# Definite Integrals Involving Combinations of Powers and Logarithmic Functions of Complicated Arguments Expressed in Terms of the Hurwitz Zeta Function

Authors: Robert Reynolds; Allan Stauffer
Journal: Journal of Mathematics
 (2021)

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## 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.


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## Body

## 1. Introduction

In this manuscript, the authors derive the definite integrals given by(1)∫01logxm+1logka/x+logkaxxdxand(2)∫01log1−xnlogka/x+logkaxxdx,in terms of the Hurwitz zeta function, where the parameters k, a, m, and n 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 by(3)ykk!=12πi∫Cewywk+1dw,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 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 y by logax and y by loga/x, respectively. Next, we add these equations followed by multiplying both sides by logxm+1/x and taking the definite integral over x∈0,1 to get(4)1k!∫01logxm+1logka/x+logkaxxdx=12πi∫01∫Caww−k−1x−w−1x2w+1logxm+1dwdx=12πi∫C∫01aww−k−1x−w−1x2w+1logxm+1dxdw=12πi∫Caww−k−3πwcscπwm−mdw,from equation (2.6.19.6) in [2] and the integral is valid for a, m, and k complex and −1<Rew<0 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 y by logax and y by loga/x, respectively. Next, we add these equations followed by multiplying both sides by log1−xn/x and taking the definite integral over x∈0,1 to get(5)1k!∫01log1−xnlogka/x+logkaxxdx=12πi∫01∫Caww−k−1x−w−1x2w+1log1−xndwdx=12πi∫C∫01aww−k−1x−w−1x2w+1log1−xndxdw=12πi∫Caww−k−3πwcotπwn−ndw,from equation (2.6.19.6) in [2] where −1<Rew<0.


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

Again, using the method in [1] and equation (3), we replace y by loga+iπ2y+1/m, multiply both sides by −2πi, replace k by k+1, and take the infinite sum of both sides over y∈0,∞ simplifying in terms of the Hurwitz zeta function to get(6)−i2πk+2i/mk+1ζ−k−1,π−imloga/2πk+1!=−12πi∑y=0∞∫C2iπw−k−1ewloga+iπ2y+1/mdw=−12πi∫C∑y=0∞2iπw−k−1ewloga+iπ2y+1/mdw=12πi∫Cπaww−k−2cscπwmdw,from equation (1.232.3) in [4] where cschix=icscx from (4.5.10) in [3] and Imw>0 for the convergence of the sum. We use equation (9.521.1) in [4], where ζs,u 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 y by loga+2iπy+1/n, multiply both sides by −2πi, replace k by k+1, and take the infinite sum of both sides over y∈0,∞ simplifying in terms of the Hurwitz zeta function to get(7)−i2πk+2i/nk+1ζ−k−1,1−inloga/2πk+1!=−12πi∑y=0∞∫C2iπw−k−1ewloga+2iπy+1/ndw=−12πi∫C∑y=0∞2iπw−k−1ewloga+2iπy+1/ndw=12πi∫Cπaww−k−2cotπwn+iπaww−k−2dw,from equation (1.232.1) in [5].


## 6. Derivation of the Additional Contours

Again, using the method in [1] and equation (3), we replace y by loga, k by k+1, multiply both sides by πi, and simplify to get(8)iπlogk+1ak+1!=12πi∫Ciπaww−k−2dw.Again, using the method in [1] and equation (3), we replace y by loga, k by k+2 and multiply both sides by −n simplify to get(9)−nlogk+2ak+2!=−12πi∫Cnaww−k−3dw.Again, using the method in [1] and equation (3), we replace y by loga, k by k+2, multiply both sides by −m, and simplify to get(10)−mlogk+2ak+2!=−12πi∫Cmaww−k−3dw.


## 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(11)∫01logxm+1logka/x+logkaxxdx=−mlogk+2ak+1k+2−i2πk+2i/mk+1ζ−k−1,π−imloga/2πk+1,(12)∫01log1−xnlogka/x+logkaxxdx=−i2πk+2i/nk+1ζ−k−1,1−inloga/2πk+1−nlogk+2ak+1k+2−iπlogk+1ak+1.


## 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(13)∫01tanh−1xmlogka/x+logkaxxdx=2πk+2i/mk2k+1mζ−k−1,π−imloga2π−ζ−k−1,1−imloga2π+iπmlogk+1a.Using equations (11) and (12) and adding them and then simplifying, we get(14)∫01logka/x+logkaxlogxm+11−xnxdx=4π2k+1k+2mn2πkk+2k+1πknimkζ−k−1,π−imloga2π+minkζ−k−1,1−inloga2π−mnlogk+1alogam+n+iπk+2.


