Note on an integral by Anatolii Prudnikov

: Closed expressions for the integral are given where the variables a , k , m and u are general complex numbers. Some of these closed expressions are given in [4]. Some special cases of the integral are derived and discussed.


Introduction
In 1986 Prudnikov et al [4] published a famous book containing a significant number of integrals. In this manuscript we focus on the integral ∞ 0 x m−1 log k (ax) which has a closed form in in terms of the Lerch function. We appeal to our method to derive a closed form solution which is continued analytically along with some special cases. A consequence of using our method with this integral is the derivation of the Mellin transform of the product of the logarithmic function and a rational function which is not found in current literature. This work also looks to expand the Table 2.5.1 in [1] where similar and relevant formula are listed. In our case the constants in the formulas are general complex numbers subject to the restrictions given below. The derivations follow the method used by us in [5], [6], [7], [8], [9] and [10]. The generalized Cauchy's integral formula is given by 2πi C e wy w k+1 dy. (1.1) This method involves using a form of Eq (1.1) then multiplies both sides by a function, then takes a definite integral of both sides. This yields a definite integral in terms of a contour integral. Then we multiply both sides of Eq (1.1) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same.

The definite integral of the contour integral
We use the method in [9]. The variable of integration in the contour integral is α = m + w. The cut and contour are in the first quadrant of the complex α-plane with 0 < (α) < 5u. The cut approaches the origin from the interior of the first quadrant and the contour goes round the origin with zero radius and is on opposite sides of the cut. Using a generalization of Cauchy's integral formula, we first replace y by log(ax) followed by multiplying both sides by from Eq (2.2.7.7) in [4], where 0 < (α) < 5u. We are able to switch the order of integration over α and x using Fubini's theorem since the integrand is of bounded measure over the space C × [0, ∞).

Derivation of the contour integral
Here we use Eq (1.1) and replacing y by y + 2iπ 3u and multiply by exp − iπ(u−4m) 6u for the first equation and then by y − 2iπ 3u and multiply by exp iπ(u−4m) 6u for the second equation. Then we subtract the two to get Next we replace y by log(a) + iπ(2y+1) u then multiply both sides by −2iπ exp iπm(2y+1) u to get Next we take the infinite sum over y ∈ [0, ∞) and multiply by 1/6u to get [3], where csch(ix) = −i csc(x) from (4.5.10) in [3] and (α) > 0.

The definite integral in terms of the Lerch function
Since the right-hand side of Eq (2.1) is equal to the sum of Eqs (3.2), (3.5) and (3.8), we may equate the left hand sides to get The Lerch function has a series representation given by where |z|< 1, v = 0, −1, .. and is continued analytically by its integral representation given by where Re(v) > 0, or |z|≤ 1, z = 1, Re(s) > 0, or z = 1, Re(s) > 1.

Derivation of Prudnikov type integral in
Using Eq (4.1), then taking the first partial derivative with respect to m, then set m = 1, k = 1 and a = 1/a on the right-hand side followed by using L'Hopital's rule on the right-hand side as u → 1, we get

Derivation of an integral of product of logarithms
Using Eq (4.1), then taking the first partial derivative with respect to m, then set m = 2, k = 1 and a = 1/a followed by using L'Hopital's rule on the right-hand side as u → 1, we get

Derivation of a Mellin transform with an example
Using Eq (4.1), we first set a = 1 and replace u by 2m and m by m/2, then take the first partial derivative with respect to k, then set k = 0 simplify to get Next we set m = 1 in Eq (9.1) simplify to get Note there is a singularity at x = 1 which is removable using the principal value of the integral.

Derivation of special cases
In this section, we will list some interesting examples by evaluating Eq (4.1) for various values of the parameters k, a, m and u in terms of fundamental constants and trigonometric functions.

Example 1
Using Eq (4.1) and setting k = a = −1, u = 1 and m = 1/2 and rationalizing the real and imaginary parts, we get

Example 2
Using Eq (4.1) and setting k = −1, a = i, u = 1 and m = 1/2 and rationalizing the real and imaginary parts, we get

Example 4
Using Eq (4.1) and setting k = −1, a = −1, and u = 1 and m = 1/4 for the first equation, then replace m = 1/2 for the second equation, followed by subtracting the two and rationalizing the real and imaginary parts, we get

Summary
In this article, we derived some interesting definite integrals in [4]. We found that we are able to achieve a wider range of computation using one formula as opposed to previous works. We will be looking at other integrals using this contour integral method for future work. The results presented were numerically verified for both real and imaginary values of the parameters in the integrals using Mathematica by Wolfram.