# Extended Adams-Wood integral

Robert Reynolds

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Robert Reynolds. Extended Adams-Wood integral. 2024. hal-04785975v2

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# Extended Adams-Wood integral

Robert Reynolds

ABSTRACT. In this work we derive a definite integral involving the product of the exponential, logarithm and a polynomial function in terms of the incomplete gamma function. Special cases of this integral were previously studied by Ramanujan, Hardy and Wood. The method used in this derivation is contour integration.

## 1. Introduction

Definite integrals play an important role in science and engineering. In this paper we look at the historical usage of one particular integral, derive its closed form solution and show a more efficient way to evaluate its special cases which were considered by many mathematicians. In 1940, Hardy [1] published work done Ramanujan and in this work on pages 195-198, contains a definite derived using the inverse formula of Laplace. Definite integrals of the form studied in this work occur in the asymptotic expansions of the solutions of heat conduction problems in regions bounded internally by a circular cylinder [2, 3]. These integrals are also used in problems on the flow of fluids through porous media [4], in electron slowing-down problems [5] and in many other fields in mathematics. In the work done by Wood [3], integer values for some of the parameters were considered, while in this work we derive and evaluate the integral for complex values of all the involved. It will

2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.

Key words and phrases. Ramanujan, definite integral, incomplete gamma function, contour integral.
be shown that some special cases of the definite integral in this work are related to the $\nu$-functions of Volterra [6, 7]. Volterra functions play an important role in the solution of certain integral equations. In this paper we derive the definite integral given by

$$ (1.1) \quad \int_0^u e^{-x} x^{-1+m} \log^k(ax) dx = \sum_{j=0}^{\infty} \frac{(-1)^{j+k} \Gamma(1+k, -((j+m)\log(au)))}{j!(j+m)^{k+1}a^{j+m}} $$

where $Re(m) > 0$. where the parameters $k, a, u, m$ are general complex numbers and $Re(m) \ge 0$. This definite integral having an upper limit as a variable over the complex plane has a wider range of evaluation for the parameters relative to previous results in the latter. The derivation follows the method used by us in [8]. This method involves using a form of the generalized Cauchy's integral formula given by

$$ (1.2) \quad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw. $$

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 (1.2) by another function of $y$ and take the infinite sum of both sides such that the contour integral of both equations are the same.

## 2. Contour integral representations

In this section we will derive the contour integral representations for the definite integrals used to derive equation (1.1). The contour integral representation for equation (1.1) is given by;

$$ (2.1) \quad \frac{1}{2\pi i} \int_0^u \int_C e^{-x} a^w w^{-k-1} x^{m+w-1} dx dw = \sum_{j=0}^{\infty} \int_C \frac{(-1)^j a^w w^{-k-1} u^{j+m+w}}{j!(j+m+w)} dw $$

where $Re(m) > 0$.
**2.1. The left-hand side contour integral.** Using a generalization of Cauchy's integral formula 1.2, we form the definite integral by replacing $y$ by $\log ax$ and multiply both sides by $e^{-x}x^{m-1}$ and take the definite integral over $x \in [0, u]$ to get;

$$ (2.2) \quad \int_0^u \frac{e^{-x} x^{m-1} \log^k(ax)}{k!} dx \\ = \frac{1}{2\pi i} \int_0^u \int_C e^{-x} w^{-k-1} x^{m-1} (ax)^w dw dx \\ = \frac{1}{2\pi i} \int_C \int_0^u e^{-x} w^{-k-1} x^{m-1} (ax)^w dxdw $$

We are able to switch the order of integration over $x$ and $w$ using Fubini's theorem for multiple integrals see page 178 in [9], since the integrand is of bounded measure over the space $\mathbb{C} \times [0, u]$.

**3. The right-hand side contour integral**

**3.1. 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

$$ (3.1) \qquad \gamma(a, z) = \int_0^z t^{a-1} e^{-t} dt $$

and

$$ (3.2) \qquad \Gamma(a, z) = \int_z^\infty t^{a-1} e^{-t} dt $$

where $Re(a) > 0$. The incomplete gamma function has a recurrence relation given by

$$ (3.3) \qquad \gamma(a, z) + \Gamma(a, z) = \Gamma(a) $$

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

$$ (3.4) \qquad \gamma(a, ze^{2m\pi i}) = e^{2\pi mia} \gamma(a, z) $$
and

$$ (3.5) \qquad \Gamma(a, ze^{2m\pi i}) = e^{2\pi m a} \Gamma(a, z) + (1 - e^{2\pi m a}) \Gamma(a) $$

