(m+2p)\pi}{b}}\right) \right. \\ \left. + \mathrm{Li}_{1-s}\left(e^{\frac{2(m+2p)\pi}{b}}\right) \right) \sec\left(\frac{\pi s}{2}\right)$$

where $Re(b) > 0$.

EXAMPLE 4.32. Product of logarithm and sine functions.

$$\begin{aligned} & (4.32) \quad \int_0^\infty \frac{\log(x) \sin(mx) \sin(px)}{(-1 + e^{bx})x} dx \\ &= \frac{1}{16} \left( -2 \log(2) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(b) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \right. \\ & \qquad \left. -2 \log(2) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(b) \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) + 2 \log(2) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\ & \qquad \left. -2 \log(b) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) + 2 \log(2) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) - 2 \log(b) \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \right. \\ & \qquad \left. -2\gamma \log(-m-p) + \log^2(-m-p) + 2\gamma \log(m-p) - \log^2(m-p) - 4\gamma \log(-(m-p)^2) \right. \\ & \qquad \left. + 2 \log(m-p) \log(-(m-p)^2) - 2 \log^2(-(m-p)^2) + 2\gamma \log(-m+p) \right. \\ & \qquad \left. + 2 \log(-(m-p)^2) \log(-m+p) - \log^2(-m+p) - 2\gamma \log(m+p) + \log^2(m+p) \right. \\ & \qquad \left. + 4\gamma \log(-(m+p)^2) - 2 \log(-m-p) \log(-(m+p)^2) - 2 \log(m+p) \log(-(m+p)^2) \right. \\ & \qquad \left. + 2 \log^2(-(m+p)^2) - 2 \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \log(\pi) - 2 \log\left(1 - e^{-\frac{2(m-p)\pi}{b}}\right) \log(\pi) \right. \\ & \qquad \left. + 2 \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \log(\pi) + 2 \log\left(1 - e^{-\frac{2(m+p)\pi}{b}}\right) \log(\pi) - 2 \mathrm{Li}_1\left(e^{\frac{2(m-p)\pi}{b}}\right) \right. \\ & \qquad \left. - 2 \mathrm{Li}_1\left(e^{\frac{2(-m+p)\pi}{b}}\right) + 2 \mathrm{Li}_1\left(e^{-\frac{2(m+p)\pi}{b}}\right) + 2 \mathrm{Li}_1\left(e^{\frac{2(m+p)\pi}{b}}\right) \right) \end{aligned}$$

where $Re(b) > 0$.

EXAMPLE 4.33. Product of cosine functions.

$$\begin{aligned} & (4.33) \quad \int_0^\infty \frac{x^{-1+s} \cos(mx) \cos(2px)}{-1 + e^{bx}} dx \\ &= \frac{1}{8} \left( \left( \frac{(-1)^{1+s} ((m-2p)^s + (-m+2p)^s)}{(m-2p)^{2s}} + \frac{(-1)^{1-s} ((-m-2p)^s + (m+2p)^s)}{(m+2p)^{2s}} \right) \Gamma(s) \right. \\ & \qquad \left. + i^{-s} b^{-s} \Gamma(s) \left( \zeta\left(s, -\frac{i(m-2p)}{b}\right) + \zeta\left(s, \frac{i(m-2p)}{b}\right) + \zeta\left(s, -\frac{i(m+2p)}{b}\right) \right. \right. \\ & \qquad \left. \left. + \zeta\left(s, \frac{i(m+2p)}{b}\right) + i^{2s} \left( \zeta\left(s, 1+\frac{i(m-2p)}{b}\right) + \zeta\left(s, \frac{b-im+2ip}{b}\right) \right) \right) \right) \end{aligned}$$
$$+\zeta\left(s, 1 - \frac{i(m+2p)}{b}\right) + \zeta\left(s, 1 + \frac{i(m+2p)}{b}\right))\right) \sec\left(\frac{\pi s}{2}\right)$$

where $Re(b) > 0$.

### EXAMPLE 4.34.

$$ (4.34) \quad \int_0^\infty \frac{\sin(mx)}{(-1 + e^{bx})(a^2 - x^2)} dx \\ = \frac{i}{4a} \left( e^{-iam}\Gamma(0, -iam) - e^{iam}\Gamma(0, iam) + e^{-\frac{2m\pi}{b}}\Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right. \\ \left. - e^{\frac{2m\pi}{b}}\Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) $$

where $Re(b) > 0, Im(a) \neq 0$.

### EXAMPLE 4.35.

$$ (4.35) \quad \int_0^\infty \frac{\sin(mx)}{(-1 + e^{bx})(a^4 - x^4)} dx \\ = -\frac{1}{8a^3} \left( \begin{aligned} & -e^{-am}E_1(-am) - ie^{-iam}E_1(-iam) + ie^{iam}E_1(iam) + e^{am}E_1(am) \\ & + e^{-\frac{2m\pi}{b}} \left( \Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 - \frac{ab}{2\pi}\right) - i\Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) \\ & - e^{\frac{2m\pi}{b}} \left( \Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 - \frac{ab}{2\pi}\right) - i\Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) \end{aligned} \right) $$

where $Re(b) > 0, Im(a) < 0, Re(a) < 0$.

### THEOREM 4.36. From Eq. (2.4.10) in [5].

$$ (4.36) \quad \int_0^\infty \frac{ie^{-imx} \left( \left(-ix + \log(a)\right)^k - e^{2imx} \left(ix + \log(a)\right)^k \right)}{1 + e^{bx}} dx \\ = -a^{-m}(-m)^{-1-k}\Gamma(1+k, -m\log(a)) \\ + 2^{k+1} \left(\frac{1}{b}\right)^{1+k} e^{\frac{m\pi}{b}} \pi^{1+k} \Phi\left(e^{\frac{2m\pi}{b}}, -k, \frac{b\left(\frac{\pi}{b} + \log(a)\right)}{2\pi}\right) $$

where $Re(b) > 0$.

### EXAMPLE 4.37.
$$+\zeta\left(s, 1 - \frac{i(m+2p)}{b}\right) + \zeta\left(s, 1 + \frac{i(m+2p)}{b}\right)\right)\sec\left(\frac{\pi s}{2}\right)$$

where $Re(b) > 0$.

### EXAMPLE 4.34.

$$\begin{aligned} & (4.34) \quad \int_0^\infty \frac{\sin(mx)}{(-1 + e^{bx})(a^2 - x^2)} dx \\ &= \frac{i}{4a} \left( e^{-iam}\Gamma(0, -iam) - e^{iam}\Gamma(0, iam) + e^{-\frac{2m\pi}{b}}\Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right. \\ & \qquad \left. - e^{\frac{2m\pi}{b}}\Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) \end{aligned}$$

where $Re(b) > 0, Im(a) \neq 0$.

### EXAMPLE 4.35.

$$\begin{aligned} & (4.35) \quad \int_0^\infty \frac{\sin(mx)}{(-1 + e^{bx})(a^4 - x^4)} dx \\ &= -\frac{1}{8a^3} \left( -e^{-am}E_1(-am) - ie^{-iam}E_1(-iam) + ie^{iam}E_1(iam) + e^{am}E_1(am) \right. \\ & \qquad \left. + e^{-\frac{2m\pi}{b}} \left( \Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 - \frac{ab}{2\pi}\right) - i\Phi\left(e^{-\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) \right. \\ & \qquad \left. - e^{\frac{2m\pi}{b}} \left( \Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 - \frac{ab}{2\pi}\right) - i\Phi\left(e^{\frac{2m\pi}{b}}, 1, 1 + \frac{iab}{2\pi}\right) \right) \right) \end{aligned}$$

where $Re(b) > 0, Im(a) < 0, Re(a) < 0$.

### THEOREM 4.36. From Eq. (2.4.10) in [5].

$$\begin{aligned} & (4.36) \quad \int_0^\infty \frac{ie^{-imx} \left( (-ix + \log(a))^k - e^{2imx}(ix + \log(a))^k \right)}{1 + e^{bx}} dx \\ &= -a^{-m}(-m)^{-1-k}\Gamma(1+k, -m\log(a)) \\ & \quad + 2^{k+1} \left(\frac{1}{b}\right)^{1+k} e^{\frac{m\pi}{b}} \pi^{1+k} \Phi\left(e^{\frac{2m\pi}{b}}, -k, \frac{b\left(\frac{\pi}{b} + \log(a)\right)}{2\pi}\right) \end{aligned}$$

where $Re(b) > 0$.

### EXAMPLE 4.37.
$$
(4.37) \quad \int_0^\infty \frac{x^k \sin(mx)}{1+e^{bx}} dx \\
= \frac{1}{4} \left( (1+(-1)^k) m^{-1-k} \Gamma(1+k) + b^{-1-k} e^{-\frac{m\pi}{b}} (2\pi)^{1+k} \left( -\Phi\left(e^{-\frac{2m\pi}{b}}, -k, \frac{1}{2}\right) \right. \right. \\
\left. \left. + e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{2m\pi}{b}}, -k, \frac{1}{2}\right) \right) \right) \sec\left(\frac{k\pi}{2}\right)
$$

where $Re(b) > 0$.

EXAMPLE 4.38.

$$
(4.38) \quad \int_0^\infty \frac{\cos(mx)}{1+e^{bx}} dx = -\frac{i}{2m} - \frac{e^{-\frac{m\pi}{b}} \Phi'\left(e^{-\frac{2m\pi}{b}}, 0, \frac{1}{2}\right) + e^{\frac{m\pi}{b}} \Phi'\left(e^{\frac{2m\pi}{b}}, 0, \frac{1}{2}\right)}{b}
$$

where $Re(b) > 0$.

EXAMPLE 4.39. Extended Fock trigonometric integral, see [15] and Eq. (2.4.12)
in [5].

$$
(4.39) \quad \int_0^\infty \frac{\sinh(xz)}{(1+e^{bx})x} dx = \int_0^z \left( -\frac{1}{2m} - \frac{e^{-\frac{im\pi}{b}} \Phi'\left(e^{-\frac{2im\pi}{b}}, 0, \frac{1}{2}\right) + e^{\frac{im\pi}{b}} \Phi'\left(e^{\frac{2im\pi}{b}}, 0, \frac{1}{2}\right)}{b} \right) dm
$$

where $Re(z) < 0$.

