Abstract

In this manuscript, the authors derive closed formula for definite integrals of combinations of powers and logarithmic functions of complicated arguments and express these integrals in terms of the Hurwitz zeta functions. These derivations are then expressed in terms of fundamental constants, elementary, and special functions. A summary of the results is produced in the form of a table of definite integrals for easy referencing by readers.

1. Introduction

In this manuscript, the authors derive the definite integrals given byandin terms of the Hurwitz zeta function, where the parameters , , , and are general complex numbers. A summary of the results is given in a table of integrals for easy reading. This work is important because the authors were unable to find similar results in current literature. Tables of definite integrals provide a useful summary and reference for readers seeking such integrals for potential use in their research. This work looks at definite integrals of the hyperbolic arctangent function and the product of logarithmic functions with complicated arguments and powers. We use our simultaneous contour integration method to aid in our derivations of the closed forms solutions in terms of the Hurwitz zeta function, which provides analytic continuation of the results.

The derivations follow the method used by us in [1]. The generalized Cauchy’s integral formula is given bywhere 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 multiplying both sides by a function, and then taking 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 different function and summing performing some substitutions so that the contour integrals are the same.

2. Derivation of the First Contour Integral

We use the method in [1]. Using a generalization of Cauchy’s integral formula equation (3), we will form two equations and add them together. For the first and second equations, replace by and by , respectively. Next, we add these equations followed by multiplying both sides by and taking the definite integral over to getfrom equation (2.6.19.6) in [2] and the integral is valid for , , and complex and where the logarithmic function is defined in equation (4.1.2) in [3].

3. Derivation of the Second Contour Integral

Using a generalization of Cauchy’s integral formula equation (3), we will form two equations and add them together. For the first and second equations, replace by and by , respectively. Next, we add these equations followed by multiplying both sides by and taking the definite integral over to getfrom equation (2.6.19.6) in [2] where .

4. Derivation of the Infinite Sum of the First Contour Integral

Again, using the method in [1] and equation (3), we replace by , multiply both sides by , replace by , and take the infinite sum of both sides over simplifying in terms of the Hurwitz zeta function to getfrom equation (1.232.3) in [4] where from (4.5.10) in [3] and for the convergence of the sum. We use equation (9.521.1) in [4], where is the Hurwitz zeta function.

5. Derivation of the Infinite Sum of the Second Contour Integral

Again, using the method in [1] and equation (3), we replace by , multiply both sides by , replace by , and take the infinite sum of both sides over simplifying in terms of the Hurwitz zeta function to getfrom equation (1.232.1) in [5].

6. Derivation of the Additional Contours

Again, using the method in [1] and equation (3), we replace by , by , multiply both sides by , and simplify to get

Again, using the method in [1] and equation (3), we replace by , by and multiply both sides by simplify to get

Again, using the method in [1] and equation (3), we replace by , by , multiply both sides by , and simplify to get

7. Derivation of the Definite Integrals in Terms of the Hurwitz Zeta Function

Since the right-hand side of equations (4) and (5) are equal to the sum of the right-hand sides of equations (6)–(10), we can equate the left-hand sides simplifying the factorials to get

8. Derivation of Logarithmic and Hyperbolic Tangent Integrals in Terms of the Hurwitz Zeta Function

Using equations (11) and (12) and taking their difference simplifying, we get

Using equations (11) and (12) and adding them and then simplifying, we get

9. Derivation of Logarithmic and Hyperbolic Arctangent Integrals in Terms of the Zeta Function

Using equations (13) and (14) and setting simplifying, we getfrom entry (2) in the table below (64 : 7) in [5].

10. Derivation of Logarithmic and Hyperbolic Arctangent Integrals in Terms of the Log Gamma Function

Using equations (13) and (14), replacing by , applying L’Hopital’s rule to the right-hand side as , respectively, and simplifying, we getfrom equation (64 : 10 : 2) in [5].

11. Derivation of Logarithmic and Hyperbolic Arctangent Integrals in Terms of the Digamma Function

Using equations (13) and (14), replacing by , applying L’Hopital’s rule to the right-hand side as , respectively, and simplifying, we getfrom equation (64 : 4:2) in [5].

12. Derivation of Logarithmic and Hyperbolic Arctangent Integrals in Terms of Fundamental Constants and Special Functions

In this section, we will derive definite integrals in terms of special functions and fundamental constants such as Euler’s constant (), Catalan’s constant (), Glaisher’s constant (), and . This section showcases just a subset of the range of evaluations of these integral formulae.

12.1. Hyperbolic Tangent Integrals

Example 1. Using equation (13), replacing by , setting , and simplifying, we getfrom equations (23.2.23) in [3] and (64 : 7:1) in [5].

Example 2. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom equation (A.11) in [6].

Example 3. Using equation (16), setting , and simplifying, we get

Example 4. Using equation (16), setting , and simplifying, we get

Example 5. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom Example 1 in Section 12.1 in [7].

Example 6. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom Example 1 in Section 12.1 in [4].

12.2. Logarithmic Integrals

Example 7. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom Example 1 in Section 12.1 in [4].

Example 8. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom equation (A.11) in [4].

Example 9. Using equation (16), setting , and simplifying, we get

Example 10. Using equation (16), setting , and simplifying, we get

Example 11. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom Example 1 in Section 12.1 in [5].

Example 12. Using equation (16), taking the first partial derivative with respect to , setting , and simplifying, we getfrom Example 1 in Section 12.1 in [5].

13. Derivation of Definite Integrals of the Logarithmic Function

Using equation (11), we take the first partial derivative with respect to and then replace by . Next, we form a second equation by replacing by in the new equation. Then, we take the difference of these two new equations simplifying to get

Repeating the steps above using equation (12) and simplifying, we get

13.1. Some Special Cases

Example 13. Using equation (31) and applying L’Hopital’s rule to the right-hand side as and simplifying, we get

Example 14. Using equation (32) and setting and simplifying, we get