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

Using equations (13) and (14) and setting a=1 simplifying, we get(15)∫01tanh−1xlogkxxdx=2−k−22k+2−1eiπkζk+2Γk+1,(16)∫01logkxlogxm+11−xnxdx=12eiπkζk+2Γk+12−2−km−k−1−2n−k−1,from 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 a by eai, applying L’Hopital’s rule to the right-hand side as k⟶−1, respectively, and simplifying, we get(17)∫01tanh−1xmxa2+log2xdx=π4alogamΓam/2π22πΓam+π/2π2,∫01logxm+11−xnxa2+log2xdx=12a−2πlogΓam+π2πΓan2π+1−am+amlogia−amlog2iπm−an+anlogia−anlog2iπn+πlogiπa−πlogi2n,from 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 a by eai, applying L’Hopital’s rule to the right-hand side as k⟶−2, respectively, and simplifying, we get(18)∫01a2−log2xtanh−1xmxa2+log2x2dx=14−mψ0am2π+1+mψ0am+π2π+πa,∫01a2−log2xlog−xm−1xn−1xa2+log2x2dx=12a−amlogia+amlogim+amlog2π+amψ0am+π2π−anlogia+anlogin+anlog2π+anψ0an2π+1−π,from 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 (γ), Catalan’s constant (C), Glaisher’s constant (A), and π. 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 a by e−iπ/2m, setting k=−3,m=1, and simplifying, we get(19)∫01π3−12πlog2xtanh−1xx4log2x+π23dx=2C−116π,from 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 k, setting k=0, and simplifying, we get(20)∫01loglogxtanh−1xxdx=124π2log16π3A36+3iπ,from equation (A.11) in [6].Example 3.
Using equation (16), setting k=1/2, and simplifying, we get(21)∫01logxtanh−1xxdx=−116i2−8πζ52.Example 4.
Using equation (16), setting k=−1/2, and simplifying, we get(22)∫01tanh−1xxlogxdx=14i2−4πζ32.Example 5.
Using equation (16), taking the first partial derivative with respect to k, setting k=1, and simplifying, we get(23)∫01logxloglogxtanh−1xxdx=18−7ζ′3+ζ3−7+7γ−7iπ−log2,from Example 1 in Section 12.1 in [7].Example 6.
Using equation (16), taking the first partial derivative with respect to k, setting k=2, and simplifying, we get(24)∫01log2xloglogxtanh−1xxdx=15ζ′48+π445−30γ+30iπ+log41440,from 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 k, setting k=0,a=−1,m=2,n=1, and simplifying, we get(25)∫01log1−xx2+1loglogxxdx=18−6ζ′2+γ−iππ2,from Example 1 in Section 12.1 in [4].Example 8.
Using equation (16), taking the first partial derivative with respect to k, setting k=0,m=n=1, and simplifying, we get(26)∫01log1−x2loglogxxdx=112π2logA12π−iπ,from equation (A.11) in [4].Example 9.
Using equation (16), setting k=1/2,m=n=1, and simplifying, we get(27)∫01logxlog1−x2xdx=−14iπ2ζ52.Example 10.
Using equation (16), setting k=−1/2,m=n=1, and simplifying, we get(28)∫01log1−x2xlogxdx=iπ2ζ32.Example 11.
Using equation (16), taking the first partial derivative with respect to k, setting k=m=n=1, and simplifying, we get(29)∫01logxlog1−x2loglogxxdx=14ζ′3+ζ31−γ+iπ−log2,from Example 1 in Section 12.1 in [5].Example 12.
Using equation (16), taking the first partial derivative with respect to k, setting k=−1/2,m=n=1, and simplifying, we get(30)∫01log1−x2loglogxxlogxdx=π2iζ′32−iζ32γ−iπ+log8,from Example 1 in Section 12.1 in [5].


## 12.1. Hyperbolic Tangent Integrals

Example 1.
Using equation (13), replacing a by e−iπ/2m, setting k=−3,m=1, and simplifying, we get(19)∫01π3−12πlog2xtanh−1xx4log2x+π23dx=2C−116π,from 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 k, setting k=0, and simplifying, we get(20)∫01loglogxtanh−1xxdx=124π2log16π3A36+3iπ,from equation (A.11) in [6].Example 3.
Using equation (16), setting k=1/2, and simplifying, we get(21)∫01logxtanh−1xxdx=−116i2−8πζ52.Example 4.
Using equation (16), setting k=−1/2, and simplifying, we get(22)∫01tanh−1xxlogxdx=14i2−4πζ32.Example 5.
Using equation (16), taking the first partial derivative with respect to k, setting k=1, and simplifying, we get(23)∫01logxloglogxtanh−1xxdx=18−7ζ′3+ζ3−7+7γ−7iπ−log2,from Example 1 in Section 12.1 in [7].Example 6.
Using equation (16), taking the first partial derivative with respect to k, setting k=2, and simplifying, we get(24)∫01log2xloglogxtanh−1xxdx=15ζ′48+π445−30γ+30iπ+log41440,from 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 k, setting k=0,a=−1,m=2,n=1, and simplifying, we get(25)∫01log1−xx2+1loglogxxdx=18−6ζ′2+γ−iππ2,from Example 1 in Section 12.1 in [4].Example 8.
Using equation (16), taking the first partial derivative with respect to k, setting k=0,m=n=1, and simplifying, we get(26)∫01log1−x2loglogxxdx=112π2logA12π−iπ,from equation (A.11) in [4].Example 9.
Using equation (16), setting k=1/2,m=n=1, and simplifying, we get(27)∫01logxlog1−x2xdx=−14iπ2ζ52.Example 10.
Using equation (16), setting k=−1/2,m=n=1, and simplifying, we get(28)∫01log1−x2xlogxdx=iπ2ζ32.Example 11.
Using equation (16), taking the first partial derivative with respect to k, setting k=m=n=1, and simplifying, we get(29)∫01logxlog1−x2loglogxxdx=14ζ′3+ζ31−γ+iπ−log2,from Example 1 in Section 12.1 in [5].Example 12.
Using equation (16), taking the first partial derivative with respect to k, setting k=−1/2,m=n=1, and simplifying, we get(30)∫01log1−x2loglogxxlogxdx=π2iζ′32−iζ32γ−iπ+log8,from 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 m and then replace m by m+1. Next, we form a second equation by replacing m by p in the new equation. Then, we take the difference of these two new equations simplifying to get(31)∫01logk+1xxm−xpxm+1+1xp+1+1dx=−2−k−12k+1−1eiπk/2ζk+2Γ⋅k+2i/m+1km+12−i/p+1kp+12.Repeating the steps above using equation (12) and simplifying, we get(32)∫01logk+1xxn−xpxn+1−1xp+1−1dx=eiπk/2ζk+2Γk+2i/p+1kp+12−i/n+1kn+12.