EXAMPLE 4.40.

$$
(4.40) \quad \int_0^\infty \frac{x^k \cos(px) \sin(mx)}{1+e^{bx}} dx \\
= \frac{1}{8} \left( \begin{aligned}
& \left( ((-1)^{-k} (1+(-1)^k) (m-p)^{-1+k} (-1)^{-k}) ((m-p)^2)^{-k} \right. \\
& \left. \qquad + \frac{(-(-m+p)^2)^{-k} ((-m-p)^k + (m+p)^k)}{m+p} \right) \Gamma(1+k) \\
& + \left( \frac{1}{b} \right)^{1+k} e^{-\frac{(m+p)\pi}{b}} (2\pi)^{1+k} \left( e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{2(m-p)\pi}{b}}, -k, \frac{1}{2}\right) - e^{\frac{2p\pi}{b}} \Phi\left(e^{\frac{2(-m+p)\pi}{b}}, -k, \frac{1}{2}\right) \right. \\
& \left. \qquad - \Phi\left(e^{-\frac{2(m+p)\pi}{b}}, -k, \frac{1}{2}\right) + e^{\frac{2(m+p)\pi}{b}} \Phi\left(e^{\frac{2(m+p)\pi}{b}}, -k, \frac{1}{2}\right) \right) \sec\left(\frac{k\pi}{2}\right)
\end{aligned} \right)
$$

where $Re(b) > 0$.
$$ (4.37) \quad \int_0^\infty \frac{x^k \sin(mx)}{1+e^{bx}} dx \\ = \frac{1}{4} \left( (1+(-1)^k) m^{-1-k} \Gamma(1+k) + b^{-1-k} e^{-\frac{m\pi}{b}} (2\pi)^{1+k} \left( -\Phi\left(e^{-\frac{2m\pi}{b}}, -k, \frac{1}{2}\right) \right. \right. \\ \left. \left. + e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{2m\pi}{b}}, -k, \frac{1}{2}\right) \right) \right) \sec\left(\frac{k\pi}{2}\right) $$

where $Re(b) > 0$.

EXAMPLE 4.38.

$$ (4.38) \quad \int_0^\infty \frac{\cos(mx)}{1+e^{bx}} dx = -\frac{i}{2m} - \frac{e^{-\frac{m\pi}{b}} \Phi'\left(e^{-\frac{2m\pi}{b}}, 0, \frac{1}{2}\right) + e^{\frac{m\pi}{b}} \Phi'\left(e^{\frac{2m\pi}{b}}, 0, \frac{1}{2}\right)}{b} $$

where $Re(b) > 0$.

EXAMPLE 4.39. Extended Fock trigonometric integral, see [16] and Eq. (2.4.12) in [5].

$$ (4.39) \quad \int_0^\infty \frac{\sinh(xz)}{(1+e^{bx})x} dx = \int_0^z \left( -\frac{1}{2m} - \frac{e^{-\frac{im\pi}{b}} \Phi'\left(e^{-\frac{2im\pi}{b}}, 0, \frac{1}{2}\right) + e^{\frac{im\pi}{b}} \Phi'\left(e^{\frac{2im\pi}{b}}, 0, \frac{1}{2}\right)}{b} \right) dm $$

where $Re(z) < 0$.

EXAMPLE 4.40.

$$ (4.40) \quad \int_0^\infty \frac{x^k \cos(px) \sin(mx)}{1+e^{bx}} dx \\ = \frac{1}{8} \left( \begin{aligned} & \left( ((-1)^{-k} (1+(-1)^k) (m-p)^{-1+k} (-1)^{-k} ((m-p)^2)^{-k} \right. \\ & \left. + \frac{(-m+p)^{-k} ((-m-p)^k + (m+p)^k)}{m+p} \right) \Gamma(1+k) \\ & + \left( \frac{1}{b} \right)^{1+k} e^{-\frac{(m+p)\pi}{b}} (2\pi)^{1+k} \left( e^{\frac{2m\pi}{b}} \Phi\left(e^{\frac{2(m-p)\pi}{b}}, -k, \frac{1}{2}\right) - e^{\frac{2p\pi}{b}} \Phi\left(e^{\frac{2(-m+p)\pi}{b}}, -k, \frac{1}{2}\right) \right. \\ & \left. - \Phi\left(e^{-\frac{2(m+p)\pi}{b}}, -k, \frac{1}{2}\right) + e^{\frac{2(m+p)\pi}{b}} \Phi\left(e^{\frac{2(m+p)\pi}{b}}, -k, \frac{1}{2}\right) \right) \sec\left(\frac{k\pi}{2}\right) \end{aligned} \right) $$

where $Re(b) > 0$.
**THEOREM 4.41.** *From Eq. (3.958.1) in [13].*

$$ (4.41) \quad \int_{-\infty}^{\infty} e^{-i(m+px)-x^2\alpha-x\beta-\gamma} x^n \left( (-i + \log(a))^k - e^{2i(m+px)}(i + \log(a))^k \right) dx \\ = \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{j=0}^{n-2l} \frac{2^{-n} \exp \left( -i \left( m - \frac{j\pi}{2} - \frac{p\beta}{2\alpha} \right) + \frac{p^2+\beta^2}{4\alpha} - \gamma \right) p^j \sqrt{\pi} \left( -\frac{1}{\alpha} \right)^n \sqrt{\frac{1}{\alpha}} \alpha^l \beta^{-j-2l+n} l! (-2l+n)!}{(-2l+n)!} \\ \times n! \binom{-2l+n}{j} \left( (-i + \log(a))^k - e^{2i(m-\frac{j\pi}{2}-\frac{p\beta}{2\alpha})} (i + \log(a))^k \right) $$

where $Re(\alpha) > 0$.

**EXAMPLE 4.42.**

$$ (4.42) \quad \int_{-\infty}^{\infty} e^{-x(ip+x\alpha+\beta)-\gamma} \left( (i(-1+a))^k + (i(1+a))^k e^{2ipx} \right) x^n dx \\ = \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{j=0}^{n-2l} \frac{2^{-n} \exp \left( -\frac{1}{2}i(j-2n)\pi - \frac{(p+i\beta)^2}{4\alpha} - \gamma \right) \left( (i(1+a))^k + (i(-1+a))^k e^{\frac{i(j\pi\alpha+p\beta)}{\alpha}} \right)}{l!(-2l+n)!} \\ \times p^j \sqrt{\pi} \alpha^{-\frac{1}{2}+l-n} \beta^{-j-2l+n} \binom{-2l+n}{j} n! $$

where $Re(\alpha) > 0$.

**THEOREM 4.43.** *From Eq. (3.978) in [13]. Extended Whittaker integral, see [16] page 279.*

$$ (4.43) \quad \int_0^\infty \exp(-mx(2i+x) - i(1+2b)\tan^{-1}(x)) (1+x^2)^{-\frac{1}{2}+b} \\ \left( e^{2i \tan^{-1}(x)} (-2ix + \log(ae^{-x^2}))^k + e^{4i(mx+b\tan^{-1}(x))} (2ix + \log(ae^{-x^2}))^k \right) dx \\ = \sum_{j=0}^\infty e^{-m} m^{-b-j} \binom{-b}{j} (1-j+k)_j \Gamma(b) \log^{-j+k} \left(\frac{a}{e}\right) \sin(b\pi) $$

where $Re(b) > 0, Re(a) > e^{2\pi}, Re(m) > \pi$ in order for the series to converge.

**EXAMPLE 4.44.**

$$ (4.44) \quad \int_0^\infty e^{-5x(2i+x)-2i \tan^{-1}(x)} \left( \frac{e^{4i(5x+\frac{1}{2}\tan^{-1}(x))}}{5-x(-2i+x)} + \frac{e^{2i \tan^{-1}(x)}}{5-x(2i+x)} \right) dx $$
THEOREM 4.41. *From Eq. (3.958.1) in [14].*

$$ (4.41) \quad \begin{aligned} & \int_{-\infty}^{\infty} e^{-i(m+px)-x^2\alpha-x\beta-\gamma} x^n \left( (-i+\log(a))^k - e^{2i(m+px)}(i+\log(a))^k \right) dx \\ & = \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{j=0}^{n-2l} \frac{2^{-n} \exp \left( -i \left( m - \frac{j\pi}{2} - \frac{p\beta}{2\alpha} \right) + \frac{p^2+\beta^2}{4\alpha} - \gamma \right) p^j \sqrt{\pi} \left( -\frac{1}{\alpha} \right)^n \sqrt{\frac{1}{\alpha}} \alpha^l \beta^{-j-2l+n} l! (-2l+n)!}{(-2l+n)!} \\ & \qquad \times n! \binom{-2l+n}{j} \left( (-i+\log(a))^k - e^{2i(m-\frac{j\pi}{2}-\frac{p\beta}{2\alpha})} (i+\log(a))^k \right) \end{aligned} $$

where $Re(\alpha) > 0$.

EXAMPLE 4.42.

$$ (4.42) \quad \begin{aligned} & \int_{-\infty}^{\infty} e^{-x(ip+x\alpha+\beta)-\gamma} \left( (i(-1+a))^k + (i(1+a))^k e^{2ipx} \right) x^n dx \\ & = \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{j=0}^{n-2l} \frac{2^{-n} \exp \left( -\frac{1}{2}i(j-2n)\pi - \frac{(p+i\beta)^2}{4\alpha} - \gamma \right) \left( (i(1+a))^k + (i(-1+a))^k e^{\frac{i(j\pi\alpha+p\beta)}{\alpha}} \right)}{l!(-2l+n)!} \\ & \qquad \times p^j \sqrt{\pi} \alpha^{-\frac{1}{2}+l-n} \beta^{-j-2l+n} \binom{-2l+n}{j} n! \end{aligned} $$

where $Re(\alpha) > 0$.

THEOREM 4.43. *From Eq. (3.978) in [14]. Extended Whittaker integral, see [17] page 279.*

$$ (4.43) \quad \begin{aligned} & \int_0^\infty \exp(-mx(2i+x) - i(1+2b)\tan^{-1}(x)) (1+x^2)^{-\frac{1}{2}+b} \\ & \qquad \left( e^{2i \tan^{-1}(x)} (-2ix + \log(ae^{-x^2}))^k + e^{4i(mx+b\tan^{-1}(x))} (2ix + \log(ae^{-x^2}))^k \right) dx \\ & = \sum_{j=0}^\infty e^{-m} m^{-b-j} \binom{-b}{j} (1-j+k)_j \Gamma(b) \log^{-j+k} \left(\frac{a}{e}\right) \sin(b\pi) \end{aligned} $$

where $Re(b) > 0, Re(a) > e^{2\pi}, Re(m) > \pi$ in order for the series to converge.

EXAMPLE 4.44.

$$ (4.44) \quad \int_0^\infty e^{-5x(2i+x)-2i \tan^{-1}(x)} \left( \frac{e^{4i(5x+\frac{1}{2}\tan^{-1}(x))}}{5-x(-2i+x)} + \frac{e^{2i \tan^{-1}(x)}}{5-x(2i+x)} \right) dx $$
$$= \sum_{j=0}^{\infty} \frac{4^{-1-j} 5^{-\frac{1}{2}-j} \sqrt{\pi} \left(-\frac{1}{j}\right) (-j)_j}{e^5}$$

## 5. The Lagrange theorem and definite integrals

In this section we apply the theorem given in equation (18) in [3] to derive the Theorems that follow.

