Deﬁnite Integral of Algebraic, Exponential and Hyperbolic Functions Expressed in Terms of Special Functions

: While browsing through the famous book of Bierens de Haan, we came across a table with some very interesting integrals. These integrals also appeared in the book of Gradshteyn and Ryzhik. Derivation of these integrals are not listed in the current literature to best of our knowledge. The derivation of such integrals in the book of Gradshteyn and Ryzhik in terms of closed form solutions is pertinent. We evaluate several of these deﬁnite integrals of the form (cid:82) ∞ 0 ( a + y ) k − ( a − y ) k e by − 1 (cid:82) ∞ dy and dy in terms of a special function where k , a and b are arbitrary complex numbers.

Abstract: While browsing through the famous book of Bierens de Haan, we came across a table with some very interesting integrals. These integrals also appeared in the book of Gradshteyn and Ryzhik. Derivation of these integrals are not listed in the current literature to best of our knowledge. The derivation of such integrals in the book of Gradshteyn and Ryzhik in terms of closed form solutions is pertinent. We evaluate several of these definite integrals of the form

Introduction
We will derive integrals as indicated in the abstract in terms of special functions which by means of analytic continuation gives a greater range to the parameters in the integral. We also derive definite integrals which yield special cases. We also noticed errors in the formula cited in [1] and we were able to derive correct formula for these integrals as motivation for this work. Some special cases of these integrals have been reported in Gradshteyn and Ryzhik [2]. In 1867, David Bierens de Haan derived Table 87 in [1] when a = 1 in the integrals above and we will consider these as well. In our case the variables in the formulas are general complex numbers subject to the restrictions given below. The derivations follow the method used by us in [3]. Generalized Cauchy's integral formula is given by where C will be defined below. This method involves using a form of Equation (1) then multiplying both sides by a function, 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) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same. This method has been used by us in previous work, [3][4][5][6][7].

Definite Integral of the Contour Integral
We use the method in [3]. Here the contour is similar to Figure 2 in [3] where the cut is along the positive imaginary axis and C is taken from positive infinity to the x-axis along the right side of the cut and zero distance from the cut, around the origin on a circle of zero radius and back to positive infinity on the left side of the cut, except we replace the vertical lines ±1 by ±Re(b). In Generalized Cauchy's integral formula we replace y by y + log(a) and −y + log(a) then subtract these two equations, followed by multiplying both sides by 1 sinh(by) to get the logarithmic function is defined in Section (4.1) in [8]. We then take the definite integral over y ∈ [0, ∞) of both sides to get from Equation (2.7.7.6) in [9] and the integral is valid for a, k and b complex and −|Re(b)| < Re(w) < |Re(b)| and Re(b) = 0.
In a similar manner we can derive the second integral formula and multiplying by 1 e by −1 we get We then take the definite integral over y ∈ [0, ∞) of both sides to get from Equation (2.3.3.12) in [9] and the integral is valid for −|Re(b)| < Re(w) < |Re(b)| and Re(b) > 0.
In a slightly different manner we can derive the fourth integral formula by adding the two equations and multiplying by 1 cosh(by) we get We then take the definite integral over y ∈ [0, ∞) of both sides to get from Equation (1.7.7.1) in [9] and the integral is valid for −|Re(b)| < Re(w) < |Re(b)| and Re(b) = 0.

