# A finite integral involving the hypergeometric function

Robert Reynolds

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# A finite integral involving the hypergeometric function

Robert Reynolds

ABSTRACT. In this paper we have established a finite integral involving the product of the hypergeometric, exponential, and rational functions in terms of an infinite series. The infinite series involves the product of the incomplete gamma function and a rational function which can be reduced to the Hurwitz-Lerch zeta function as a special case. Other results are possible for various special cases of the parameters involved. The finite integral will be used to derive formulae involving elliptic functions, product of logarithm functions and a few other special case examples which are new to best of our knowledge along with errata for examples in some well known textbooks. The method used to derive this integral is contour integration. The parameters involved are valid over the complex plane unless stated otherwise.

## 1. Preliminaries

Throughout this work we will use the following functions; Lerch's Transcendent, given in [DLMF, 25.14], Complete Elliptic Integrals, given in [DLMF, 15.9], the Hypergeometric Function given in [DLMF, 15], Pochhammer's Symbol, given in [DLMF, 5.2(iii)], the Incomplete gamma function, given in [DLMF, 8.2].

**1.1. The Incomplete Gamma Function.** The incomplete gamma functions are given in equation [DLMF,8.4.13], $\gamma(a, z)$ and $\Gamma(a, z)$, are defined by

$$ (1.1) \qquad \gamma(a, z) = \int_{0}^{z} t^{a-1}e^{-t}dt $$

and

$$ (1.2) \qquad \Gamma(a, z) = \int_{z}^{\infty} t^{a-1}e^{-t}dt $$

<footnote n="2020">2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.</footnote>

<footnote>Key words and phrases. Hypergeometric function, hurwitz-Lerch zeta function, elliptic integral.</footnote>
where $Re(a) > 0$. The incomplete gamma function has a recurrence relation given by

$$ (1.3) \qquad \gamma(a, z) + \Gamma(a, z) = \Gamma(a) $$

where $a \neq 0, -1, -2, ...$ The incomplete gamma function is continued analytically by

$$ (1.4) \qquad \gamma(a, ze^{2m\pi i}) = e^{2\pi m\alpha} \gamma(a, z) $$

and

$$ (1.5) \qquad \Gamma(a, ze^{2m\pi i}) = e^{2\pi m\alpha} \Gamma(a, z) + (1 - e^{2\pi m\alpha}) \Gamma(a) $$

where $m \in \mathbb{Z}$, $\gamma^*(a, z) = \frac{z^{-a}}{\Gamma(a)}\gamma(a, z)$ is entire in $z$ and $a$. When $z \neq 0$, $\Gamma(a, z)$ is an entire function of $a$ and $\gamma(a, z)$ is meromorphic with simple poles at $a = -n$ for $n = 0, 1, 2, ...$ with residue $\frac{(-1)^n}{n!}$. These definitions are listed in [DLMF,8.2(i)] and [DLMF,8.2(ii)]. The incomplete gamma functions are particular cases of the more general hypergeometric and Meijer G functions see section (5.6) and equation (6.9.2) in [1]. Some Meijer G representations we will use in this work are given by;

$$ (1.6) \qquad \Gamma(a, z) = \Gamma(a) - G_{1,2}^{1,1} \begin{pmatrix} z \\ a, 0 \end{pmatrix} $$

and

$$ (1.7) \qquad \Gamma(a, z) = G_{1,2}^{2,0} \begin{pmatrix} z \\ 0, a \end{pmatrix} $$

from equations (2.4) and (2.6a) in [2]. We will also use the derivative notation given by;

$$ (1.8) \qquad \frac{\partial \Gamma(a, z)}{\partial a} = \Gamma(a, z) \log(z) + G_{2,3}^{3,0} \begin{pmatrix} z \\ 0, 0, a \end{pmatrix} $$

In the work by Alzer et al. [6], series representations were derived from definite integrals. The method used is called the $\lambda$-method and involves a generalized definite integral and the double sum of a definite integral. In this work we have expanded on the work by Alzer et al. by deriving series representations in terms of definite integrals over a finite interval. The approach differs from Alzer et al. in that we are able to iteratively generate integrands over a finite domain and evaluate

<footnote n="8">8 from equations (2.19a) in [2], (9.31.3) in [3] and equations (5.11.1), (6.2.11.1) and</footnote>

<footnote n="9">9 (6.2.11.2) in [4], and (6.36) in [5].</footnote>
them in terms of finite or infinite series. The work by Alzer et al. involved definite integrals over [0, 1], while this current work involves finite integrals over [0, b] where $Re(b) > 0$. Finite integrals of kernels involving the hypergeometric function have studied by Choi et al. [7], Love [8], Ta Li [9], Buschman [10], Srivastava [11], Higgins [12], Wimp [13] Jayarama [14], Adamchik [15], Biró [16] and Erdélyi [17]. Tables listing these types of integrals are found in Prudnikov volume III Table 2.21.1 [31], Gradshteyn and Ryzhik sections 8 and 9 [3], Brychkov section (4.24) [19]. Many techniques exist for evaluating definite integrals. One very powerful tool is contour integration [20].