### THEOREM 5.1. *The Laguerre polynomial.*

$$ (5.1) \qquad \frac{e^{-\frac{a(1+x)}{-1+x}} (-1+x)^2}{1-2(1+b)x+x^2} = \sum_{n=0}^{\infty} e^{a+b(m+n)} \left(e^{\frac{b(1+x)}{-1+x}}\right)^{m+n} x^n L_n^{-1}(-2(a+b(m+n))) \\ \text{where } |Re(a)|<1, |Re(b)|<1, |Re(x)|<1 \text{ and } f(x) = x^m e^{\frac{(x+1)(a+bn)}{1-x}} $$

### THEOREM 5.2.

$$ (5.2) \qquad \frac{e^{\frac{(a+bm)(1+cx)}{c}}}{(-1+bx)(-1+c^2x^2)} \\ = \sum_{n=0}^{\infty} \frac{\left(ce^{-\frac{b(1+cx)}{c}}x\right)^n \left((-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right)\right)}{2\Gamma(1+n)} \\ \text{where } |Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1, |Re(x)|<1 \text{ and } f(x) = \frac{x^m e^{ax+bxn}}{1-c^2x^2}. $$

### EXAMPLE 5.3.

$$ (5.3) \qquad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log(x)}{(1-bx)(-1+c^2x^2)} dx \\ = \sum_{n=0}^{\infty} \frac{c^n e^{-\frac{bn}{c}} \left((-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right)\right)}{2(1+n)\Gamma(2+n)} \\ \times {}_2F_2(1+n, 1+n; 2+n, 2+n; -bn) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

### EXAMPLE 5.4.
$$= \sum_{j=0}^{\infty} \frac{4^{-1-j} 5^{-\frac{1}{2}-j} \sqrt{\pi} \left(-\frac{1}{j}\right) (-j)_j}{e^5}$$

## 5. The Lagrange theorem and definite integrals

In this section we apply the theorem given in equation (18) in [3] to derive the Theorems that follow.

### THEOREM 5.1. *The Laguerre polynomial.*

$$ (5.1) \qquad \frac{e^{-\frac{a(1+x)}{-1+x}} (-1+x)^2}{1-2(1+b)x+x^2} = \sum_{n=0}^{\infty} e^{a+b(m+n)} \left(e^{\frac{b(1+x)}{-1+x}}\right)^{m+n} x^n L_n^{-1}(-2(a+b(m+n))) \\ \text{where } |Re(a)|<1, |Re(b)|<1, |Re(x)|<1 \text{ and } f(x) = x^m e^{\frac{(x+1)(a+bn)}{1-x}} $$

### THEOREM 5.2.

$$ (5.2) \qquad \frac{e^{\frac{(a+bm)(1+cx)}{c}}}{(-1+bx)(-1+c^2x^2)} \\ = \sum_{n=0}^{\infty} \frac{\left(ce^{-\frac{b(1+cx)}{c}}x\right)^n \left((-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right)\right)}{2\Gamma(1+n)} \\ \text{where } |Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1, |Re(x)|<1 \text{ and } f(x) = \frac{x^m e^{ax+bxn}}{1-c^2x^2}. $$

### EXAMPLE 5.3.

$$ (5.3) \qquad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log(x)}{(1-bx)(-1+c^2x^2)} dx \\ = \sum_{n=0}^{\infty} \frac{c^n e^{-\frac{bn}{c}} \left((-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right)\right)}{2(1+n)\Gamma(2+n)} \\ \times {}_2F_2(1+n, 1+n; 2+n, 2+n; -bn) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

### EXAMPLE 5.4.
$$
(5.4) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^2(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} \frac{c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^3 \Gamma(1+n)} \\
\times {}_3F_3(1+n, 1+n, 1+n; 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.5.

$$
(5.5) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^3(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} - \frac{3c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^4 \Gamma(1+n)} \\
\times {}_4F_4(1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.6.

$$
(5.6) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^4(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} \frac{12c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^5 \Gamma(1+n)} \\
\times {}_5F_5(1+n, 1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.7.

$$
(5.7) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^5(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= -\sum_{n=0}^{\infty} \frac{60c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^6 \Gamma(1+n)} \\
\times {}_6F_6(1+n, 1+n, 1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n, 2+n, 2+n; -bn)
$$
$$
(5.4) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^2(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} \frac{c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^3 \Gamma(1+n)} \\
\times {}_3F_3(1+n, 1+n, 1+n; 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.5.

$$
(5.5) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^3(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} - \frac{3c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^4 \Gamma(1+n)} \\
\times {}_4F_4(1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.6.

$$
(5.6) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^4(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= \sum_{n=0}^{\infty} \frac{12c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^5 \Gamma(1+n)} \\
\times {}_5F_5(1+n, 1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n, 2+n; -bn)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

EXAMPLE 5.7.

$$
(5.7) \quad \int_0^1 \frac{e^{\frac{(a+bm)(1+cx)}{c}} \log^5(x)}{(-1+bx)(-1+c^2x^2)} dx \\
= -\sum_{n=0}^{\infty} \frac{60c^n e^{-\frac{bn}{c}} \left( (-1)^n \Gamma\left(1+n, -\frac{a+b(m+n)}{c}\right) + e^{\frac{2(a+b(m+n))}{c}} \Gamma\left(1+n, \frac{a+b(m+n)}{c}\right) \right)}{(1+n)^6 \Gamma(1+n)} \\
\times {}_6F_6(1+n, 1+n, 1+n, 1+n, 1+n, 1+n; 2+n, 2+n, 2+n, 2+n, 2+n, 2+n; -bn)
$$
where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

### THEOREM 5.8.

$$ (5.8) \quad \frac{e^{ax}(1-x)^{1-d}x^m}{1+(-1+c+b(-1+x))x} \\ = \sum_{n=0}^{\infty} \frac{(-1)^n (e^{-bx}(1-x)^{-c}x)^{m+n} \Gamma(1-d+c(m+n))}{\Gamma(1+n)\Gamma(1-d-n+c(m+n))} \\ \times {}_1F_1(-n; 1-d+m+(-1+c)(m+n); a+b(m+n)) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Im(d)| \neq 0, -Re(x)-i1, |Re(m)|<1$ and
$f(x) = x^m e^{ax+bnx} (1-x)^{cn-d}$

### THEOREM 5.9.

$$ (5.9) \quad \frac{e^{(a+bm)x}}{1-bx} = \sum_{n=0}^{\infty} \frac{(a+b(m+n))^n (e^{-bx}x)^n}{\Gamma(1+n)} $$

where $|Re(a)|<1, |Re(b)|<1, |Re(m)|<1, |Re(x)|<1$. where $f(x) = x^m e^{x(a+bn)}$

### EXAMPLE 5.10. Sine integral function.

$$ (5.10) \quad \mathrm{Si}((a+bm)x) + bx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (a+b(m+n))^n x^n \sin\left(\frac{1}{2}n(\pi+2bx)\right)}{n^2 \Gamma(n)} $$

where $|Re(a)|<1, |Re(b)|<1, |Re(m)|<1, |Re(x)|<1$.

### THEOREM 5.11.

$$ (5.11) \quad \frac{e^{ax}(1-x)^{1-d}x^m}{1+(-1+c+b(-1+x))x} \\ = \sum_{n=0}^{\infty} \frac{(-1)^n (e^{-bx}(1-x)^{-c}x)^{m+n} \Gamma(1-d+c(m+n))}{\Gamma(1+n)\Gamma(1-d-n+c(m+n))} \\ \times {}_1F_1(-n; 1-d+m+(-1+c)(m+n); a+b(m+n)) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Im(d)| \neq 0, -Re(x)-i1, |Re(m)|<1$ and
$f(x) = x^m e^{ax+bnx} (1-x)^{cn-d}$
where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1$.

THEOREM 5.8.

$$ (5.8) \quad \frac{e^{ax}(1-x)^{1-d}x^m}{1+(-1+c+b(-1+x))x} = \sum_{n=0}^{\infty} \frac{(-1)^n (e^{-bx}(1-x)^{-c}x)^{m+n} \Gamma(1-d+c(m+n))}{\Gamma(1+n)\Gamma(1-d-n+c(m+n))} \times {}_1F_1(-n; 1-d+m+(-1+c)(m+n); a+b(m+n)) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Im(d)|\neq 0, -Re(x)-i1, |Re(m)|<1$ and
$f(x) = x^m e^{ax+bnx} (1-x)^{cn-d}$

THEOREM 5.9.

$$ (5.9) \quad \frac{e^{(a+bm)x}}{1-bx} = \sum_{n=0}^{\infty} \frac{(a+b(m+n))^n (e^{-bx}x)^n}{\Gamma(1+n)} $$

where $|Re(a)|<1, |Re(b)|<1, |Re(m)|<1, |Re(x)|<1$. where $f(x) = x^m e^{x(a+bn)}$

EXAMPLE 5.10. Sine integral function.

$$ (5.10) \quad \mathrm{Si}((a+bm)x) + bx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (a+b(m+n))^n x^n \sin\left(\frac{1}{2}n(\pi+2bx)\right)}{n^2 \Gamma(n)} $$

where $|Re(a)|<1, |Re(b)|<1, |Re(m)|<1, |Re(x)|<1$.

THEOREM 5.11.

$$ (5.11) \quad \frac{e^{ax}(1-x)^{1-d}x^m}{1+(-1+c+b(-1+x))x} = \sum_{n=0}^{\infty} \frac{(-1)^n (e^{-bx}(1-x)^{-c}x)^{m+n} \Gamma(1-d+c(m+n))}{\Gamma(1+n)\Gamma(1-d-n+c(m+n))} \times {}_1F_1(-n; 1-d+m+(-1+c)(m+n); a+b(m+n)) $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Im(d)|\neq 0, -Re(x)-i1, |Re(m)|<1$ and
$f(x) = x^m e^{ax+bnx} (1-x)^{cn-d}$
### THEOREM 5.12.

$$ (5.12) \quad \frac{\log(1 - x^2 z^2)}{\sqrt{1+x}} = \sum_{n=0}^{\infty} \frac{\sqrt{\pi} x^{2+n} z \left( {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; -z\right) - {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; z\right) \right)}{\Gamma\left(-\frac{1}{2}-n\right) \Gamma(2+n)} $$ 
where $|Re(x)|<1, |Re(z)|<1$ where $f(x) = \frac{\log(1-x^2 z^2)}{\sqrt{x+1}}$.

### EXAMPLE 5.13.

$$ (5.13) \quad \int_0^1 \frac{e^{-bx} \log(1 - x^2 z^2)}{\sqrt{1+x}} dx = \sum_{n=0}^{\infty} \frac{2^{-3-2n} b^{-3-n} z (\cos(n\pi)\Gamma(-\frac{1}{2}-n)\Gamma(5+2n) + 2^{3+2n}(3+2n)\sqrt{\pi}\Gamma(3+n,b))}{(3+2n)\Gamma(-\frac{1}{2}-n)\Gamma(2+n)} \\ \times \left( {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; -z\right) - {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; z\right) \right) $$ 
where $|Re(b)|>\pi, |Im(b)|>\pi, |Re(z)|<1$.

### THEOREM 5.14.

$$ (5.14) \quad \frac{(1-x)^{1+c} x^m \log(1-ax)}{1+(-1+b)x} = \sum_{n=0}^{\infty} \frac{(-1)^n a n ((1-x)^{-b} x)^{m+n} \Gamma(1+c+b(m+n))}{(m+n)! \Gamma(2+c-n+b(m+n))} \\ \times {}_3F_2(1, 1, 1-n; 2, 2+c+bm-n+bn; a)(1+n)_m $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1, |Re(x)|<1$ where $f(x) = x^m \log(1-ax)(1-x)^{bn+c}$.