Infinite Sum of the Contour Integral
Again, using the method in [3], replacing y with 2πi(p + 1)/(2b) + log(a) and multiplying both sides by followed by taking the infinite sum of both sides of Equation (10) from (1.232.1) in [2] where tanh(ix) = i tan(x) from (4.5.9) in [8] and Im(w) > 0 for the convergence of the sum and if the Re(k) < 0 then the argument of the sum over p cannot be zero for some value of p. We use (9.521) in [2] where ζ(s, u) is the Hurwitz Zeta function. Similarly, using the method in [3], replacing y with 2πi(p + 1)/b + log(a) and multiplying both sides by 2πi followed by taking the infinite sum of both sides of Equation (12) with respect to p over [0, ∞) to get from Equation (1.232.1) in [2] where ctnh(iw) = −i cot(w), tan(w + π/2) = − cot(w) and where Re(z) < 0, which implies that Im(w) > 0. Similarly, using the method in [3], replacing y with πi(2p + 1)/b + log(a) and multiplying both sides by followed by taking the infinite sum of both sides of Equation (15) with respect to p over [0, ∞) to get e w(πi(2p+1)/b+log(a)) w k+1 a w dw from Equation (1.232.3) in [2] Similarly, using the method in [3], replacing y with πi(2p + 1)/(2b) + log(a) and multiply both sides by (−1) p 2πi b to yield followed by taking the infinite sum of both sides of Equation (17) with respect to p over [0, ∞) to get from Equation (1.232.2) in [2].
To obtain the first contour integral in the last line of Equations (5) and (7) we use the Cauchy formula by replacing y by log(a), k by k + 1, and multiplying both sides by −1 and simplifying we get To obtain the first contour integral in the last line in Equations (11) and (13) we use the Cauchy formula by replacing y by log(a) and multiplying both sides by π ib and simplifying we get Since the right hand-side of Equation (3) is equal to the addition of (11) and (20), we can equate the left hand-sides and simplify to get ∞ 0 (y + log(a)) k − (−y + log(a)) k sinh(by) dy where Re(b) = 0 and k and a are general complex numbers. We can write down an equivalent formula for the corresponding Hurwitz Zeta function for the second integral using Equations (5), (13), (19) and (20), where Re(b) > 0 and k and a are general complex numbers. We can write down an equivalent formula for the corresponding Hurwitz Zeta function for the second integral using Equations (7), (16) and (19), ∞ 0 (y + log(a)) k − (−y + log(a)) k e by + 1 where Re(b) > 0 and k and a are general complex numbers. We can write down an equivalent formula for the corresponding Hurwitz Zeta function for the second integral using Equations (9) and (18), ∞ 0 (y + log(a)) k + (−y + log(a)) k cosh(by) dy where Re(b) = 0 and k and a are general complex numbers.

Derivation of Integrals
In this section, we derive the entries of Table 87 [1] in terms of the Hurwitz zeta and zeta functions which will analytically continue these integrals. The integrals in the Table can be achieved by using the Equation (12.11.17) in [10] given by 3.1. Using Equation (23) and Setting a = e i , b = π and Replacing k with 2k We Get from Equation (9.534) in [2]. This result is Equation (1) Table 87 in [1] where this result is in error. (23) and Setting a = e i , b = π and Replacing k with 2k − 1 We Get

Using Equation
from Equation (9.534) in [2]. This result is Equation (2) Table 87 in [1] where this result is in error. (22) and Setting a = e i , b = π and Replacing k with 2k − 1 We Get

Using Equation
from Equation (9.534) in [2]. This result is Equation (3) Table 87 in [1] where this result is in error. (22) and Setting a = e i , b = 2π and Replacing k with 2k We Get

Using Equation
from Equation (9.534) in [2]. This result is Equation (4) Table 87 in [1] where this result is in error. (22) and Setting a = e i , b = 2π and Replacing k with 2k − 1 We Get

Using Equation
from Equation (9.534) in [2]. This result is Equation (5) Table 87 in [1] where this result is in error.
3.6. Using Equation (24) and Setting a = e i , b = π/2 and Replacing k with 2k − 1 We Get from Equation (9.534) in [2]. This result is Equation (6) Table 87 in [1] where this result is in error. (21) and Setting a = e i , b = π/2 and Replacing k with 2k We Get

Using Equation
This result is Equation (7) Table 87 in [1] where this result is in error. (21) and Setting a = e i , b = π/2 and Replacing k with 2k − 1 We Get

Using Equation
This result is Equation (8) Table 87 in [1] where this result is in error.

Generalizations and Table of Integrals
In this section we summarized the integrals evaluated in this work in the form of a table (see Table 1).

Special Cases of the Definite Integrals
In this ection we will look at a few of the integrals derived in this work and evaluate special cases in terms of the log-gamma function and π.

When a Is Replaced by e mi
We take the first derivative of Equation (24) with respect to k then set k = 0, factorize and simplify the the log terms to get ∞ 0 log m 2 + y 2 sech(by)dy from (3.10) in [11] where Re(b) > 0. This is the same as (4.373.1) in [2].

When a Is Replaced by e ni
We take the first derivative of Equation (21) with respect to k then set k = 0 to get from (3.10) in [11] where Re(b) > 0.

Summary
In this article we derived the definite logarithmic algebraic functions in terms of the Hurwitz Zeta function. We were able to produce a closed form solution for integrals tabled in Bierens de Haan [1,2] not previously derived. 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.
Author Contributions: Writing-review and editing, original draft preparation, conceptualization, R.R.; Writing-review and editing, supervision, funding acquisition, A.S. All authors have read and agreed to the published version of the manuscript.