In this work using contour integration, we will derive and evaluate the finite integral given by;

$$ (2.1) \quad \int_0^b e^{dx^\nu} x^m (1+cx^\mu)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma)(-\log(ax))^k dx \\ = \frac{1}{a^{1+m}} \sum_{h=0}^\infty \sum_{j=0}^\infty \sum_{l=0}^\infty \frac{a^{-j-hv-l\mu} c^l d^h (1+j+m+hv+l\mu)^{-1-k} \sigma^j \binom{\lambda}{l} (\alpha)_j (\beta)_j}{h! j! (\gamma)_j} \\ \times \Gamma\left(1+k, (1+j+m+hv+l\mu) \log\left(\frac{1}{ab}\right)\right) $$

where $Re(\mu) > 0, Re(v) > 0$. The derivation involves two definite integrals and follows the method used by us in [21]. This method involves using a form of the generalized Cauchy's integral formula given by

$$ (2.2) \qquad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw. $$

where C is in general an open contour in the complex plane where the bilinear concomitant has the same value at the end points of the contour. We then multiply both sides by a function of x, then take a definite integral of both sides. This yields a definite integral in terms of a contour integral. Then we multiply both sides of Equation (2.2) by another function of y and take the infinite sum of both sides such that the contour integral of both equations are the same. The contour integral representation form involving the function is given by;

$$ (2.3) \quad \frac{1}{2\pi i} \int_0^b \int_C a^w e^{dx^\nu} w^{-1-k} x^{m+w} (1+cx^\mu)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dx dw \\ = \frac{1}{2\pi i} \int_C \sum_{h=0}^\infty \sum_{j=0}^\infty \sum_{l=0}^\infty \frac{a^w b^{1+j+m+hv+w+l\mu} c^l d^h w^{-1-k} \sigma^j \binom{\lambda}{l} (\alpha)_j (\beta)_j}{(1+j+m+hv+w+l\mu)h! j! (\gamma)_j} dw $$

<footnote n="11">11 where $Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(z) < 1$.</footnote>
**2.1. Left-hand side contour integral representation.** Using a generalization of Cauchy's integral formula 2.2, we form the definite integral by replacing $y$ by $\log ax$ and multiply both sides by

$$ (2.4) \qquad \frac{c^l d^h \sigma^j(\lambda_l)(\alpha)_j(\beta)_j x^{hv+j+l\mu+m}}{h! j! (\gamma)_j} $$

then take the infinite sum of both sides over $h \in [0, \infty)$, $j \in [0, \infty)$, $l \in [0, \infty)$ to get;

$$ \begin{align} (2.5) \quad & \int_0^b \frac{x^m e^{dx^v} \log^k(ax) (cx^\mu + 1)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma)}{k!} dx \nonumber \\ &= \frac{1}{2\pi i} \sum_{h=0}^\infty \sum_{j=0}^\infty \sum_{l=0}^\infty \int_C \int_0^b \frac{c^l d^h \sigma^j w^{-k-1} (ax)^w (\lambda_l)(\alpha)_j(\beta)_j x^{hv+j+l\mu+m}}{h! j! (\gamma)_j} dxdw \nonumber \\ &= \frac{1}{2\pi i} \int_C \int_0^b \sum_{h=0}^\infty \sum_{j=0}^\infty \sum_{l=0}^\infty \frac{c^l d^h \sigma^j w^{-k-1} (ax)^w (\lambda_l)(\alpha)_j(\beta)_j x^{hv+j+l\mu+m}}{h! j! (\gamma)_j} dxdw \nonumber \\ &= \frac{1}{2\pi i} \int_C \int_0^b w^{-k-1} x^m (ax)^w e^{dx^v} (cx^\mu + 1)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dxdw \nonumber \\ &= \frac{1}{2\pi i} \int_0^b \int_C w^{-k-1} x^m (ax)^w e^{dx^v} (cx^\mu + 1)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dxdw \end{align} $$

where $Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(z) < 1$. We are able to switch the order

of integration over $t$ and $w$ using Fubini's theorem for multiple integrals see page

178 in [22], since the integrand is of bounded measure over the space $\mathbb{C} \times [0, b]$.

**2.2. Right-hand side contour integral.** Using a generalization of Cauchy's integral formula 2.2, we replace $y \to y + xi + \log(a)$ and multiply both sides by $e^{mxi}$. Then form a second equation by replacing $a \to -a$ and take their difference to get;

$$ (2.6) \quad \frac{i e^{-imx} \left( (\log(a) - ix + y)^k - e^{2imx} (\log(a) + ix + y)^k \right)}{2k!} \\ = \frac{1}{2\pi i} \int_C a^w w^{-k-1} e^{wy} \sin(x(m+w)) dw $$

Next we take the indefinite integral with respect to $y$ to get;

$$ (2.7) \quad \frac{i(-1)^{k+1} a^{-m} m^{-k-1} \left( \Gamma(k+1, -m(-ix+y+\log(a))) - \Gamma(k+1, -m(ix+y+\log(a))) \right)}{2k!} \\ = \frac{1}{2\pi i} \int_C \frac{a^w w^{-k-1} e^{y(m+w)} \sin(x(m+w))}{m+w} dw $$

from equation (2.325.6(12i)) in [3]. Finally we replace $y \to -\gamma, x \to \beta$, multiply both sides by $2\pi$ and simplify to get;
$$ (2.8) \frac{i\pi(-1)^{k+1}a^{-m}m^{-k-1}(\Gamma(k+1, m(-i\beta + \gamma - \log(a))) - \Gamma(k+1, m(i\beta + \gamma - \log(a))))}{k!} = \frac{1}{2\pi i} \int_C \frac{2\pi a^w w^{-k-1}e^{-(\gamma(m+w))}\sin(\beta(m+w))}{m+w}dw $$