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 [10]. Some Meijer G representations we will use in this work are given by;

$$ (3.6) \qquad \Gamma(a, z) = \Gamma(a) - G_{1,2}^{1,1} \begin{pmatrix} z \\ a, 0 \end{pmatrix} $$

and

$$ (3.7) \qquad \Gamma(a, z) = G_{1,2}^{2,0} \begin{pmatrix} z \\ 0, a \end{pmatrix} $$

from equations (2.4) and (2.6a) in [11]. We will also use the derivative notation given by;

$$ (3.8) \qquad \frac{\partial \Gamma(a, z)}{\partial a} = \Gamma(a, z) \log(z) + G_{2,3}^{3,0} \begin{pmatrix} z \\ 0, 0, a \end{pmatrix} $$

from equations (2.19a) in [11], (9.31.3) in [12] and equations (5.11.1), (6.2.11.1) and (6.2.11.2) in [13], and (6.36) in [14].

## 3.2. Incomplete gamma function in terms of the contour integral.
In this section, we will once again use Cauchy's generalized integral formula, equation (1.2), and take the infinite integral to derive equivalent sum representations for the contour integrals. We proceed using equation (1.2) and replace $y$ by $\log(a) + x$ and multiply both sides by $e^{mx}$ and replace $m \to m + j$ and simplify. Next, take the
infinite integral over $x \in [0, \infty)$ and simplify in terms of the incomplete gamma function using equation (3.382.4) in [12] and finally take the infinite sum of both sides over $j \in [0, \infty)$ to obtain

$$ (3.9) \quad -\sum_{j=0}^{\infty} \frac{(-1)^j (-j-m)^{-k-1} u^{j+m} (au)^{-j-m} \Gamma(k+1, -((j+m)\log(au)))}{j! k!} \\ = \frac{1}{2\pi i} \sum_{j=0}^{\infty} \int_C \frac{(-1)^j w^{-k-1} (au)^w u^{j+m}}{j! (j+m+w)} dw \\ = \frac{1}{2\pi i} \sum_{j=0}^{\infty} \int_C \frac{(-1)^j w^{-k-1} (au)^w u^{j+m}}{j! (j+m+w)} dw $$

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 [9], since the summand is of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

**THEOREM 3.1.** For all $Re(m) > 0$ then,

$$ (3.10) \quad \int_0^u e^{-x} x^{-1+m} \log^k(ax) dx = \sum_{j=0}^{\infty} \frac{(-1)^{j+k} \Gamma(1+k, -((j+m)\log(au)))}{j! (j+m)^{k+1} a^{j+m}} $$

**PROOF.** Since the right-hand sides of equations (2.2) and (3.9) are equal relative to equation (2.1), we may equate the left-hand sides and simplify the gamma function to yield the stated result. $\square$

## 4. Evaluations and extended forms

In this section we look at special case evaluations and extended closed forms of the definite integral stated in (3.10).

**EXAMPLE 4.1.** In this example we use equation (3.10) and set $u = 1, a = 1$ and replace $m \to s$ to form a second equation and take their difference. Next apply l'Hopital's rule as $k \to -1$ and replace $m \to m+1, s \to s+1$ and simplify.

$$ (4.1) \quad \int_0^1 \frac{e^{-x} (x^m - x^s)}{\log(x)} dx = \sum_{j=0}^{\infty} \frac{(-1)^j}{j!} \log \left( \frac{1 + j + m}{1 + j + s} \right) $$
where $Re(m) > 0, Re(s) > 0$.

**EXAMPLE 4.2.** In this example we use equation (3.10) and set $k = -1, a \to e^a$. Next take the $n$-th derivative with respect $a$ and simplify. Similar forms are in (4.285) in [12] and 321(6) in [15].

$$ (4.2) \quad \int_0^u \frac{e^{-x} x^{-1+m}}{(a + \log(x))^{v+1}} dx \\ = \sum_{j=0}^{\infty} \frac{(-1)^{j+v} e^{-a(j+m)} (j+m)^v \Gamma(-v, -((j+m)(a+\log(u))))}{\Gamma(1+j)} $$

where $Re(m) > 0$.