### THEOREM 5.15.

$$ (5.15) \quad \frac{b \log \left( \frac{a-b}{b(-1+ax)} \right)}{a(-1+bx)} = \sum_{n=0}^{\infty} \frac{a^n x^n {}_2F_1 \left( 1, 1+n; 2+n; \frac{a}{b} \right)}{1+n} $$ 
where $|Re(a)|<1, |Re(b)|<1, |Re(x)|<1$ where $f(x) = \frac{\log(1-ax)}{1-bx}$.
THEOREM 5.12.

$$ (5.12) \quad \frac{\log(1 - x^2 z^2)}{\sqrt{1+x}} = \sum_{n=0}^{\infty} \frac{\sqrt{\pi} x^{2+n} z \left( {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; -z\right) - {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; z\right) \right)}{\Gamma\left(-\frac{1}{2}-n\right) \Gamma(2+n)} $$

where $|Re(x)|<1, |Re(z)|<1$ where $f(x) = \frac{\log(1-x^2 z^2)}{\sqrt{x+1}}$.

EXAMPLE 5.13.

$$ (5.13) \quad \int_0^1 \frac{e^{-bx} \log(1 - x^2 z^2)}{\sqrt{1+x}} dx = \sum_{n=0}^{\infty} \frac{2^{-3-2n} b^{-3-n} z (\cos(n\pi)\Gamma(-\frac{1}{2}-n)\Gamma(5+2n) + 2^{3+2n}(3+2n)\sqrt{\pi}\Gamma(3+n,b))}{(3+2n)\Gamma(-\frac{1}{2}-n)\Gamma(2+n)} \times \left( {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; -z\right) - {}_3F_2\left(1, 1, -1-n; 2, -\frac{1}{2}-n; z\right) \right) $$

where $|Re(b)|>\pi, |Im(b)|>\pi, |Re(z)|<1$.

THEOREM 5.14.

$$ (5.14) \quad \frac{(1-x)^{1+c} x^m \log(1-ax)}{1+(-1+b)x} = \sum_{n=0}^{\infty} \frac{(-1)^n a n \left((1-x)^{-b} x\right)^{m+n} \Gamma(1+c+b(m+n))}{(m+n)! \Gamma(2+c-n+b(m+n))} \times {}_3F_2(1,1,1-n;2,2+c+bm-n+bn;a)(1+n)_m $$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(m)|<1, |Re(x)|<1$ where $f(x) = x^m \log(1-ax)(1-x)^{bn+c}$.

THEOREM 5.15.

$$ (5.15) \quad \frac{b \log \left( \frac{a-b}{b(-1+ax)} \right)}{a(-1+bx)} = \sum_{n=0}^{\infty} \frac{a^n x^n {}_2F_1 \left( 1, 1+n; 2+n; \frac{a}{b} \right)}{1+n} $$

where $|Re(a)|<1, |Re(b)|<1, |Re(x)|<1$ where $f(x) = \frac{\log(1-ax)}{1-bx}$.
THEOREM 5.16.

$$
(5.16) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^2}
$$

THEOREM 5.17.

$$
(5.17) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^3}
$$

THEOREM 5.18.

$$
(5.18) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^4}
$$

THEOREM 5.19. *The incomplete beta function.*

$$
(5.19) \qquad \frac{\log\left(\frac{b(-1+ax)}{a-b}\right)}{-1+bx} = \sum_{n=0}^{\infty} b^n x^n B_{\frac{a}{b}}(1+n, 0)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{x^4 \log(1-ax)}{1-bx}
$$

THEOREM 5.20.
THEOREM 5.16.

$$
(5.16) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^2}
$$

THEOREM 5.17.

$$
(5.17) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^3}
$$

THEOREM 5.18.

$$
(5.18) \qquad \frac{bx \left(ax(1-bx) + (-1+ax)(-2+bx) \log\left(1-\frac{a}{b}\right) - (-1+ax)(-2+bx) \log(1-ax)\right)}{(-1+ax)(-1+bx)^2} \\
\qquad\qquad = \sum_{n=1}^{\infty} a^n (1+n) x^n \Phi\left(\frac{a}{b}, 1, n\right)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{\log(1-ax)}{(1-bx)^4}
$$

THEOREM 5.19. The *incomplete beta function*.

$$
(5.19) \qquad \frac{\log\left(\frac{b(-1+ax)}{a-b}\right)}{-1+bx} = \sum_{n=0}^{\infty} b^n x^n B_{\frac{a}{b}}(1+n, 0)
$$

$$
\text{where } |\mathrm{Re}(a)|<1, |\mathrm{Re}(b)|<1, |\mathrm{Re}(x)|<1 \text{ where } f(x) = \frac{x^4 \log(1-ax)}{1-bx}
$$

THEOREM 5.20.
$$
\begin{equation} \tag{5.20}
\begin{aligned}
& \frac{1}{a(b-x)(-c+x)(-d+x)} ((c-d)(c-x)(d-x)\log(1-ab) \\
& \qquad + (-b+x)((b-d)(d-x)\log(1-ac) + (b-c)(-c+x)\log(1-ad)) \\
& \qquad + (b-c)(b-d)(-c+d)\log(1-ax)) \\
& = \sum_{n=0}^{\infty} a^n x^n \left( -\left(\frac{1}{ab}\right)^n (ab)^n (c-d)\Phi(ab, 1, 1+n) \right. \\
& \qquad \left. + \left(\frac{1}{ac}\right)^n (ac)^n (b-d)\Phi(ac, 1, 1+n) \right. \\
& \qquad \left. + (-b+c) \left(\frac{1}{ad}\right)^n (ad)^n \Phi(ad, 1, 1+n) \right)
\end{aligned}
\end{equation}
$$

where $|Re(a)|< 1, |Re(b)|< 1, |Re(c)|< 1, |Re(d)|< 1, |Re(x)|< 1$ where $f(x) = \frac{\log(1-ax)}{(x-b)(x-c)(x-d)}$

THEOREM 5.21.

$$
\begin{equation} \tag{5.21}
\begin{aligned}
& \frac{c(c-d)de^{(a+bm)x}}{(c-x)(-d+x)(-1+bx)} \\
& = \sum_{n=0}^{\infty} \frac{c^{-n}d^{-n}(e^{-bx}x)^n}{\Gamma(1+n)} \left( \begin{aligned}[t]
& -d^{1+n}e^{c(a+b(m+n))}\Gamma(1+n,c(a+b(m+n))) \\
& +c^{1+n}e^{d(a+b(m+n))}\Gamma(1+n,d(a+b(m+n)))
\end{aligned} \right)
\end{aligned}
\end{equation}
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(d)|<1, |Re(m)|<1, |Re(x)|<1$ where $f(x) = \frac{x^m e^{x(a+bn)}}{(x-c)(x-d)}$

THEOREM 5.22.

$$
(5.22) \quad - \frac{e^{ax^2}(1+x)}{-1+2bx^2(1+x)} = \sum_{n=0}^{\infty} \left( \frac{e^{-bx^2}x}{1+x} \right)^n {}_2F_2 \left( \frac{1}{2} - \frac{n}{2}, -\frac{n}{2}; \frac{1}{2}; 1+a+bn \right)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(x)|<1$ where $f(x) = (x+1)^n e^{ax^2+bnx^2}$

THEOREM 5.23.

$$
(5.23) \qquad - \frac{e^{ax^2}(c+x)}{2bx^3 + c(-1 + 2bx^2)} = \sum_{n=0}^{\infty} \left( \frac{e^{-bx^2}x}{c+x} \right)^n {}_2F_2 \left( \frac{1}{2} - \frac{n}{2}, -\frac{n}{2}; \frac{1}{2}; c^2(a+bn) \right)
$$
$$
\begin{equation} \tag{5.20}
\begin{aligned}
& \frac{1}{a(b-x)(-c+x)(-d+x)} ((c-d)(c-x)(d-x) \log(1-ab) \\
& \qquad + (-b+x)((b-d)(d-x) \log(1-ac) + (b-c)(-c+x) \log(1-ad)) \\
& \qquad + (b-c)(b-d)(-c+d) \log(1-ax)) \\
& = \sum_{n=0}^{\infty} a^n x^n \left( - \left(\frac{1}{ab}\right)^n (ab)^n (c-d) \Phi(ab, 1, 1+n) \right. \\
& \qquad \left. + \left(\frac{1}{ac}\right)^n (ac)^n (b-d) \Phi(ac, 1, 1+n) \right. \\
& \qquad \left. + (-b+c) \left(\frac{1}{ad}\right)^n (ad)^n \Phi(ad, 1, 1+n) \right)
\end{aligned}
\end{equation}
$$

where $|Re(a)|< 1, |Re(b)|< 1, |Re(c)|< 1, |Re(d)|< 1, |Re(x)|< 1$ where $f(x) = \frac{\log(1-ax)}{(x-b)(x-c)(x-d)}$

THEOREM 5.21.

$$
\begin{equation} \tag{5.21}
\begin{aligned}
& \frac{c(c-d)de^{(a+bm)x}}{(c-x)(-d+x)(-1+bx)} \\
& = \sum_{n=0}^{\infty} \frac{c^{-n}d^{-n}(e^{-bx}x)^n}{\Gamma(1+n)} \left( \begin{aligned}[t]
& -d^{1+n}e^{c(a+b(m+n))}\Gamma(1+n,c(a+b(m+n))) \\
& +c^{1+n}e^{d(a+b(m+n))}\Gamma(1+n,d(a+b(m+n)))
\end{aligned} \right)
\end{aligned}
\end{equation}
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(d)|<1, |Re(m)|<1, |Re(x)|<1$ where
$f(x) = \frac{x^m e^{x(a+bn)}}{(x-c)(x-d)}$

THEOREM 5.22.

$$
(5.22) \quad - \frac{e^{ax^2}(1+x)}{-1+2bx^2(1+x)} = \sum_{n=0}^{\infty} \left( \frac{e^{-bx^2}x}{1+x} \right)^n {}_2F_2 \left( \frac{1}{2} - \frac{n}{2}, -\frac{n}{2}; \frac{1}{2}; 1; a+bn \right)
$$

where $|Re(a)|<1, |Re(b)|<1, |Re(x)|<1$ where $f(x) = (x+1)^n e^{ax^2+bnx^2}$

THEOREM 5.23.

$$
(5.23) \qquad - \frac{e^{ax^2}(c+x)}{2bx^3 + c(-1+2bx^2)} = \sum_{n=0}^{\infty} \left( \frac{e^{-bx^2}x}{c+x} \right)^n {}_2F_2 \left( \frac{1}{2} - \frac{n}{2}, -\frac{n}{2}; \frac{1}{2}; 1; c^2(a+bn) \right)
$$
where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(x)|<1$ where $f(x) = (c+x)^n e^{ax^2+bnx^2}$.

## 6. Finite quadruple series with examples

In this section we look at applying the contour integral method in [1] to a few integrals listed in [8]. This application yields extended definite integral forms in terms of special functions. A few examples in terms of special cases of the parameters involved are evaluated.

