Derivation of Some Entries in the Tables of David Bierens De Haan and Anatolii Prudnikov: An Exercise in Integration Theory

It is always useful to improve the catalogue of deﬁnite integrals available in tables. In this paper we use our previous work on Lobachevsky integrals to derive entries in the tables by Bierens De Haan and Anatolli Prudnikov featuring errata results and new integral formula for interested readers. In this work we derive a deﬁnite integral given by in terms of the Lerch function. The importance of this work lies in the derivation of known and new results not presently found in current literature. We used our contour integral method and applied it to an integral in Prudnikov and used it to derive a closed form solution in terms of a special function. The advantage of using a special function is the added beneﬁt of analytic continuation which widens the range of computation of the parameters. Special functions have signiﬁcance in mathematical analysis, functional analysis, geometry, physics, and other applications. Special functions are used in the solutions of differential equations or integrals of elementary functions. Special functions are linked to the theory of Lie groups and Lie algebras, as well as certain topics in mathematical physics.


Introduction
In 1867 David Bierens De Haan [3] and 1990 Anatolii Prudnikov [5] both produced famous books on definite integrals. In this work, the authors expand on their work in [7] and their contour integral method and applied it to an interesting integral in the book of Prudnikov et al. [5] and expressed its closed form in terms of the Lerch function. This derived integral formula was then used to provide formal derivations for some definite integrals in Table 356 in [3] and 2.4.7 in [5], along with some new formula and errata in certain cases. The Lerch function being a special function has the fundamental property of analytic continuation, which enables us to widen the range of evaluation for the parameters involved in our definite integral.
The definite integral derived in this manuscript is given by where the parameters k, a, m, c and t are general complex numbers. This work is important because the authors were unable to find similar derivations in current literature. The derivation of the definite integral follows the method used by us in [6] which involves Cauchy's integral formula. The generalized Cauchy's integral formula is given by 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. This method involves using a form of equation (3) then multiply both sides by a function, then take a definite integral of both sides. This yields a definite integral in terms of a contour integral. A second contour integral is derived by multiplying equation (3) by a function and performing some substitutions so that the contour integrals are the same.

Definite integral of the contour integral
We use the method in [6]. The variable of integration in the contour integral is z = m + w. The cut and contour are in the second quadrant of the complex z-plane. The cut approaches the origin from the interior of the second 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 will form two equations and add them. For the first equation replace y by x + log(a) then multiply by e mx . To form the second equation we replace x by −x in the first and add both equations followed by multiplying both sides by 1 2(cosh 2 (cx)+sinh 2 (t)) . Next we take the infinite integral over x ∈ [0, ∞) to get from equation (2.4.6.27) in [5] where −1 < Re(w + m) < 0 and 2Re(c) > |Re(w + m)|. The logarithmic function is given for example in section (4.1) in [2]. We are able to switch the order of integration over w + m and x using Fubini's theorem since the integrand is of bounded measure over the space C × [0, ∞).

Infinite sum of the contour integral
In this section we will again use Cauchy's integral formula (3) and taking the infinite sum to derive equivalent sum representations for the contour integrals. For the first equation we replace y by y + t and multiply both sides by e mt . Next we replace t by −t and take the difference of these two equations followed by replacing t by t/c simplifying to get Derivation of some entries in the tables of David Bierens De Haan and Anatolii Prudnikov: An exercise in integration theory Next we replace y by log(a)+ iπ(2y+1) 2c and multiply both sides by i c πcsch(2t)e iπmy c + iπm 2c and take the infinite over y ∈ [0, ∞) and simplify using the Lerch function to get [4] and Im(m + w) > 0 for convergence of the sum.

Definite integral in terms of the Lerch function
Proof. Since the right-hand side of equation (4) is equal to the right-hand side of (8) we can equate the left-hand sides and simplify the factorials to achieve the stated formula.

Main results
6 The Fourier Cosine transform in terms of the Lerch function Proof. Use equation (9) Proof. Use equation (10) and set a = 1 and simplify the left-hand side.
Proof. Use equation (10) take the first partial derivative with respect to t and multiply both sides by − 1 2 csch(2t) and simplify the left-hand side.
8 Derivation of entry BI(356)(16) in [3] Theorem 5. For t ∈ C, Proof. We will form three equations using equation (12) and add to yield the stated result. For the first equation we use equation (12) and set m = 0, c = 1/2, then take the first partial derivative with respect to k and set k = 2, multiply by −3 cos(p) and simplify using entry (4) in Next we derive the second equation by using (12) and setting m = 1, c = 1/2 then take the first partial derivative with respect to k and set k = 2 and multiply by −3 and simplify using entry (4) in Finally we derive the third equation using (12) by setting m = 1, c = 1/2 followed by taking the first partial derivative with respect to k and setting k = 2 and simplify using entry (4) in Add equation (14), (15) and (16) and replace t by ti/2, p by t and simplify. Note the result in equation (4.376.10) in [4] is in error.
9 Derivation of entry BI(356)(17) in [3] Theorem 6. For all k, t ∈ C, Proof. We will derive three equations using (12). To derive first equation we set m = 0, c = 1/2, t = it/2, followed by taking the first partial derivative with respect to k and setting k = k − 1 and multiplying by k cos(t) 2 and using entry (4) in Table below (64:12:7) in [1] and simplify to get Mathematics and Statistics 9(4): 588-602, 2021 593 To derive the second equation we set m = 1, c = 1/2, t = it/2, followed by taking the first partial derivative with respect to k and setting k = k − 1 and multiplying by k 2 and using entry (4) in Table below (64:12:7) in [1] and simplify to get To derive the third equation we set m = 1, c = 1/2t = it/, k = k − 1, followed by taking the first partial derivative with respect to m and multiplying by − k 2 and using entry (4) in Add equations (18), (19) and (20) and simplify.
10 Derivation of entry BI(356)(18) in [3] In this derivation the authors recognized a typo in the denominator of the integrand listed in [3], corrected it and proceeded with the following evaluation.
12 Derivation of entry BI(356)(11) in [4] Theorem 9. For all a ∈ C Proof. Derive the first equation. Use equation (11) and set m = 0, t = ti/2, c = q/2 then set t = π/2 followed by multiplying both sides by 2a+1 2 then taking the first partial derivative with respect to k and setting k = 2a simplify using entry (4) in Table below (64:12:7) in [1] to get Derive the second equation. Use equation (12) and set c = q/2, t = ti/2, m = q then set t = π/2 followed by multiplying both sides by −q/2 then taking the first partial derivative with respect to k and setting k = 2a simplify using entry (4) in Table below (64:12:7) in [1] to get Add equations (27) and (28) and simplify for the stated result. Note: Use L'Hopital's rule to evaluate the right-hand for q ∈ 1 2 Z.