**EXAMPLE 4.3.** In this example we use equation (3.10) and set $a = 1, u = 1$ and form a second equation by replacing $m \to s$ and take their difference. Next take the first partial derivative with respect to $k$ and apply l'Hopital's rule as $k \to -1$ and simplify.

$$ (4.3) \quad \int_0^1 \frac{e^{-x} (x^s - x^m) \log(\log(x))}{\log(x)} dx \\ = \sum_{j=0}^{\infty} \frac{(-1)^j \log\left(\frac{1+j+m}{1+j+s}\right) (-2i\pi + \log(\exp(2\gamma)(1+j+m)(1+j+s)))}{2\Gamma(1+j)} $$

where $Re(m) > 0, Re(s) > 0$.

**EXAMPLE 4.4.** Use equation (3.10) and set $a = 1, u = 1$, then take the first partial derivative with respect to $k$ and set $k \to -1/2$ and simplify.

$$ (4.4) \quad \int_0^1 \frac{e^{-x} x^{m-1} \log\left(\log\left(\frac{1}{x}\right)\right)}{\sqrt{\log\left(\frac{1}{x}\right)}} dx = -\sqrt{\pi} \sum_{j=0}^{\infty} \frac{(-1)^j \log(4 \exp(\gamma)(j+m))}{\sqrt{j+mj!}} $$

where $Re(m) > 0$.

**EXAMPLE 4.5.** Use equation (3.10) and set $a = 1, u = 1, m = 1$, then take the first partial derivative with respect to $k$ and apply l'Hopital's rule as $k \to 0$ and
simplify.

$$ (4.5) \quad \int_0^1 e^{-x} \log(\log(x)) dx = - \sum_{j=0}^{\infty} \frac{(-1)^j \log(-\exp(\gamma)(1+j))}{\Gamma(j+2)} $$

EXAMPLE 4.6. Use equation (3.10) and set $u = 1$ then take the first partial derivative with respect to $k$ and set $k = 0$ and simplify.

$$ (4.6) \quad \int_0^1 e^{-x} x^{-1+m} \log(\log(ax)) dx = \sum_{j=0}^{\infty} \frac{(-1)^j a^{-j-m} (E_1(-(j+m)\log(a))) + a^{j+m}(i\pi - \log(j+m) + \log(-(j+m)\log(a))))}{(j+m)!} $$

where $Re(m) > 0$.

EXAMPLE 4.7. Extended McLachlan [16] form over $[0, 1]$. Use equation (3.10) and replace $k = -1, m = 1, e \to e^a$ and form a second equation by replacing $a \to -a$ and take the difference and simplify.

$$ (4.7) \quad \int_0^1 \frac{e^{-x\mu}}{a^2 + \log^2(x)} dx = -\frac{i}{2a} \sum_{j=0}^{\infty} \frac{(-\mu)^j (e^{-ia(1+j)}\Gamma(0, -ia(1+j)) - e^{ia(1+j)}\Gamma(0, ia(1+j)))}{\Gamma(j+1)} $$

where $Re(\mu) > 0$.

EXAMPLE 4.8. Use equation (3.10) and set $k = 1, m \to m + 1, x \to bx, a \to 1/b, u \to b, m \to v - 1, b \to \mu$ and simplify.

$$ (4.8) \quad \int_1^\infty e^{-x\mu} x^{-1+v} \log(x) dx = \frac{\Gamma(v)(\psi(v) - \log(\mu))}{\mu^v} + \frac{{}_2F_2(v, v; 1+v, 1+v; -\mu)}{v^2} $$

where $Re(v) > 0$.

EXAMPLE 4.9. Derivation of equation (4.359.2(11)) in [12], (2.6.27.1) in [18]. Use equation (3.10) and replace $m \to m + 1, x \to bx, u \to b, a \to 1/b$. Next form a second equation by replacing $m \to s$ and take their difference. Next take the limit
as $k \to -1$ using l'Hopital's rule and replace $b \to \mu$ and simplify.

$$ (4.9) \quad \int_0^1 \frac{e^{-x\mu} (x^s - x^m)}{\log(x)} dx = \sum_{j=0}^{\infty} \frac{(-\mu)^j}{j!} \log \left( \frac{1+j+s}{1+j+m} \right) $$

where $Re(m) > 0, Re(s) > 0$.