Use equation (2.7) and set $y = 0$, replace $m \to hv + j + l\mu + m + 1$ and multiply both sides by

$$ (2.9) \qquad \frac{c^l d^h \sigma^j(\lambda_l)(\alpha)_j(\beta)_j b^{hv+j+l\mu+m+1}}{h! j! (\gamma)_j}. $$

Next take the infinite sum of both sides over $h \in [0, \infty)$, $j \in [0, \infty)$, $l \in [0, \infty)$ and simplify to get;

$$ (2.10) \sum_{h=0}^{\infty} \sum_{j=0}^{\infty} \sum_{l=0}^{\infty} \frac{b^{1+j+m+hv+l\mu} c^l d^h \sigma^j(\lambda_l) \Gamma(1+k, -((1+j+m+hv+l\mu)\log(ab)))(\alpha)_j(\beta)_j}{h! j! k! (\gamma)_j (-1-j-m-hv-l\mu)^{1+k} (ab)^{1+j+m+hv+l\mu}} \\ = -\frac{1}{2\pi i} \sum_{h=0}^{\infty} \sum_{j=0}^{\infty} \sum_{l=0}^{\infty} \int_C \frac{b^{1+j+m+hv+l\mu} (ab)^w c^l d^h w^{-1-k} \sigma^j(\lambda_l)(\alpha)_j(\beta)_j}{(1+j+m+hv+w+l\mu)h! j! (\gamma)_j} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{h=0}^{\infty} \sum_{j=0}^{\infty} \sum_{l=0}^{\infty} \frac{b^{1+j+m+hv+l\mu} (ab)^w c^l d^h w^{-1-k} \sigma^j(\lambda_l)(\alpha)_j(\beta)_j}{(1+j+m+hv+w+l\mu)h! j! (\gamma)_j} dw $$

where $Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(z) < 1$. from equation [Wolfram, 07.27.02.0002.01] where $|Re(b)|< 1$. We are able to switch the order of integration and summation over $w$ using Tonellii's theorem for integrals and sums see page 177 in [22], since the summand is of bounded measure over the space $\mathbb{C} \times [0, \infty)$

**THEOREM 2.1.** *Generalized Riemann-Louville integral form involving the hypergeometric function. See page 308 in [23], page 518 in [24] and page 185 in [25].*

$$ (2.11) \quad \int_0^b e^{dx^\nu} x^m (1+cx^\mu)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma)(-\log(ax))^k dx \\ = \frac{1}{a^{1+m}} \sum_{h=0}^{\infty} \sum_{j=0}^{\infty} \sum_{l=0}^{\infty} \frac{a^{-j-hv-l\mu} c^l d^h (1+j+m+hv+l\mu)^{-1-k} \sigma^j(\lambda_l)(\alpha)_j(\beta)_j}{h! j! (\gamma)_j} \\ \times \Gamma\left(1+k, (1+j+m+hv+l\mu)\log\left(\frac{1}{ab}\right)\right) $$

where $Re(\mu) > 0, Re(v) > 0$.

**PROOF.** Since the right-hand sides of equations (2.5) and (2.10) are equal relative to equation (2.3), we may equate the left-hand sides and simplify the gamma function to yield the stated result. $\square$
### 3. Evaluations of the finite integral

**EXAMPLE 3.1.** Product of logarithm functions. Use equation (2.11) and set $\alpha \to 1, \beta \to 1, \gamma \to 2, \lambda \to -1, v \to 0, d \to 1, \mu \to 1$ and simplify;

$$ (3.1) \quad \int_0^b x^{-1+m} \log(1+x\alpha) \log(1+x\beta) dx \\ = -b^{1+m}\alpha \sum_{l=0}^{\infty} \frac{(-b\alpha)^l (b\beta {}_2F_1(1,2+l+m;3+l+m;-b\beta) - \log((1+b\beta)^{2+l+m}))}{(1+l)(1+l+m)(2+l+m)} $$

24 where $Re(m) > 0, Re(b) > 0$.

**EXAMPLE 3.2.** Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, m \to m-1$ and simplify. Special case forms are found in [Wolfram, 01], [Wolfram, 01], [DLMF, 15.9], [DLMF, 15.4(iii)], in Gradshteyn and Ryzhik (for complete elliptic integrals, see 8.113 1 and 8.114 1; for integrals of Bessel functions, see 6.574 1, 3, 6.576 2–5, 6.621 1–3; for Legendre polynomials, see 8.911 and 8.916. (All these hypergeometric series terminate; that is, these series are finite sums); for Legendre functions, see 8.820 and 8.837; for associated Legendre functions, see 8.702, 8.703, 8.751, 8.77, 8.852, and 8.853; for Chebyshev polynomials, see 8.942 1; for Jacobi's polynomials, see 8.962; for Gegenbauer polynomials, see 8.932; for integrals of parabolic cylinder functions, see 7.725 6. for Lerch function $\Phi(z, s, v)$; see 9.559 1.

$$ (3.2) \quad \int_0^b x^{-1+m} (1+cx^\mu)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dx \\ = \sum_{j=0}^{\infty} \frac{b^{j+m} \sigma^j {}_2F_1\left(-\lambda, \frac{j}{\mu} + \frac{m}{\mu}; 1 + \frac{j}{\mu} + \frac{m}{\mu}; -b^\mu c\right) (\alpha)_j (\beta)_j}{(j+m)j! (\gamma)_j} $$

25 where $Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(\mu) > 0, |Re(m)| < 1$.