**THEOREM 6.1.** *From eq. (1.7.7.3) in [8].*

$$ (6.1) \quad \int_0^\infty \frac{(-1)^{-k} e^{-imx} (x-a)^k + e^{imx} (x+a)^k}{\cosh^{2n}(x\beta)} dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{-1+2n-q-r} e^{\frac{m\pi}{2\beta}} m^{1+q+r-s} \pi^{1+k-s}}{(-1+2n)!} \\ \times \beta^{-2-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \Phi\left(e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) (1+k-s)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} \\ \text{where } \text{Re}(\beta) > \text{Re}(m). $$

**EXAMPLE 6.2.**

$$ (6.2) \quad \int_0^\infty (e^{-ik\pi} (-a+x)^k + (a+x)^k) \cos(mx) \cosh^{-2n}(x\beta) dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{-2+2n-q-r} e^{-\frac{m\pi}{2\beta}} (-m)^{-s} m^{1-s} \pi^{1+k-s}}{\Gamma(2n)} \\ \times \beta^{-2-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \left( (-m)^{q+r} m^s \Phi\left(e^{-\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) \right. \\ \left. - e^{-\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) \right) (1+k-s)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} $$

**EXAMPLE 6.3.** Mellin transform.

$$ (6.3) \quad \int_0^\infty x^{t-1} \frac{\cos(mx)}{\cosh^{2n}(x\beta)} dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-q+r-t} 2^{-3+2n-q-r} e^{-\frac{m\pi}{2\beta}} (-m)^{-s} m^{1-s} \pi^{-s+t}}{\Gamma(2n)} $$
where $|Re(a)|<1, |Re(b)|<1, |Re(c)|<1, |Re(x)|<1$ where $f(x) = (c+x)^n e^{ax^2+bnx^2}$.

## 6. Finite quadruple series with examples

In this section we look at applying the contour integral method in [1] to a few integrals listed in [9]. This application yields extended definite integral forms in terms of special functions. A few examples in terms of special cases of the parameters involved are evaluated.

**THEOREM 6.1.** *From eq. (1.7.7.3) in [9].*

$$ (6.1) \quad \int_0^\infty \frac{(-1)^{-k} e^{-imx} (x-a)^k + e^{imx} (x+a)^k}{\cosh^{2n}(x\beta)} dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{-1+2n-q-r} e^{\frac{m\pi}{2\beta}} m^{1+q+r-s} \pi^{1+k-s}}{(-1+2n)!} \\ \times \beta^{-2-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \Phi\left(e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) (1+k-s)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} \\ \text{where } \mathrm{Re}(\beta) > \mathrm{Re}(m). $$

**EXAMPLE 6.2.**

$$ (6.2) \quad \int_0^\infty (e^{-ik\pi} (-a+x)^k + (a+x)^k) \cos(mx) \cosh^{-2n}(x\beta) dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{-2+2n-q-r} e^{-\frac{m\pi}{2\beta}} (-m)^{-s} m^{1-s} \pi^{1+k-s}}{\Gamma(2n)} \\ \times \beta^{-2-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \left( (-m)^{q+r} m^s \Phi\left(e^{-\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) \right. \\ \left. - e^{\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2}+\frac{ia\beta}{\pi}\right) \right) (1+k-s)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} $$

**EXAMPLE 6.3.** Mellin transform.

$$ (6.3) \quad \int_0^\infty x^{t-1} \frac{\cos(mx)}{\cosh^{2n}(x\beta)} dx \\ = \sum_{j=0}^{n-1} \sum_{l=0}^{n-1} \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{1+q+r} \frac{(-1)^{-j-l+2n+q} i^{-q+r-t} 2^{-3+2n-q-r} e^{-\frac{m\pi}{2\beta}} (-m)^{-s} m^{1-s} \pi^{-s+t}}{\Gamma(2n)} $$
$$ \times \beta^{-1-q-r+s-t} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \left(i + \cot\left(\frac{\pi t}{2}\right)\right) \left((-m)^{q+r} m^s \Phi\left(e^{-\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right) - e^{\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(e^{\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right)\right) (-s+t)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} $$

EXAMPLE 6.4. From equation (2.3.18) in [4].

$$ (6.4) \qquad \Gamma(s) = - \frac{2^{-5+3s}\pi^{-1+s}(-1+s)\left(\zeta\left(2-s, \frac{1}{4}\right) - \zeta\left(2-s, \frac{3}{4}\right)\right) \sec\left(\frac{\pi s}{2}\right)}{\zeta\left(-1+s, \frac{1}{4}\right) - \zeta\left(-1+s, \frac{3}{4}\right)} $$

THEOREM 6.5. From eq. (1.7.7.4) in [8].

$$ \begin{aligned} & (6.5) \quad \int_0^\infty \frac{e^{-imx} \cosh^{-1-2n}(x\beta) ((-ix + \log(a))^k + e^{2imx}(ix + \log(a))^k)}{2k!} dx \\ & = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-2)^{-j-l+2n} e^{\frac{m\pi}{2\beta}} m^{q+r-s} \pi^{1+k-s} \left(-\frac{i}{\beta}\right)^q \left(\frac{i}{\beta}\right)^r \left(\frac{1}{\beta}\right)^{1+k-s} \binom{j}{q} \binom{l}{r} \binom{q+r}{s}}{(2n)!(k-s)!} \\ & \qquad \times \Phi\left(-e^{\frac{m\pi}{\beta}}, -k+s, \frac{\beta\left(\frac{\pi}{2\beta} + \log(a)\right)}{\pi}\right) S_n^{(j)} S_n^{(l)} \end{aligned} $$

EXAMPLE 6.6.

$$ \begin{aligned} & (6.6) \quad \int_0^\infty (-1)^{-k} e^{-imx} ((-a+x)^k + (-1)^k e^{2imx} (a+x)^k) \cosh^{-1-2n}(x\beta) dx \\ & = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{1-j-l+2n} e^{\frac{m\pi}{2\beta}} m^{q+r-s} \pi^{1+k-s}}{(2n)!} \\ & \qquad \times \beta^{-1-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{q+r}{s} \Phi\left(-e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2} + \frac{ia\beta}{\pi}\right) (1+k-s)_s S_n^{(j)} S_n^{(l)} \end{aligned} $$

EXAMPLE 6.7.

$$ \begin{aligned} & (6.7) \quad \int_0^\infty x^{-1+t} \cosh^{-1-2n}(x\beta) \sin(mx) dx \\ & = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-1)^{-j-l+2n+q} i^{1+q+r-t} 2^{-1-j-l+2n} e^{-\frac{m\pi}{2\beta}} (-m^2)^{-s} \pi^{-s+t}}{(2n)!} \\ & \qquad \times \beta^{-q-r+s-t} \binom{j}{q} \binom{l}{r} \binom{q+r}{s} \left(i + \cot\left(\frac{1}{2}\pi(-1+t)\right)\right) \\ & \qquad \left((-m)^{q+r} m^s \Phi\left(-e^{-\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right) - e^{\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(-e^{\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right)\right) \end{aligned} $$
$$ \times \beta^{-1-q-r+s-t} \binom{j}{q} \binom{l}{r} \binom{1+q+r}{s} \left(i + \cot\left(\frac{\pi t}{2}\right)\right) \left((-m)^{q+r} m^s \Phi\left(e^{-\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right) - e^{\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(e^{\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right)\right) (-s+t)_s S_{-1+n}^{(j)} S_{-1+n}^{(l)} $$

EXAMPLE 6.4. From equation (2.3.18) in [4].

$$ (6.4) \qquad \Gamma(s) = - \frac{2^{-5+3s}\pi^{-1+s}(-1+s)\left(\zeta\left(2-s, \frac{1}{4}\right) - \zeta\left(2-s, \frac{3}{4}\right)\right) \sec\left(\frac{\pi s}{2}\right)}{\zeta\left(-1+s, \frac{1}{4}\right) - \zeta\left(-1+s, \frac{3}{4}\right)} $$

THEOREM 6.5. From eq. (1.7.7.4) in [9].

$$ (6.5) \qquad \int_0^\infty \frac{e^{-imx} \cosh^{-1-2n}(x\beta) ((-ix + \log(a))^k + e^{2imx}(ix + \log(a))^k)}{2k!} dx \\ = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-2)^{-j-l+2n} e^{\frac{m\pi}{2\beta}} m^{q+r-s} \pi^{1+k-s} \left(-\frac{i}{\beta}\right)^q \left(\frac{i}{\beta}\right)^r \left(\frac{1}{\beta}\right)^{1+k-s} \binom{j}{q} \binom{l}{r} \binom{q+r}{s}}{(2n)! (k-s)!} \\ \times \Phi\left(-e^{\frac{m\pi}{\beta}}, -k+s, \frac{\beta\left(\frac{\pi}{2\beta} + \log(a)\right)}{\pi}, \frac{1}{\pi}\right) S_n^{(j)} S_n^{(l)} $$

EXAMPLE 6.6.

$$ (6.6) \qquad \int_0^\infty (-1)^{-k} e^{-imx} ((-a+x)^k + (-1)^k e^{2imx} (a+x)^k) \cosh^{-1-2n}(x\beta) dx \\ = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-1)^{-j-l+2n+q} i^{-k+q+r} 2^{1-j-l+2n} e^{\frac{m\pi}{2\beta}} m^{q+r-s} \pi^{1+k-s}}{(2n)!} \\ \times \beta^{-1-k-q-r+s} \binom{j}{q} \binom{l}{r} \binom{q+r}{s} \Phi\left(-e^{\frac{m\pi}{\beta}}, -k+s, \frac{1}{2} + \frac{ia\beta}{\pi}, (1+k-s)_s S_n^{(j)} S_n^{(l)} \right) $$

EXAMPLE 6.7.

$$ (6.7) \qquad \int_0^\infty x^{-1+t} \cosh^{-1-2n}(x\beta) \sin(mx) dx \\ = \sum_{j=0}^n \sum_{l=0}^n \sum_{q=0}^j \sum_{r=0}^l \sum_{s=0}^{q+r} \frac{(-1)^{-j-l+2n+q} i^{1+q+r-t} 2^{-1-j-l+2n} e^{-\frac{m\pi}{2\beta}} (-m^2)^{-s} \pi^{-s+t}}{(2n)!} \\ \times \beta^{-q-r+s-t} \binom{j}{q} \binom{l}{r} \binom{q+r}{s} \left(i + \cot\left(\frac{1}{2}\pi(-1+t)\right)\right) \\ \left((-m)^{q+r} m^s \Phi\left(-e^{-\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right) - e^{\frac{m\pi}{\beta}} (-m)^s m^{q+r} \Phi\left(-e^{\frac{m\pi}{\beta}}, 1+s-t, \frac{1}{2}\right)\right) $$
$$(-s + t)_s S_n^{(j)} S_n^{(l)}$$

EXAMPLE 6.8. Gamma function in terms of quotient special functions.

$$ (6.8) \qquad \Gamma(t) = \frac{2^{-1+2t} e^{-\frac{i\pi(a+b\tau)}{2b}} \pi^t \left( \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-t, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}} \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-t, \frac{1}{2}\right) \right) (-i + \tan\left(\frac{\pi t}{2}\right))}{\zeta\left(t, \frac{3}{4} - \frac{a}{4b}\right) - \zeta\left(t, \frac{1}{4}\left(3 + \frac{a}{b}\right)\right) - \zeta\left(t, \frac{-a+b}{4b}\right) + \zeta\left(t, \frac{a+b}{4b}\right)} $$

EXAMPLE 6.9. Gamma function in terms of the finite product of quotient special functions.

$$ (6.9) \quad n^{\frac{1}{2}-nz}(2\pi)^{\frac{1}{2}(-1+n)}\Gamma(nz) \\ = \prod_{k=0}^{-1+n} \frac{2^{-1+2(\frac{k}{n}+z)}e^{-\frac{i\pi(a+b(\frac{k}{n}+z))}{2b}}\pi^{\frac{k}{n}+z}}{\zeta(\frac{k}{n}+z, \frac{3}{4}-\frac{a}{4b}) - \zeta(\frac{k}{n}+z, \frac{1}{4}(3+\frac{a}{b})) - \zeta(\frac{k}{n}+z, -\frac{a+b}{4b}) + \zeta(\frac{k}{n}+z, \frac{a+b}{4b})} \\ \times \left( \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-\frac{k}{n}-z, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}}\Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-\frac{k}{n}-z, \frac{1}{2}\right) \right) \\ \left(-i + \tan\left(\frac{1}{2}\pi\left(\frac{k}{n}+z\right)\right)\right) $$

where $Re(z) > 1, Re(b) > Re(a)$.