**EXAMPLE 4.10.** Use equation (3.10) and replace $x \to bt^c, u \to bu^c, a = 1/b, m \to m/c, t \to x$ and simplify.

$$ (4.10) \quad \int_0^u e^{-bx^c} x^{-1+m} \log^k(x) dx = \sum_{j=0}^{\infty} \frac{(-1)^{j+k} b^j (cj + m)^{-1-k} \Gamma(1+k, -((cj + m) \log(u)))}{j!} $$

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

**EXAMPLE 4.11.** Derivation of generalized form of equations (4.361.1-2) in [12].
Use equation (3.10) and replace $x \to bt, u \to ub, a \to a/b$. Then set $k = -1, a \to e^a$.
Next form a second equation by replacing $a \to -a$ and take their difference and simplify.

$$ (4.11) \quad \int_0^u \frac{e^{-bt} t^{-1+m}}{a^2 + \log^2(t)} dt = -\frac{1}{2a} \sum_{j=0}^{\infty} \frac{i(-b)^j}{j!} \left( e^{-ia(j+m)} \Gamma(0, -((j+m)(ia + \log(u)))) - e^{ia(j+m)} \Gamma(0, i(j+m)(a + i \log(u))) \right) $$

where $Re(m) > 0$.

**EXAMPLE 4.12.** Use equation (4.11) and form two equations by setting $m = 1, 2$ respectively, add and simplify.

$$ (4.12) \quad \int_0^1 \frac{e^{-x\mu}(1+x)}{\pi^2 + \log^2(x)} dx = \sum_{j=0}^{\infty} \frac{i\mu^j (E_1(-i(1+j)\pi) - E_1(-i(2+j)\pi) + e^{2ij\pi}(-E_1(i(1+j)\pi) + E_1(i(2+j)\pi)))}{2\pi j!} $$
where $Re(\mu) > 0$.

EXAMPLE 4.13. Use equation (4.11) and set $m = 1, t \to x, u = 1, b \to \mu, a = \pi$ then integrate both sides with respect to $\mu \in [-u, u]$ and simplify.

$$ (4.13) \quad \int_0^1 \frac{\sinh(ux)}{x(\pi^2 + \log^2(x))} dx \\ = \sum_{j=0}^{\infty} \frac{i(-1)^j (1 + (-1)^j) u^{1+j} \Gamma(0, -i(1+j)\pi) - e^{2ij\pi} \Gamma(0, i(1+j)\pi)}{4\pi\Gamma(2+j)} $$

where $Re(u) > 0$.

EXAMPLE 4.14. In this example we use equation (3.10) and replace $x \to bt, u \to ub, a \to a/b, a \to e^a$ and simplify.

$$ (4.14) \quad \int_0^u e^{-bt} t^{m-1} \log^k(at) dt = \sum_{j=0}^{\infty} \frac{(-1)^{j+k} a^{-j-m} b^j \Gamma(1+k, -((j+m)\log(au)))}{j!(j+m)^{k+1}} $$

where $Re(m) > 0$.

EXAMPLE 4.15. This integral is a closed form solution for equation (1) in [3]. Use equation (4.14) and replace $u \to 1/a$ and simplify the incomplete gamma function as $u \to \infty$ and simplify in terms of the gamma function.

$$ (4.15) \quad \int_0^\infty e^{-bt} t^{m-1} (-\log(at))^k dt = \frac{\Gamma(1+k)}{a^m} \sum_{j=0}^\infty \frac{(-\frac{b}{a})^j}{j!(j+m)^{k+1}} $$

where $Re(t) > 0, |Im(b)| \ge 1$.

EXAMPLE 4.16. Use equation (3.10) and form a second equation by replacing $m \to s$ and take their difference. Next set $a = 1$ and take the first partial derivative with respect to $k$ and apply l'Hopital's rule as $k \to -1$ and simplify the incomplete gamma function.

$$ (4.16) \quad \int_0^\infty \frac{e^{-bt} (-t^m + t^s) \log(\log(t))}{t \log(t)} dt $$
$$= \sum_{j=0}^{\infty} \frac{(-b)^j \log\left(\frac{j+m}{j+s}\right) (-2i\pi + \log(\exp(2\gamma)(j+m)(j+s)))}{2\Gamma(1+j)}$$

where $Re(m) > 0, Re(b) \ge 0$.

**EXAMPLE 4.17.** In this example we use equation (3.10) and take the first partial derivative with respect to $k$ and set $k = 0, a \to e^{\alpha+i\beta}$. Next form a second equation by replacing $\beta \to -\beta$ and add these two equations and simplify the logarithm function.

$$ (4.17) \quad \int_0^\infty e^{-bt} t^{-1+m} \log(\alpha^2 + \beta^2 + 2\alpha \log(t) + \log^2(t)) dt \\ = \sum_{j=0}^\infty \frac{2(-b)^j e^{-(j+m)\alpha} (-\cos((j+m)\beta)(\gamma + \log(j+m)) + \pi \sin((j+m)\beta))}{(j+m)j!} $$

where $Re(m) > 0, Re(\beta) \ge 0$.