**EXAMPLE 3.3.** Derivation of Ramanujan-Berndt-Shen definite integral form [26]. Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, x \to \sin^2(t), \mu \to 1, c \to -1, \lambda \to -\frac{1}{2}, m \to -\frac{1}{2}$ and simplify;

$$ (3.3) \quad \int_0^\phi {}_2F_1(\alpha, \beta; \gamma; \sigma \sin^2(t)) dt = \sum_{j=0}^{\infty} \frac{b^{\frac{1}{2}+j} \sigma^j {}_2F_1\left(\frac{1}{2}, \frac{1}{2}+j; \frac{3}{2}+j; b\right) (\alpha)_j (\beta)_j}{(1+2j)j! (\gamma)_j} $$

26 where $Re(m) > 0, Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(\sigma) > 0, \phi = \sin^{-1}(\sqrt{b})$.

**EXAMPLE 3.4.** Derivation of Theorem A in [26]. Use equation (3.3) and set $\alpha \to \frac{1}{3}, \beta \to \frac{2}{3}, \gamma \to \frac{1}{2}, b \to \sin^2(\phi)$ and simplify;
$$ (3.4) \quad \int_0^\phi \frac{\cos\left(\frac{1}{3}\sin^{-1}\left(\sqrt{\sigma}\sin(t)\right)\right)}{\sqrt{1 - \sigma\sin^2(t)}} dt \\ = \sum_{j=0}^\infty \frac{\sigma^j {}_2F_1\left(\frac{1}{2}, \frac{1}{2} + j; \frac{3}{2} + j; \sin^2(\phi)\right) \binom{1}{3} \binom{2}{3} \sin^{1+2j}(\phi)}{(1+2j)j! \binom{1}{2}_j} $$

27 where $0 < Re(\sigma) < 1, 0 < Re(\phi) < 1$.

EXAMPLE 3.5. Derivation of Theorem B in [26]. Use equation (3.3) and set $\alpha \to \frac{1}{2}, \beta \to \frac{1}{2}, \gamma \to \frac{1}{2}, b \to \sin^2(\phi)$ and simplify;

$$ (3.5) \quad \int_0^\phi \frac{1}{\sqrt{1 - \sigma \sin^2(t)}} dt = \sum_{j=0}^\infty \frac{\sigma^j {}_2F_1\left(\frac{1}{2}, \frac{1}{2} + j; \frac{3}{2} + j; \sin^2(\phi)\right) \binom{1}{2} \sin^2(\phi)^{\frac{1}{2}+j}}{(1+2j)j!} $$

28 where $0 < Re(\sigma) < 1, 0 < Re(\phi) < 1$.

EXAMPLE 3.6. Derivation of Theorem C in [26]. Use equation (3.3) and set $\alpha \to \frac{1}{4}, \beta \to \frac{3}{4}, \gamma \to \frac{1}{2}, b \to \sin^2(\phi)$ and simplify;

$$ (3.6) \quad \int_0^\phi \frac{\sqrt{1 + \sqrt{1 - \sigma \sin^2(t)}}}{\sqrt{2}\sqrt{1 - \sigma \sin^2(t)}} dt = \sum_{j=0}^\infty \frac{\sigma^j {}_2F_1\left(\frac{1}{2}, \frac{1}{2} + j; \frac{3}{2} + j; \sin^2(\phi)\right) \binom{1}{4} \binom{3}{4} \sin^{1+2j}(\phi)}{(1+2j)j! \binom{1}{2}_j} $$

29 where $0 < Re(\sigma) < 1, 0 < Re(\phi) < 1$.

EXAMPLE 3.7. Use equation (3.3) and set $b \to \sin^2(b)$ and form a second equation by replacing $b \to c$ take their difference and simplify;

$$ (3.7) \quad \int_b^c {}_2F_1(\alpha, \beta; \gamma; \sigma \sin^2(t)) dt = - \sum_{j=0}^\infty \frac{\sigma^j (\alpha)_j (\beta)_j}{(1+2j)j! (\gamma)_j} \\ \times \left( {}_2F_1\left(\frac{1}{2}, \frac{1}{2} + j; \frac{3}{2} + j; \sin^2(b)\right) \sin^{1+2j}(b) \right. \\ \left. - {}_2F_1\left(\frac{1}{2}, \frac{1}{2} + j; \frac{3}{2} + j; \sin^2(c)\right) \sin^{1+2j}(c) \right) $$

30 where $Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0, Re(\sigma) > 0$.

EXAMPLE 3.8. Derivation of generalized form of entry (1.15.2.1) in [31]. Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, \mu \to 1, m \to 0$. Then form two equations by replacing $b \to -b$ and $b \to 1$ and take their difference and simplify;

$$ (3.8) \quad \int_b^1 (1 - cx)^\lambda {}_2F_1(\alpha, \beta; \gamma; -x\sigma) dx $$
$$ = \sum_{j=0}^{\infty} \frac{\sigma^j (e^{-ij\pi} {}_2F_1(1+j, -\lambda; 2+j; c) - (-\frac{1}{b})^{-j} b {}_2F_1(1+j, -\lambda; 2+j; bc)) (\alpha)_j (\beta)_j}{(1+j)j! (\gamma)_j} $$

31 where $Re(\alpha + \beta + \gamma + \lambda) > 0, Re(b) > 0$.

EXAMPLE 3.9. Derivation of entry (1.15.2.1) in [31]. Use equation (3.8) and set $\sigma \to -1, c \to 1$ and simplify;

$$ (3.9) \quad \int_b^1 (1-x)^\lambda {}_2F_1(\alpha, \beta; \gamma; x) dx = \frac{\Gamma(1+\lambda) {}_3F_2(1, \alpha, \beta; \gamma, 2+\lambda; 1)}{\Gamma(2+\lambda)} \\ - b \sum_{j=0}^{\infty} \frac{b^j {}_2F_1(1+j, -\lambda; 2+j; b)(\alpha)_j (\beta)_j}{(1+j)j! (\gamma)_j} $$

32 where $Re(\alpha + \beta + \gamma + \lambda) > 0, Re(b) > 0$.