EXAMPLE 6.10. Combining the definition of the gamma function.

$$ (6.10) \qquad \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-s, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}} \Phi\left(-e^{\frac{i\alpha\pi}{b}}, 1-s, \frac{1}{2}\right) = \frac{i2^{-4+s}e^{\frac{i\alpha\pi}{2b}}(-1+s)}{\pi\left(\zeta\left(-1+s, \frac{1}{4}\right) - \zeta\left(-1+s, \frac{3}{4}\right)\right)} \\ \times \left(\zeta\left(2-s, \frac{1}{4}\right) - \zeta\left(2-s, \frac{3}{4}\right)\right) \left(\zeta\left(s, \frac{3}{4} - \frac{a}{4b}\right) - \zeta\left(s, \frac{1}{4}\left(3+\frac{a}{b}\right)\right) \right. \\ \left. - \zeta\left(s, \frac{-a+b}{4b}\right) + \zeta\left(s, \frac{a+b}{4b}\right)\right) $$

where $Re(b) > Re(a)$.

References

[1] Reynolds, R.; Stauffer, A. A Method for Evaluating Definite Integrals in Terms of Special Functions with Examples. *Int. Math. Forum* (2020), 15, 235–244, doi:10.12988/imf.2020.91272
$$(-s + t)_s S_n^{(j)} S_n^{(l)}$$

EXAMPLE 6.8. Gamma function in terms of quotient special functions.

$$ (6.8) \qquad \Gamma(t) = \frac{2^{-1+2t} e^{-\frac{i\pi(a+b)t}{2b}} \pi^t \left( \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-t, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}} \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-t, \frac{1}{2}\right) \right) (-i + \tan\left(\frac{\pi t}{2}\right))}{\zeta\left(t, \frac{3}{4} - \frac{a}{4b}\right) - \zeta\left(t, \frac{1}{4}\left(3 + \frac{a}{b}\right)\right) - \zeta\left(t, \frac{-a+b}{4b}\right) + \zeta\left(t, \frac{a+b}{4b}\right)} $$

EXAMPLE 6.9. Gamma function in terms of the finite product of quotient special functions.

$$ (6.9) \quad n^{\frac{1}{2}-nz}(2\pi)^{\frac{1}{2}(-1+n)}\Gamma(nz) \\ = \prod_{k=0}^{-1+n} \frac{2^{-1+2(\frac{k}{n}+z)}e^{-\frac{i\pi(a+b(\frac{k}{n}+z))}{2b}}\pi^{\frac{k}{n}+z}}{\zeta(\frac{k}{n}+z, \frac{3}{4}-\frac{a}{4b}) - \zeta(\frac{k}{n}+z, \frac{1}{4}(3+\frac{a}{b})) - \zeta(\frac{k}{n}+z, -\frac{a+b}{4b}) + \zeta(\frac{k}{n}+z, \frac{a+b}{4b})} \\ \times \left( \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-\frac{k}{n}-z, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}}\Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-\frac{k}{n}-z, \frac{1}{2}\right) \right) \\ \left(-i + \tan\left(\frac{1}{2}\pi\left(\frac{k}{n}+z\right)\right)\right) $$

where $Re(z) > 1, Re(b) > Re(a)$.

EXAMPLE 6.10. Combining the definition of the gamma function.

$$ (6.10) \qquad \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-s, \frac{1}{2}\right) - e^{\frac{i\alpha\pi}{b}} \Phi\left(-e^{-\frac{i\alpha\pi}{b}}, 1-s, \frac{1}{2}\right) = -\frac{i2^{-4+s}e^{\frac{i\alpha\pi}{2b}}(-1+s)}{\pi\left(\zeta\left(-1+s, \frac{1}{4}\right) - \zeta\left(-1+s, \frac{3}{4}\right)\right)} \\ \times \left(\zeta\left(2-s, \frac{1}{4}\right) - \zeta\left(2-s, \frac{3}{4}\right)\right) \left(\zeta\left(s, \frac{3}{4} - \frac{a}{4b}\right) - \zeta\left(s, \frac{1}{4}\left(3+\frac{a}{b}\right)\right) \right. \\ \left. - \zeta\left(s, \frac{-a+b}{4b}\right) + \zeta\left(s, \frac{a+b}{4b}\right)\right) $$

where $Re(b) > Re(a)$.

## 7. Extended Magnus definite integral

The formulae listed in this Table (3.238) in [14], are used in various areas of mathematics; some examples are the generation of a wavelet-type transform by the
[2] E. R. Hansen (1975) *A Table of Series and Products*. Prentice-Hall, Englewood Cliffs, N.J.

[3] J. Riordan (1958) *An Introduction to Combinatorial Analysis*. John Wiley & Sons Inc., New York.

[4] Brychkov, Y., Marichev, O., & Savischenko, N. (2018). *Handbook of Mellin Transforms* (1st ed.). Chapman and Hall/CRC. https://doi.org/10.1201/9780429434259

[5] A. Erdélyi, W. Magnus, F. Oberhettinger, and F. G. Tricomi (1954a) *Tables of Integral Transforms*. Vol. I. McGraw-Hill Book Company, Inc., New York-Toronto-London.

[6] Terazawa K (1915), *On deep-sea water waves caused by a local disturbance on or beneath the surface*, Proc. R. Soc. Lond. A9257-81 http://doi.org/10.1098/rspa.1915.0053

[7] A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1986a) *Integrals and Series: Elementary Functions*, Vol. 1. Gordon & Breach Science Publishers, New York.

[8] Fritz Oberhettinger, *Tables of Fourier Transforms and Fourier Transforms of Distributions*, (1990), Springer-Verlag Berlin Heidelberg

[9] Catalan, E., *Journal de mathématiques pures et appliquées*, Liouville, Joseph, (1809-1882); Centre national de la recherche scientifique (France).

[10] J.W.L. Glaisher. *The Messenger of Mathematics*, vol.ll. Macmillan and Co., London and Cambridge, (1873)

[11] F. Oberhettinger (1990) *Tables of Fourier Transforms and Fourier Transforms of Distributions*. Springer-Verlag, Berlin.

[12] D. Bierens de Haan (1939) *Nouvelles Tables d'Intégrales Définies*. G.E. Stechert & Co., New York.

[13] I. S. Gradshteyn and I. M. Ryzhik (2000) *Table of Integrals, Series, and Products*. 6th edition, Academic Press Inc., San Diego, CA.

[14] A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (1992a) *Integrals and Series: Direct Laplace Transforms*, Vol. 4. Gordon and Breach Science Publishers, New York.

[15] Fock, V. Über die Wärmeleitung in mehrphasigen Kabeln. Archiv f. Elektrotechnik 16, 331-340 (1926). https://doi.org/10.1007/BF01656280

[16] Whittaker, E. T. and Watson, G. N., *Modern Analysis*, 4th ed. Cambridge University Press, London, (1927), part II, 1934. pp. 279

[17] Reynolds R, Stauffer A (2023) *Chebyshev series: Derivation and evaluation*. PLoS ONE 18(3): e0282703. https://doi.org/10.1371/journal.pone.0282703

(Robert Reynolds) DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO, ON, CANADA, M3J1P3

Email address, Corresponding author: milver73@gmail.com
beta-semigroup [20], and the study of parabolic wavelet transforms associated with
heat operators [21].

THEOREM 7.1. From Eq. (3.238.3) in [14] and [1].

$$
(7.1) \quad \int_{\alpha}^{\beta} \frac{(x - \alpha)^{-m} (-x + \beta)^{-1+m} \log^k \left( \frac{a(-x+\beta)}{x-\alpha} \right)}{|u-x|} dx \\
= (-1)^{m+1} (2\pi i)^{1+k} |u-\alpha|^{-m} |u-\beta|^{-1+m} \Phi \left( e^{2im\pi}, -k, -\frac{i \left( i\pi + \log(a) + \log \left( \frac{|u-\beta|}{|u-\alpha|} \right) \right)}{2\pi} \right)
$$
where $\mathrm{Re}(m) > 0$.

EXAMPLE 7.2. The degenerate case.

$$
(7.2) \quad \int_{\alpha}^{\beta} \frac{(x - \alpha)^{-m} (-x + \beta)^{-1+m}}{|u-x|} dx = \pi |u - \alpha|^{-m} |u - \beta|^{-1+m} \csc(m\pi)
$$

EXAMPLE 7.3.

$$
(7.3) \quad \int_{\alpha}^{\beta} \frac{1}{\sqrt{x-\alpha}\sqrt{-x+\beta}u-x\log\left(\frac{a(-x+\beta)}{x-\alpha}\right)} dx \\
= \frac{i\left(-\psi^{(0)}\left(\frac{3\pi - i \log(a) - i \log\left(\frac{|u-\beta|}{|u-\alpha|}\right)}{4\pi}\right) + \psi^{(0)}\left(\frac{\pi - i \log\left(\frac{a|u-\beta|}{|u-\alpha|}\right)}{4\pi}\right)\right)}{2\sqrt{|u-\alpha|}\sqrt{|u-\beta|}}
$$

EXAMPLE 7.4.

$$
(7.4) \quad \int_{\alpha}^{\beta} \frac{1}{\sqrt{x-\alpha}\sqrt{-x+\beta}u-x\left(a+ib+\log\left(\frac{-x+\beta}{x-\alpha}\right)\right)} dx \\
= \frac{i\left(-\psi^{(0)}\left(\frac{-i(a+ib)+3\pi-i\log\left(\frac{|u-\beta|}{|u-\alpha|}\right)}{4\pi}\right) + \psi^{(0)}\left(\frac{\pi-i(a+ib+\log\left(\frac{|u-\beta|}{|u-\alpha|}\right))}{4\pi}\right)\right)}{2\sqrt{|u-\alpha|}\sqrt{|u-\beta|}}
$$

EXAMPLE 7.5.

$$
(7.5) \quad \int_{\alpha}^{\beta} \frac{\log\left(\log\left(\frac{a(-x+\beta)}{x-\alpha}\right)\right)}{\sqrt{x-\alpha}\sqrt{-x+\beta}|u-x|} dx =
$$
$$ \frac{2\pi}{|\sqrt{u-\alpha}\sqrt{u-\beta}|} \log \left( \frac{2e^{\frac{i\pi}{4}} \sqrt{\pi\Gamma}\left(\frac{3\pi - i \log(a) - i \log\left(\frac{u-\beta}{(u-\alpha)\operatorname{sgn}(u-\alpha)}\right)}{4\pi}\right)}{\Gamma\left(\frac{\pi - i \log\left(\frac{a(u-\beta)\operatorname{sgn}(u-\alpha)}{(u-\alpha)\operatorname{sgn}(u-\beta)}\right)}{4\pi}\right)} \right) $$

where $Im(a) \ge 0$.