**EXAMPLE 4.18.** In this example we integrate equation (4.11) over $b \in [-b, b]$. Then set $m = 1, a = \pi$ and replace $t \to t/c, u \to u/c, b \to bc$ and finally $u = 1/2$ and simplify.

$$ (4.18) \quad \int_0^{\frac{1}{2}} \frac{\sinh(bt)}{t\left(\pi^2 + \log^2\left(\frac{t}{c}\right)\right)} dt \\ = \sum_{j=0}^{\infty} \frac{i(1+(-1)^j)(bc)^{1+j}}{4\pi\Gamma(2+j)} \left( \Gamma\left(0, -\left((1+j)\left(i\pi + \log\left(\frac{1}{2c}\right)\right)\right)\right) \right. \\ \left. - e^{2ij\pi} \Gamma\left(0, (1+j)\left(i\pi + \log(2) - \log\left(\frac{1}{c}\right)\right)\right) \right) $$

where $Re(b) > 0$.

**EXAMPLE 4.19.** Derivation of a closed form solution for equation (6c) in [3]. In this example we integrate equation (4.11) over $b \in [0, b]$. Then set $m = 1, t \to t/d, a = \pi, u \to 1/(2d), b \to d$ and finally replace $t \to v, d \to t$ and simplify.

$$ (4.19) \quad \int_0^{\frac{1}{2}} \frac{1 - e^{-v}}{v \left( \pi^2 + \log^2 \left( \frac{v}{t} \right) \right)} dv $$
$$ = \frac{it^{1+j}}{2\pi} \sum_{j=0}^{\infty} \frac{(\Gamma(0, -((1+j)(i\pi - \log(2t)))) - \Gamma(0, (1+j)(i\pi + \log(2t))))}{\Gamma(2+j)} $$

where $Re(t) > 0$.

EXAMPLE 4.20. Representation of the exponential integral function through hypergeometric functions. Use equation (4.14) and set $a \to 1, u \to 1, k \to 1, b \to -u, m \to j+1$ and simplify. Next replace $t \to x, j \to n$ and simplify the series in terms of the generalized hypergeometric function. Next use this equation and substitute into equation (4.353.2) in [12] and simplify using [DLMF,8.19.16] and [Wolfram,01]. A special case of this example, is in Theorem 55 on pp. 235. in [17].

(4.20)

$$ E_v(z) = \frac{z_2 F_2(2 - v, 2 - v; 3 - v, 3 - v; -z)}{(v - 2)^2} + \frac{2 F_2(1 - v, 1 - v; 2 - v, 2 - v; -z)}{v - 1} + z^{v-1} \Gamma(1 - v) $$

where $Re(v) > 0, Re(z) > 0$.

EXAMPLE 4.21. Derivation of equation (4.331.3) in [12]. Use equation (4.14) and replace $t \to x, b \to -u, k = 1, a = 1, m = 1, u = 1$ and simplify.

(4.21) $$ \int_0^1 e^{ux} \log(x) dx = \frac{-\gamma - \Gamma(0, -u) - \log(-u)}{u} $$

where $Re(u) > 0$.

EXAMPLE 4.22. Use equation (4.14) and set $b \to (2j+1)u$ and multiply both sides by $(-1)^j$ and take the infinite sum of both sides over $j \in [0, \infty)$ and simplify using equation (1.232.2) in [12].

(4.22) $$ \int_0^\alpha \frac{t^{-1+m} \log^k(at)}{\cosh(tu)} dt = -\sum_{n=0}^\infty \frac{2^{1+2n} e^{-i(k-n)\pi} (m+n)^{-1-k} u^n a^{-m-n}}{n!} $$ $$ \Gamma(1+k, -((m+n)\log(a\alpha))) \left( \zeta\left(-n, \frac{1}{4}\right) - \zeta\left(-n, \frac{3}{4}\right) \right) $$

where $Re(m) < 1, Re(u) > 0$.
**5. Conflict of Interest**

The author has no conflicts to disclose.

**6. Data Availability Statements**

Data sharing not applicable – no new data generated

**7. Conclusion**

In this paper, we produced a definite integral and its special cases. The integral derived in this work was previously studied by Ramanujan using the inverse Laplace transform, then Wood studied this integral and its applications in science and engineering. These previous works on this definite integral did not yield a closed form solution, like the one produced in this work. We will be using our contour integral to derive closed form formulae for other definite integrals which are used in applied mathematics. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram.

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

Email address, Corresponding author: milver73@gmail.com