33 **4. Derivation of Table (2.21.1) in [31]**

EXAMPLE 4.1. Generalized form of Entry (2.21.1.4-9) in [31]. Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, \mu \to 1, m \to m-1$ and simplify;

$$ (4.1) \quad \int_0^b x^{-1+m} (1+cx)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dx \\ = \sum_{j=0}^{\infty} \frac{b^{j+m} \sigma^j {}_2F_1(j+m, -\lambda; 1+j+m; -bc)(\alpha)_j (\beta)_j}{(j+m)j! (\gamma)_j} $$

34 where $Re(\alpha + \beta + \gamma + \lambda) > 0, Re(b) > 0$.

EXAMPLE 4.2. Entry (2.21.1.12) in [31]. Errata. Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, \mu \to \frac{1}{2}, \beta \to \alpha + 1, \gamma \to 2, \sigma \to \frac{1}{\sigma}, c \to \frac{1}{\sqrt{b}}, \sigma \to b, m \to -\frac{1}{2}, \lambda \to 2\alpha$. Then form a second equation by replacing $\sigma \to b$. Next replace $\alpha \to a, b \to y$ and simplify;

$$ (4.2) \quad \int_0^y \frac{(\sqrt{x} + \sqrt{y})^{2a} {}_2F_1\left(a, 1+a; 2; \frac{x}{y}\right)}{\sqrt{x}} dx \\ = \sum_{j=0}^{\infty} \frac{2y^{\frac{1}{2}+a} {}_2F_1\left(-2a, 1+2j; 2+2j; -1\right) (a)_j (1+a)_j}{(1+2j)j! (2)_j} \neq \frac{y^{a+\frac{1}{2}}}{(1-a)2^{2a+1}} $$

35 where $Re(y) > 2Re(a), |Re(a)| < 1, |Re(y)| < 1$.

EXAMPLE 4.3. Similar form to Entry (2.21.3.6) in [31]. Use equation (2.11) and set $k \to 0, a \to 1, v \to \frac{1}{2}, d \to id, \mu \to 1$ and form a second equation by replacing $d \to -d$ and take their difference. Next replace $c \to -1/b, d \to \theta$ and simplify;

$$ (4.3) \quad \int_0^b (b-x)^\lambda x^m {}_2F_1(\alpha, \beta; \gamma; x\sigma) \sin(\sqrt{x}\theta) dx $$
$$ = \sum_{h=0}^{\infty} \frac{i^{h+1} (-1 + (-1)^h) b^{1+\frac{h}{2}+m+\lambda} \theta^h \Gamma(\frac{1}{2}(2+h+2m)) \Gamma(1+\lambda)}{2\Gamma(1+h)\Gamma(\frac{1}{2}(4+h+2m+2\lambda))} \times {}_3F_2\left(1+\frac{h}{2}+m, \alpha, \beta; \gamma, 2+\frac{h}{2}+m+\lambda; b\sigma\right) $$

36 where $Re(\alpha + \beta + \gamma + \lambda) > 0, Re(b) > 0$.

### 5. Derivation of entries in Table 9.121 in [3]

EXAMPLE 5.1. From 9.121.26 in [3]. Use equation (2.11) and set $\alpha \to \frac{1}{2}, \beta \to \frac{1}{2}, \gamma \to \frac{3}{2}, v \to 0, d \to 1, a \to 1, b \to 1, \lambda \to -1, \mu \to 2$ and replace $c \to z$ and simplify;

$$ (5.1) \quad \int_0^1 \frac{x^{-\frac{1}{2}+m} \sin^{-1}(\sqrt{x}\sqrt{\sigma}) \log^k\left(\frac{1}{x}\right)}{1+x^2 z} dx \\ = \sum_{j=0}^{\infty} \frac{2^{-1-k} \sigma^{\frac{1}{2}+j} \Gamma\left(\frac{1}{2}(1+2j)\right) \Gamma(1+k) \Phi(-z, 1+k, \frac{1}{2}(1+j+m))}{(1+2j)\sqrt{\pi} \Gamma(1+j)} $$

38 where $Re(z) < 1$.

EXAMPLE 5.2. From 9.121.29 in [3]. Use equation (2.11) and set $\alpha \to \frac{n+1}{2}, \beta \to \frac{1-n}{2}, \gamma \to \frac{3}{2}, v \to 0, d \to 1, \lambda \to -\frac{1}{2}, \mu \to 2, m \to 0, k \to -1$ and form a second equation by replacing $a \to -a$ and take their difference and simplify.

$$ (5.2) \quad \int_0^b \frac{\sin(n \sin^{-1}(\sqrt{x}\sqrt{\sigma}))}{\sqrt{x}\sqrt{1+cx^2}(a^2+\log^2(x))} dx \\ = \sum_{j=0}^{\infty} \sum_{l=0}^{\infty} \frac{i2^{-1+2j}c^l e^{-ia(1+j+2l)} n\sigma^{\frac{1}{2}+j} \left(-\frac{1}{l}\right) \cos\left(\frac{n\pi}{2}\right) \Gamma\left(\frac{1}{2}+j-\frac{n}{2}\right) \Gamma\left(\frac{1}{2}+j+\frac{n}{2}\right)}{a\pi\Gamma(2+2j)} \\ \times \left(e^{2ia(1+j+2l)} \Gamma(0, i(1+j+2l)(a+i\log(b))) - \Gamma(0, -((1+j+2l)(ia+\log(b))))\right) $$

39 where $|Re(c)|<1, Re(b)>0, Re(\sigma)>0$.