EXAMPLE 7.6. From [Wolfram,7]

$$ (7.6) \qquad \int_{\alpha}^{\beta} \frac{1}{\sqrt[4]{x-\alpha}(-x+\beta)^{3/4}|u-x|\log^2\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)} dx = \frac{(-1)^{3/4}\left(C + \frac{i\pi^2}{48}\right)}{2\pi^4\sqrt{|u-\alpha||u-\beta|}^{3/4}} $$

EXAMPLE 7.7. From [Wolfram,7]

$$ (7.7) \qquad \int_{\alpha}^{\beta} \frac{1}{(x-\alpha)^{3/4} \sqrt[4]{-x+\beta} |u-x| \log^2\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)} dx = \frac{(-1)^{3/4} (48iC + \pi^2)}{96\pi |u-\alpha|^{3/4} \sqrt[4]{|u-\beta|}} $$

EXAMPLE 7.8. From [Wolfram,8]

$$ (7.8) \qquad \int_{\alpha}^{\beta} \frac{\log\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right) \log\left(\log\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)\right)}{\sqrt{x-\alpha}\sqrt{-x+\beta}|u-x|} dx = \frac{i\pi^2 \log\left(\frac{4\sqrt[3]{2e\pi}}{iA^{12}}\right)}{\sqrt{|(u-\alpha)(u-\beta)|}} $$

EXAMPLE 7.9. From [Wolfram,10]

$$ (7.9) \qquad \int_{\alpha}^{\beta} \frac{\log\left(\frac{(-x+\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right) \log\left(\log\left(\frac{-x+\beta}{(x-\alpha)|u-\beta|}\right)\right)}{\sqrt{x-\alpha}\sqrt{-x+\beta}|u-x|} dx = -\frac{4iC\pi}{\sqrt{|(u-\alpha)(u-\beta)|}} $$

EXAMPLE 7.10.

$$ (7.10) \qquad \int_{\alpha}^{\beta} \frac{\log\left(\log\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)\right)}{\sqrt{x-\alpha}\sqrt{-x+\beta}|u-x|} dx = \frac{\pi(i\pi + \log(16))}{2\sqrt{|(u-\alpha)(u-\beta)|}} $$

EXAMPLE 7.11.

$$ (7.11) \qquad \int_{\alpha}^{\beta} \frac{\log\left(\log\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)\right)}{\sqrt{x-\alpha}\sqrt{-x+\beta}|u-x|\log\left(\frac{(x-\beta)|u-\alpha|}{(x-\alpha)|u-\beta|}\right)} dx = \frac{\log(2)\log\left(\frac{2i\pi\sqrt{2}}{\exp(\gamma)}\right)}{i\sqrt{|(u-\alpha)(u-\beta)|}} $$
### EXAMPLE 7.12.

$$ (7.12) \quad \int_{\alpha}^{\beta} \frac{\log\left(\log\left(\frac{a(-x+\beta)}{x-\alpha}\right)\right)}{\sqrt[4]{x-\alpha}(-x+\beta)^{3/4}|u-x|} dx \\ = \frac{\sqrt{2\pi}\left(\log(2i\pi) - (1-i)\left(\log\left(\frac{\Gamma\left(\frac{\pi-i\log\left(\frac{a|u-\beta|}{|u-\alpha|}\right)}{8\pi}\right)}{2\Gamma\left(\frac{5}{8}-\frac{i\log\left(\frac{a|u-\beta|}{|u-\alpha|}\right)}{8\pi}\right)}\right) + i\log\left(\frac{\Gamma\left(\frac{3}{8}-\frac{i\log\left(\frac{a|u-\beta|}{|u-\alpha|}\right)}{8\pi}\right)}{2\Gamma\left(\frac{7}{8}-\frac{i\log\left(\frac{a|u-\beta|}{|u-\alpha|}\right)}{8\pi}\right)}\right)\right)}{\sqrt[4]{|u-\alpha||u-\beta|^{3/4}}} $$

## 8. Extended entries in Table (3.277) in [14]

An example of where integral formulae in Table (3.277) is in the perimeter of a rose [22]. In Butler's paper, the question regarding the length of the lemniscate was an increasing function. A systematic treatment, based on Green's functions and integral equations, is given to the analytical and numerical methods and results for a great number of 3-D contact problems for elastic bodies [23].

### 8.1. An extended Erdélyi integral.

#### THEOREM 8.1. From Eq. (3.277.6) in [14] and [1].

$$ (8.1) \quad \int_{0}^{\frac{\pi}{4}} \cot^m(t) \left( \log^k(a \cot(t)) + 4^m \log^k(a \tan(t)) \sin^2(t) \sin^{-2m}(2t) \right) dt \\ = i^k e^{\frac{im\pi}{2}} \pi^{1+k} \Phi \left( -e^{im\pi}, -k, \frac{1}{2} - \frac{i \log(a)}{\pi} \right) \\ \text{where } |\mathrm{Re}(m)|<1, |\mathrm{Im}(m)|<1. $$

### EXAMPLE 8.2.

$$ (8.2) \quad \int_{0}^{\frac{\pi}{4}} \left( \frac{1}{\log(a \cot(t))} + \frac{1}{\log(a \tan(t))} \right) dx \\ = \frac{1}{2} i \left( \psi^{(0)}\left(\frac{\pi - 2i \log(a)}{4\pi}\right) - \psi^{(0)}\left(\frac{3}{4} - \frac{i \log(a)}{2\pi}\right) \right) $$

### EXAMPLE 8.3.

$$ (8.3) \quad \int_{0}^{\frac{\pi}{4}} \frac{-\cot^m(t) (-1 + \tan^{2m}(t)) + \cot^s(t) (-1 + \tan^{2s}(t))}{\log(\tan(t))} dx $$
$$= 2i \left( \tan^{-1} \left( e^{\frac{im\pi}{2}} \right) - \tan^{-1} \left( e^{\frac{i\pi s}{2}} \right) \right)$$

### EXAMPLE 8.4.

$$ (8.4) \qquad \int_{0}^{\frac{\pi}{4}} \frac{\cot^{\frac{1}{2}}(t) - \cot^{-\frac{1}{2}}(t)}{\log(\cot(t))} dt = \coth^{-1}(\sqrt{2}) $$

### EXAMPLE 8.5.

$$ (8.5) \qquad \int_{0}^{\frac{\pi}{4}} \frac{\cot(t) \log(\log(a \cot(t))) + \log(\log(a \tan(t)))}{\sqrt{\cot(t)}} dt \\ = \frac{\pi \left( i\pi + 2\log(\pi) - (2+2i) \left( \log\left(\frac{\Gamma\left(\frac{1}{4}\left(\frac{1}{2}-\frac{i\log(a)}{\pi}\right)\right)}{2\Gamma\left(\frac{1}{4}\left(\frac{5}{2}-\frac{i\log(a)}{\pi}\right)\right)}\right) - i\log\left(\frac{\Gamma\left(\frac{1}{4}\left(\frac{3}{2}-\frac{i\log(a)}{\pi}\right)\right)}{2\Gamma\left(\frac{1}{4}\left(\frac{7}{2}-\frac{i\log(a)}{\pi}\right)\right)}\right) \right)}{2\sqrt{2}} $$

### EXAMPLE 8.6.

$$ (8.6) \qquad \int_{0}^{\frac{\pi}{4}} \frac{(\cot(t) + 1) \log(\log(\tan(t)))}{\sqrt{\cot(t)}} dt \\ = \frac{\pi \left( 2i\pi + 2i \coth^{-1}(\sqrt{2}) + \log(16) + 2\log(\pi) + (2+2i) \log\left(\frac{\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}\right) - (2-2i) \log\left(\frac{\Gamma(\frac{3}{8})}{\Gamma(\frac{7}{8})}\right) \right)}{2\sqrt{2}} $$

### EXAMPLE 8.7.

$$ (8.7) \qquad \int_{0}^{\frac{\pi}{4}} \frac{\log(\log(a \cot(t))) + \cot(t) \log(\log(a \tan(t)))}{\sqrt{\cot(t)}} dt \\ = \frac{\pi \left( i\pi + \log(16) + 2\log(\pi) - (2-2i) \log\left(\frac{\Gamma(\frac{\pi-2i\log(a)}{8\pi})}{\Gamma(\frac{5}{8}-\frac{i\log(a)}{4\pi})}\right) - (2+2i) \log\left(\frac{\Gamma(\frac{3}{8}-\frac{i\log(a)}{4\pi})}{\Gamma(\frac{7}{8}-\frac{i\log(a)}{4\pi})}\right) \right)}{2\sqrt{2}} $$

### EXAMPLE 8.8.

$$ (8.8) \qquad \int_{0}^{\frac{\pi}{4}} (\log(i \cot(t)) \log(\log(i \cot(t))) + \log(\log(i \tan(t))) \log(i \tan(t))) dt \\ = -\frac{\pi^3}{8} + \frac{1}{4} i \pi^2 \log \left( \frac{2\sqrt{2}e\pi}{A^{12}} \right) $$

### EXAMPLE 8.9.
$$ (8.9) \qquad \int_0^{\frac{\pi}{4}} (\log^2(i \cot(t)) \log(\log(i \cot(t))) + \log(\log(i \tan(t))) \log^2(i \tan(t))) dt = \frac{7\pi\zeta(3)}{4} $$

EXAMPLE 8.10.

$$ (8.10) \qquad \int_0^{\frac{\pi}{4}} \log(\log(\cot(t))) \log^2(\tan(t)) dt = \frac{1}{16}\pi^3 \left( \log(2\pi) + 32 \left( \zeta'\left(-2, \frac{1}{4}\right) - \zeta'\left(-2, \frac{3}{4}\right) \right) \right) $$

EXAMPLE 8.11.

$$ (8.11) \qquad \int_0^{\frac{\pi}{4}} \log(\log(\tan(t))) dt = \frac{1}{2}\pi \log \left( \frac{i\sqrt{2\pi}\Gamma\left(-\frac{1}{4}\right)}{3\Gamma\left(-\frac{3}{4}\right)} \right) $$

EXAMPLE 8.12.

$$ (8.12) \qquad \int_0^{\frac{\pi}{4}} \log \left( \frac{a^2 \pi^2}{4} + \log^2(\tan(t)) \right) dt = \pi \log \left( \frac{\sqrt{2\pi}\Gamma\left(\frac{3+a}{4}\right)}{\Gamma\left(\frac{1+a}{4}\right)} \right) $$

## 8.2. An extended Erdélyi integral.

THEOREM 8.13. From Eq. (3.277.2) in [14].

$$ (8.13) \qquad \int_0^{\frac{\pi}{2}} \log^k \left(ab^2 \tan(t)(\sec(t) + \tan(t))\right) \sec(t) \left(b^2 \tan(t)(\sec(t) + \tan(t))\right)^{-1+m} dt \\ = -2^{1+k-m} b^{-2+2m} e^{im\pi} (i\pi)^{1+k} \Phi \left( e^{2im\pi}, -k, -\frac{i\left(i\pi + \log(a) + \log\left(\frac{b^2}{2}\right)\right)}{2\pi} \right) $$

where $0 < \mathrm{Re}(m) < 1$.