### 13.1. Some Special Cases

Example 13.
Using equation (31) and applying L’Hopital’s rule to the right-hand side as k⟶−2 and simplifying, we get(33)∫01xm−xpxm+1+1xp+1+1logxdx=12logm+1p+1.Example 14.
Using equation (32) and setting k=1 and simplifying, we get(34)∫01log2xxn−xpxn+1−1xp+1−1dx=2ζ31n+13−1p+13.


## 13.1. Some Special Cases

Example 13.
Using equation (31) and applying L’Hopital’s rule to the right-hand side as k⟶−2 and simplifying, we get(33)∫01xm−xpxm+1+1xp+1+1logxdx=12logm+1p+1.Example 14.
Using equation (32) and setting k=1 and simplifying, we get(34)∫01log2xxn−xpxn+1−1xp+1−1dx=2ζ31n+13−1p+13.


## 14. Table of Integrals

Table of definite integrals provided in Table1Table 1
Table of definite integrals.
fx∫01fxdxtanh−1xlogkx/x2−k−22k+2−1eiπkζk+2Γk+1logkxlogxm+11−xn/x1/2eiπkζk+2Γk+12−2−km−k−1−2n−k−1π3−12πlog2xtanh−1x/x4log2x+π232C−1/16πtanh−1xm/xa2+log2xπ/4alogamΓam/2π2/2πΓam+π/2π2loglogxtanh−1x/x1/24π2log16π3/A36+3iπlogxtanh−1x/x−1/16i2−8πζ5/2tanh−1x/xlogx1/4i2−4πζ3/2logxloglogxtanh−1x/x1/8−7ζ′3+ζ3−7+7γ−7iπ−log2log2xloglogxtanh−1x/x15ζ′4/8+π445−30γ+30iπ+log4/1440log1−xx2+1loglogx/x1/8−6ζ′2+γ−iππ2log1−x2loglogx/x1/12π2logA12/π−iπlogxlog1−x2/x−1/4iπ/2ζ5/2log1−x2/xlogxiπ/2ζ3/2logxlog1−x2loglogx/x1/4ζ′3+ζ31−γ+iπ−log2log1−x2loglogx/xlogxπ/2iζ′3/2−iζ3/2γ−iπ+log8logk+1xxm−xp/xm+1+1xp+1+1−2−k−12k+1−1eiπk/2ζk+2Γk+2i/m+1k/m+12−i/p+1k/p+12xm−xp/xm+1+1xp+1+1logx1/2logm+1/p+1logk+1xxn−xp/xn+1−1xp+1−1eiπk/2ζk+2Γk+2i/p+1k/p+12−i/n+1k/n+12


## 15. Discussion

In this work, the authors looked at deriving definite integrals of combinations of logarithmic functions of complicated arguments and powers and expressed them in terms of the Hurwitz zeta function. One of the interesting properties of these integrals is that, by adding them, we were able to get the integral of the product of the hyperbolic arctangent function and the logarithmic function. The authors formally derived a few integrals in terms of fundamental constants and special functions. One of our goals was to supply a table of definite integrals (Table1) for easy reading by researchers and to have these results added to existing textbooks.The results presented were numerically verified for both real and imaginary values of the parameters in the integrals using Mathematica by Wolfram. We considered various ranges of these parameters for real, integer, negative, and positive values. We compared the evaluation of the definite integral to the evaluated special function and ensured agreement.


## 16. Conclusion

In this paper, the authors used our method to evaluate definite integrals using the Hurwitz zeta function. The contour we used was specific to solving integral representations in terms of the Hurwitz zeta function. The author expects that other contours and integrals can be derived using this method.



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