### 6. Extended Sofo definite integrals

EXAMPLE 6.1. Derivation of equation (1) in [27]. Use equation (2.11) and set $v \to 0, d \to 1, a \to 1, \lambda \to -1, \mu \to 1, \alpha \to 1, \beta \to 2, \sigma \to -1, \gamma \to k+2, k \to 2m-1, m \to 1, c \to -1, b \to 1$ and simplify;

$$ (6.1) \quad \int_0^1 \frac{x_2 F_1(1, 2; 2+k; -x) \log^{-1+2m}\left(\frac{1}{x}\right)}{1-x} dx \\ = \sum_{j=0}^{\infty} \frac{(-1)^j \Gamma(2+j) \Gamma(2+k) \Gamma(2m) \zeta(2m, 2+j)}{\Gamma(2+j+k)} $$

41 where $Re(m) > 1$.
EXAMPLE 6.2. The Hurwitz-Lerch zeta function. Use equation (2.11) and set $v \to 0, d \to 1, a \to 1, \lambda \to -1, \mu \to 1, \alpha \to 1, \beta \to r+1, \sigma \to -1, \gamma \to r+2, k \to 2m-1, m \to 1, c \to -1, b \to 1$ and simplify;

$$ (6.2) \quad \int_0^1 \frac{x\Phi(-x, 1, 1+r) \log^{-1+2m}\left(\frac{1}{x}\right)}{1-x} dx = \sum_{j=0}^{\infty} \frac{(-1)^j \Gamma(2m)\zeta(2m, 2+j)}{1+j+r} $$

42 where $Re(m) > 1$.

EXAMPLE 6.3. The complete elliptic integral of the first kind. Use equation (2.11) and set $a \to 1, b \to 1, p \to 1, \alpha \to \frac{1}{2}, \beta \to \frac{1}{2}, \gamma \to 1, \lambda \to -1, d \to 1, v \to 0, m \to m-1$ and simplify;

$$ (6.3) \quad \int_0^1 \frac{x^{-1+m} K(x\sigma) \log^k\left(\frac{1}{x}\right)}{1-zx^\mu} dx = \frac{\Gamma(1+k)}{2\mu^{k+1}} \sum_{j=0}^{\infty} \sigma^j \left( \frac{\Gamma\left(\frac{1}{2}+j\right)}{\Gamma(1+j)} \right)^2 \Phi\left(z, 1+k, \frac{j+m}{\mu}\right) $$

43 where $|Re(m)|<1, 0 < Re(\sigma) < 1$.

EXAMPLE 6.4. Reciprocal logarithm function. Use equation (2.11) and set $v \to 0, d \to 1, a \to 1, \lambda \to 0, c \to 1, b \to 1, k \to -1$ and form a second equation by replacing $m \to s$ and take their difference and simplify;

$$ (6.4) \quad \int_0^1 \frac{(x^m - x^s) {}_2F_1(\alpha, \beta; \gamma; x\sigma)}{\log(x)} dx = \sum_{j=0}^{\infty} \frac{\sigma^j \Gamma(j+\alpha)\Gamma(j+\beta)\Gamma(\gamma)}{\Gamma(1+j)\Gamma(\alpha)\Gamma(\beta)\Gamma(j+\gamma)} \log\left(\frac{1+j+m}{1+j+s}\right) $$

44 where $Re(\sigma) > 0$.

EXAMPLE 6.5. Finite Mellin transform of the complete elliptic integral of the first kind. Use equation (2.11) and set $k \to 0, v \to 0, d \to 1, a \to 1, \alpha \to \frac{1}{2}, \beta \to \frac{1}{2}, \gamma \to 1, \lambda \to \frac{1}{2}, c \to \sigma, \mu \to 1, m \to m-1$ and replace $\sigma \to -\sigma$ and simplify using $K(-m) = \frac{1}{\sqrt{m+1}} K\left(\frac{m}{m+1}\right)$;

$$ (6.5) \quad \int_0^b x^{-1+m} K\left(\frac{x\sigma}{1+x\sigma}\right) dx = \sum_{l=0}^{\infty} \frac{b^{l+m} \pi \sigma^l \binom{\frac{1}{2}}{l} {}_3F_2\left(\frac{1}{2}, \frac{1}{2}, l+m; 1, 1+l+m; -b\sigma\right)}{2(l+m)} \\ = \sum_{j=0}^{\infty} \frac{(-1)^j b^{j+m} \pi \sigma^j {}_2F_1\left(-\frac{1}{2}, j+m; 1+j+m; -b\sigma\right) \binom{\frac{1}{2}}{j}^2}{2(j+m)j! (1)_j} $$

45 where $Re(\sigma) > 0$.