EXAMPLE 8.14. The degenerate case.

$$ (8.14) \qquad \int_0^{\frac{\pi}{2}} \sec(t) \left(b^2 \tan(t)(\sec(t) + \tan(t))\right)^{-1+m} dt = 2^{-m} b^{-2+2m} \pi \csc(m\pi) $$

EXAMPLE 8.15.

$$ (8.15) \qquad \int_0^{\frac{\pi}{2}} \frac{\sec(t)}{\log(ab^2 \tan(t)(\sec(t) + \tan(t))) \sqrt{\tan(t)(\sec(t) + \tan(t))}} dt $$
$$= \frac{i \left( \psi^{(0)} \left( \frac{\pi - i \log\left(\frac{a}{2}\right)}{4\pi} - i \log(b^2) \right) - \psi^{(0)} \left( \frac{3\pi + i \log(2) - i \log(a) - i \log(b^2)}{4\pi} \right) \right)}{2\sqrt{2}}$$

### EXAMPLE 8.16.

$$\begin{aligned} & (8.16) \quad \int_0^{\frac{\pi}{2}} \frac{\log\left(\log\left(ab^2 \tan(t)(\sec(t) + \tan(t))\right)\right) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt \\ &= \frac{\pi}{2\sqrt{2}} \left( i\pi + \log(16) + 2\log(\pi) - 4\log\left(-3\pi - i\log\left(\frac{a}{2}\right) - i\log(b^2)\right) \right. \\ & \qquad \left. + 4\log\left(-\pi + i\log(2) - i\log(a) - i\log(b^2)\right) - 4\log\Gamma\left(-\frac{3\pi + i\log\left(\frac{a}{2}\right) + i\log(b^2)}{4\pi}\right) \right. \\ & \qquad \left. + 4\log\Gamma\left(-\frac{i\left(-\pi + \log\left(\frac{a}{2}\right) + \log(b^2)\right)}{4\pi}\right) \right) \end{aligned}$$

### EXAMPLE 8.17.

$$ (8.17) \quad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-2 \tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = \frac{\pi(i\pi + \log(16))}{2\sqrt{2}} $$

### EXAMPLE 8.18. Cauchy principal value of an integral.

$$\begin{aligned} & (8.18) \quad \int_0^{\frac{\pi}{2}} \frac{\log(\log(\tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt \\ &= \frac{\pi \left( -4\log\Gamma\left(-\frac{3}{4} + \frac{i\log(2)}{4\pi}\right) + 4\log\left(\frac{(1+i)\sqrt{2\pi}\Gamma\left(-\frac{1}{4} + \frac{i\log(2)}{4\pi}\right)(\pi - i\log(2))}{3\pi - i\log(2)}\right) \right)}{2\sqrt{2}} \end{aligned}$$

<figure><img src="image_3.png" /><figcaption>FIGURE 1. Plot $Re(\frac{\sec(t) \log(\log(\tan(t)(\tan(t)+\sec(t))))}{\sqrt{\tan(t)(\tan(t)+\sec(t))}})$, where $t \in [0, \pi/2]$</figcaption></figure>
<figure><img src="image_4.png" /><figcaption>FIGURE 2. Plot Im($\frac{\sec(t) \log(\log(\tan(t)(\tan(t)+\sec(t))))}{\sqrt{\tan(t)(\tan(t)+\sec(t))}}$), where $t \in [0, \pi/2]$</figcaption></figure>

EXAMPLE 8.19.

$$ (8.19) \quad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-4 \tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt \\ = \frac{\pi}{2\sqrt{2}} \left( 3i\pi + \log(16) + 2\log(\pi) + 4\log(\log(2)) + \log\left(\frac{1}{(-2i\pi + \log(2))^4}\right) \right. \\ \left. \qquad + 4\log\Gamma\left(-\frac{i\log(2)}{4\pi}\right) - 4\log\Gamma\left(-\frac{1}{2} - \frac{i\log(2)}{4\pi}\right) \right) $$

EXAMPLE 8.20.

$$ (8.20) \quad \int_0^{\frac{\pi}{2}} \frac{\log\left(\log\left(ab^2 \tan(t)(\sec(t) + \tan(t))\right)\right) \sec(t)}{\left(b^2 \tan(t)(\sec(t) + \tan(t))\right)^{3/4}} dt \\ = \frac{\pi}{2^{3/4}b^{3/2}} \left( i\pi + \log(64) + 2\log(\pi) - (2 - 2i) \log \right. \\ \left. \left( \frac{\Gamma\left(\frac{\pi - i \log\left(\frac{a}{2}\right) - i \log(b^2) }{8\pi}\right)}{\Gamma\left(\frac{5\pi - i \log\left(\frac{a}{2}\right) - i \log(b^2) }{8\pi}\right)} \right) \right. \\ \left. - (2 + 2i) \log \left( \frac{\Gamma\left(\frac{3\pi - i \log\left(\frac{a}{2}\right) - i \log(b^2) }{8\pi}\right)}{\Gamma\left(\frac{7\pi - i \log\left(\frac{a}{2}\right) - i \log(b^2) }{8\pi}\right)} \right) \right) $$

EXAMPLE 8.21. The Cauchy principal value of an integral.

$$ (8.21) \quad \int_0^{\frac{\pi}{2}} \frac{\log(\log(\tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\left(i \tan(t)(\sec(t) + \tan(t))\right)^{3/4}} dt $$
$$ = \frac{1}{2}\sqrt{1+i\pi} \left( \pi - 2i \log(\pi) + \log \left( 64^{-i} \left( \frac{\Gamma\left(\frac{\pi+i\log(2)}{8\pi}\right)}{\Gamma\left(\frac{5}{8} + \frac{i\log(2)}{8\pi}\right)} \right)^{2+2i} \left( \frac{\Gamma\left(\frac{3}{8} + \frac{i\log(2)}{8\pi}\right)}{\Gamma\left(\frac{7}{8} + \frac{i\log(2)}{8\pi}\right)} \right)^{-2+2i} \right) \right) $$

<figure><img src="image_5.png" /><figcaption>FIGURE 4. Plot Im($\frac{\sec(t) \log(\log(ab^2 \tan(t)(\tan(t)+\sec(t))))}{(b^2 \tan(t)(\tan(t)+\sec(t)))^{3/4}}$), where $t \in [0, \pi/2]$</figcaption></figure>

<figure><img src="image_6.png" /></figure>

EXAMPLE 8.22.

$$ (8.22) \qquad \int_0^{\frac{\pi}{2}} \frac{\log(\log(2 \tan(t)(\sec(t) + \tan(t)))) \log(2 \tan(t)(\sec(t) + \tan(t))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = -2i\sqrt{2C\pi} $$

EXAMPLE 8.23.
$$
(8.23) \qquad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-2 \tan(t)(\sec(t) + \tan(t)))) \log(-2 \tan(t)(\sec(t) + \tan(t))) \sec(t)}{\sqrt{2 \tan(t)(\sec(t) + \tan(t))}} dt \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = i\pi^2 \log\left(\frac{2e^{\frac{i\pi}{4}}\sqrt{2}\sqrt{e\pi}}{A^6}\right)
$$

EXAMPLE 8.24.

$$
(8.24) \qquad \int_0^{\frac{\pi}{2}} \frac{\csc(t) \left( -(b^2 \tan(t)(\sec(t) + \tan(t)))^m + (b^2 \tan(t)(\sec(t) + \tan(t)))^s \right)}{\log(b^2 \tan(t)(\sec(t) + \tan(t))) (\sec(t) + \tan(t))} dt \\
\qquad = 2^{-m} b^{2m} e^{im\pi} \Phi \left( e^{2im\pi}, 1, \frac{\pi - i \log\left(\frac{b^2}{2}\right)}{2\pi} \right) - 2^{-s} b^{2s} e^{i\pi s} \Phi \left( e^{2i\pi s}, 1, \frac{\pi - i \log\left(\frac{b^2}{2}\right)}{2\pi} \right)
$$

EXAMPLE 8.25.

$$
(8.25) \qquad \int_0^{\frac{\pi}{2}} \log^k \left( -2 \tan(t)(\sec(t) + \tan(t)) \right) \sec(t) \left( b^2 \tan(t)(\sec(t) + \tan(t)) \right)^{-1+m} dt \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = -2^{1+k-m} b^{-2+2m} e^{-im\pi} (i\pi)^{1+k} \mathrm{Li}_{-k} (e^{2im\pi})
$$

EXAMPLE 8.26.

$$
(8.26) \qquad \int_0^{\frac{\pi}{2}} \frac{\log^k \left(-2 \tan(t)(\sec(t) + \tan(t))\right) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = -i^k 2^{\frac{1}{2}+k} (-1+2^{1+k}) \pi^{1+k} \zeta(-k)
$$

EXAMPLE 8.27.

$$
(8.27) \qquad \int_0^{\frac{\pi}{2}} \frac{\sec(t)}{\log(-2 \tan(t)(\sec(t) + \tan(t))) \sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = -\frac{i \log(2)}{\sqrt{2}}
$$

EXAMPLE 8.28.

$$
(8.28) \quad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-2 \tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = \frac{\pi(i\pi + \log(16))}{2\sqrt{2}}
$$
EXAMPLE 8.29.

$$
(8.29) \qquad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-2 \tan(t)(\sec(t) + \tan(t)))) \sec(t)}{\log(-2 \tan(t)(\sec(t) + \tan(t))) \sqrt{\tan(t)(\sec(t) + \tan(t))}} dt = \frac{i \log(2) \log\left(\frac{e^\gamma}{2i\sqrt{2\pi}}\right)}{\sqrt{2}}
$$

$$
(8.30) \qquad \int_0^{\frac{\pi}{2}} \frac{\log(\log(-2 \tan(t)(\sec(t) + \tan(t)))) \log(-2 \tan(t)(\sec(t) + \tan(t))) \sec(t)}{\sqrt{\tan(t)(\sec(t) + \tan(t))}} dt \\
\qquad\qquad = - \frac{\pi^2 \left( \pi - 2i \log\left(\frac{4\sqrt{2}e\pi}{A^{12}}\right) \right)}{2\sqrt{2}}
$$

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(Robert Reynolds) DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO, ON, CANADA, M3J1P3

Email address, Corresponding author: milver73@gmail.com