EXAMPLE 6.6. Integral of the complete elliptic integral of the first kind. Note the series involving the sum over $j \in [0, \infty)$ yields a more accurate result for the
integral over $b \in [0, 1]$. This could be due to the higher convergence rate of the series. Use equation (6.5) and set $m = b = 1, \sigma = -1$ and simplify.

$$ (6.6) \qquad \int_0^1 K \left( \frac{x}{x-1} \right) dx = \sum_{j=0}^{\infty} \frac{e^{2ij\pi} \sqrt{\pi\Gamma(\frac{1}{2}+j)}}{(3+4j(2+j))\Gamma(1+j)} = \frac{\pi^2}{8} $$

EXAMPLE 6.7. Derivation of Bateman's fractional integral formula see [28] and page 184 in [29]. Use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, \mu \to 1, c \to -\frac{1}{b}, m \to \gamma - 1, \sigma \to 1$ and simplify.

$$ (6.7) \qquad \int_0^b (b-x)^\lambda x^{-1+\gamma} {}_2F_1(\alpha, \beta; \gamma; x) dx = \frac{b^{\gamma+\lambda}\Gamma(\gamma)\Gamma(1+\lambda) {}_2F_1(\alpha, \beta; 1+\gamma+\lambda; b)}{\Gamma(1+\gamma+\lambda)} $$

where $Re(b) > 0, Re(\alpha) > 0, Re(\beta) > 0, Re(\gamma) > 0$.

EXAMPLE 6.8. Use equation (2.11) and set $v \to 0, d \to 1, k \to 0, a \to 1, \gamma \to \alpha, \lambda \to -\frac{1}{2}, \beta \to \frac{1}{2}, \mu \to 1$ and simplify using [DLMF, 15.4].

$$ (6.8) \qquad \int_0^b \frac{x^m}{\sqrt{1+cx}\sqrt{1-x\sigma}} dx = \sum_{j=0}^{\infty} \frac{\left(\frac{1}{b}\right)^{-1-j-m} \sigma^j {}_2F_1\left(\frac{1}{2}, 1+j+m; 2+j+m; -bc\right) \left(\frac{1}{2}\right)_j}{(1+j+m)j!} \\ = \sum_{l=0}^{\infty} \frac{\left(\frac{1}{b}\right)^{-1-l-m} c^l \left(-\frac{1}{2}\right)_2 {}_2F_1\left(\frac{1}{2}, 1+l+m; 2+l+m; b\sigma\right)}{1+l+m} $$

where $|Re(m)|< 1, Re(b) > 0$.

EXAMPLE 6.9. Table 34 in [30]. Some relationships between elliptic integrals, elliptic functions in Legendre notation, and Jacobian elliptic functions (from Abramowitz and Stegun, Page 566). In the following. Use equation (6.8) and simplify;

$$ (6.9) \qquad \int_0^{\sin(\phi)} \frac{1}{\sqrt{1+c^2x^2}\sqrt{1-x^2\sigma^2}} dx \\ = \sum_{j=0}^{\infty} \frac{\sin^{1+2j}(\phi)\sigma^{2j} {}_2F_1\left(\frac{1}{2}, \frac{1}{2}+j; \frac{3}{2}+j; -\sin^2(\phi)c^2\right) \left(\frac{1}{2}\right)_j}{j!+2jj!} \\ = \sum_{l=0}^{\infty} \frac{\sin^{1+2l}(\phi)c^{2l}\left(-\frac{1}{2}\right)_2 {}_2F_1\left(\frac{1}{2}, \frac{1}{2}+l; \frac{3}{2}+l; \sin^2(\phi)\sigma^2\right)}{1+2l} $$

where $Re(c) > 0, Re(\sigma) > 0$.

## 7. Infinite integrals involving the hypergeometric function

In this section we derive an infinite definite integral using equation (2.11) and [Wolfram,01]. This integral will be used to derive series forms for infinite integrals in
52 current literature. Entries in [3] which can be derived using equation (7.1) include (7.522.1,3,4) (7.524.1), (7.527) (7.531).

53

EXAMPLE 7.1. In this example we use equation (2.11) and set $k = 0, a = 1$ and simplify using equation [Wolfram,01] for the generalized incomplete gamma function.

$$ (7.1) \quad \int_0^\infty e^{-dx^\nu} x^m (1+cx^\mu)^\lambda {}_2F_1(\alpha, \beta; \gamma; x\sigma) dx \\ = \sum_{j=0}^\infty \sum_{l=0}^\infty \frac{(-1)^{-\frac{1+j+m+l\mu}{v}} c^l (-d)^{-\frac{1+j+m+l\mu}{v}} \sigma^j \binom{\lambda}{l} \Gamma(j+\alpha)\Gamma(j+\beta)\Gamma(\gamma)\Gamma\left(\frac{1+j+m+l\mu}{v}\right)}{v\Gamma(1+j)\Gamma(\alpha)\Gamma(\beta)\Gamma(j+\gamma)} $$

54 where $Re(v) \ge \pi/2, Im(d) \ne 0$.

EXAMPLE 7.2. Derivation of series form for equation [Wolfram,01]. Here we use equation (7.1) and set $v \to 1, \lambda \to 0, \mu \to 0, c \to 1, m \to \alpha - 1, \alpha \to a, \beta \to b, \gamma \to c, \sigma \to -1, x \to t$ and simplify.

$$ (7.2) \quad \int_0^\infty e^{-dt} t^{-1+\alpha} {}_2F_1(a, b; c; -t) dt \\ = \sum_{j=0}^\infty \frac{(-1)^{-\alpha} (-d)^{-j-\alpha} \Gamma(c) \Gamma(a+j) \Gamma(b+j) \Gamma(j+\alpha)}{\Gamma(a) \Gamma(b) \Gamma(1+j) \Gamma(c+j)} \\ = \frac{(\Gamma(c)\Gamma(a-\alpha)\Gamma(b-\alpha)\Gamma(\alpha)) {}_2F_2(\alpha, 1-c+\alpha; 1-a+\alpha, 1-b+\alpha; d)}{\Gamma(a)\Gamma(b)\Gamma(c-\alpha)} \\ + \frac{(\Gamma(c)\Gamma(a-b)\Gamma(\alpha-b)) d^{b-\alpha} {}_2F_2(b, b-c+1; 1-a+b, b-\alpha+1; d)}{\Gamma(a)\Gamma(c-b)} \\ + \frac{(\Gamma(c)\Gamma(b-a)\Gamma(\alpha-a)) d^{a-\alpha} {}_2F_2(a, a-c+1; a-b+1, a-\alpha+1; d)}{\Gamma(b)\Gamma(c-a)} $$

55 where $Re(\alpha) > |Im(\beta)|, Re(\mu + 2\nu) > 0$.

EXAMPLE 7.3. Derivation of equation (7.522.2) in [3]. Here we use equation (7.1) and set $\lambda \to 0, \mu \to 0, c \to 1$ then $d \to \lambda, v \to 1, m \to m-1, \sigma \to -\frac{1}{2}, \alpha \to v + \frac{1}{2}, \beta \to \frac{1}{2} - v, \gamma \to a$ and simplify.

$$ (7.3) \quad \int_0^\infty e^{-bx} x^{-1+a} {}_2F_1\left(\frac{1}{2}-v, \frac{1}{2}+v; a; -\frac{x}{2}\right) dx \\ = \sum_{j=0}^\infty \frac{(-1)^{-a} 2^{-j} (-b)^{-a-j} \cos(\pi v) \Gamma(a) \Gamma\left(\frac{1}{2}+j-v\right) \Gamma\left(\frac{1}{2}+j+v\right)}{\pi \Gamma(1+j)} \\ = \frac{2^a \exp(b) \Gamma(a) (2b)^{\frac{1}{2}-a} K_v(b)}{\sqrt{\pi}} $$

56 where $Re(\alpha) > 0, Re(b) > 0$.
EXAMPLE 7.4. Derivation of equation (2.21.2.4) in [31]. Here we use equation (7.1) and set $\alpha \to \frac{1}{2}, \beta \to 1, \gamma \to c, \sigma \to -w, m \to c$ and simplify;

$$ (7.4) \quad \int_0^\infty e^{-px} x^{-1+c} {}_2F_1\left(\frac{1}{2}, 1; c; -wx\right) dx \\ = \frac{p^{-c} \sum_{j=0}^\infty \left(-\frac{w}{p}\right)^j \Gamma(c)\Gamma\left(\frac{1}{2}+j\right)}{\sqrt{\pi}} = \frac{\sqrt{\frac{\pi}{w}} \Gamma(c) \exp\left(\frac{w}{p}\right) \operatorname{erfc}\left(\sqrt{\frac{w}{p}}\right)}{p^{c-\frac{1}{2}}} $$

57 where $Re(c) > 0, Re(p) > 0, |\arg w| < \pi$.

EXAMPLE 7.5. Derivation of equation (2.21.2.8) in [31]. Here we use equation (2.11) and set $\lambda \to 0, \mu \to 0, c \to 1, d \to p, v \to \frac{1}{2}, m \to m-1, \sigma \to -w, \alpha \to a, \beta \to b, \gamma \to c$ and simplify;

$$ (7.5) \quad \int_0^\infty e^{-p\sqrt{x}} x^{-1+m} {}_2F_1(a, b; c; -wx) dx \\ = \frac{(2p^{-2m}\Gamma(c))}{\Gamma(a)\Gamma(b)} \sum_{j=0}^\infty \left(-\frac{w}{p^2}\right)^j \frac{\Gamma(a+j)\Gamma(b+j)\Gamma(2(j+m))}{\Gamma(1+j)\Gamma(c+j)} $$

58 where $Re(c) > 0, Re(p) > 0, |\arg w| < \pi$.

EXAMPLE 7.6. Derivation of equation (7.513) in [3]. Errata. Here we use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, c \to -1, x \to t^2$ then $\sigma \to 1, t \to x, b \to 1$ and simplify;

$$ (7.6) \quad \int_0^1 x^{-1+s} (1-x^2)^v {}_2F_1(a, -n; b; x^2) dx \\ = \sum_{j=0}^\infty \frac{\Gamma(b)\Gamma(a+j)\Gamma(j-n)\Gamma\left(j+\frac{s}{2}\right)\Gamma(1+v)}{2\Gamma(a)\Gamma(1+j)\Gamma(b+j)\Gamma(-n)\Gamma\left(1+j+\frac{s}{2}+v\right)} \\ \neq \frac{1}{2}\binom{v+1}{\frac{s}{2}} {}_3F_2\left(-n, a, \frac{s}{2}; b, v+1+\frac{s}{2}; 1\right) $$

59 where $Re(s) > 0, Re(v) > -1$.

EXAMPLE 7.7. Derivation of equation (7.512.6(12)) in [3]. Here we use equation (2.11) and set $k \to 0, a \to 1, v \to 0, d \to 1, c \to -1, b \to 1$ and simplify;

$$ (7.7) \quad \int_0^1 (1-x)^{-1+\beta-\lambda} x^{-1+\lambda} {}_2F_1(\alpha, \beta; \gamma; xz) dx = \frac{\Gamma(\beta-\lambda)\Gamma(\lambda) {}_2F_1(\alpha, \lambda; \gamma; z)}{\Gamma(\beta)} $$

60 where $Re(\beta) > Re(\lambda), |\arg(1-z)| < \pi$.

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

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Email address, Corresponding author: milver73@gmail.com