# The Mellin Transform of Logarithmic and Rational Quotient Function in terms of the Lerch Function

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

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## Abstract

Upon reading the famous book on integral transforms volume II by Erdeyli et al., we encounter a formula which we use to derive a Mellin transform given by∫0∞xm−1logkax/β2+x2γ+xdx, where the parameters a,k,β, and γ are general complex numbers. This Mellin transform will be derived in terms of the Lerch function and is not listed in current literature to the best of our knowledge. We will use this transform to create a table of definite integrals which can be used to extend similar tables in current books featuring such formulae.


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## Body

## 1. Introduction

The well-known book on Mellin transforms by Brychkov et al. [1] contains a vast number of integral formulae. In this work, we focus on adding to section (2.5.1) featuring “logarithmic and algebraic functions.” In particular, this work will derive a closed form solution for the integral given by(1)∫0∞xm−1logkaxβ2+x2γ+xdx,in terms of the Lerch function. This is achieved by applying our contour integration method to an integral transform from Erdeyli’s book [2] to yield a definite integral in terms of the Lerch function. We have also produced other formulae in terms of the Lerch function using our method in [3]. We use this new integral transform in terms of the Lerch function to derive new formulae for Brychkov’s book [1] thereby extending the current tables in section (2.5.1). We will also derive new Malmsten-type loglogx integrals in terms of special functions and provide new formula also extending the work done by Malmsten [4]. The derivations of the formulae in this work follow the method used by us in [5]. This method involves using the generalized Cauchy’s integral formula given by(2)ykk!=12πi∫Cewywk+1dy,and multiplying both sides of (2) by a function, followed by taking a definite integral on both sides which yields a definite integral in terms of a contour integral. Then, we multiply both sides of equation (2) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same. Here, 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.


## 2. Definite Integral of the Contour Integral

We use the method in [5]. 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<Reα<u. 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 logax followed by multiplying both sides by(3)xm−1β2+x2γ+x,and then taking the definite integral with respect to x∈0,∞ to obtain(4)1k!∫0∞xmlogkaxβ2+x2γ+xdx=12πi∫0∞∫Caww−k−1xm+wβ2+x2γ+xdwdx=12πi∫C∫0∞aww−k−1xm+wβ2+x2γ+xdxdw=12πi∫Cπaww−k−12β2+γ2γβm+w−1sec12πm+w+βm+wcsc12πm+w−2γm+wcscπm+wdw,from equation (3.263) in [6], where −1<Rem+w<2 and Reβ>0,argγ<π. We are able to switch the order of integration over w and x using Fubini’s theorem since the integrand is of bounded measure over the space ℂ×0,∞.


## 3. The Lerch Function

The Hurwitz–Lerch function [7] has a series representation given by(5)Φz,s,v=∑n=0∞v+n−szn,where v∈ℂ∖ℤ0−; s∈ℂ when z<1;Res>1 when z=1 and is continued analytically by its integral representation given by(6)Φz,s,v=1Γs∫0∞ts−1e−vt1−ze−tdt=1Γs∫0∞ts−1e−v−1tet−zdt,where Rev>0 and either z≦1,z≠1,Res>0; z=1,Res>1.


## 4. Infinite Sum of the Contour Integrals

### 4.1. Derivation of the First Contour Integral

In this section, we will derive the Lerch function representation for the first contour integral given by(7)12πi∫Cπaww−k−1βm+wcsc1/2πm+w2β2+γ2dw.Using equation (2) and replacing y by loga+logβ+1/2iπ2y+1 and then multiplying both sides by(8)−iπβme1/2iπm2y+1β2+γ2,we obtain(9)−ii/2kπk+1βmeiπmy+iπm/2k!β2+γ2−2ilogaπ−2ilogβπ+2y+1k=−12πi∫Ciπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(10)−iπk+1eiπm/2βmΦeimπ,−k,−2iloga−2ilogβ+π/2πk!β2+γ2=−12πi∑y=0∞∫Ciπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=−12πi∫C∑y=0∞iπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=12πi∫Cπaww−k−1βm+wcsc1/2πm+w2β2+γ2dw,from (1.232.3) in [6], where Imm+w>0 in order for the sum to converge, and cschix=−icscx from (4.5.10) in [8].


### 4.2. Derivation of the Second Contour Integral

In this section, we will derive the Lerch function representation for the second contour integral given by(11)−12πi∫Cπaww−k−1γm+wcscπm+wβ2+γ2dw.Using equation (2) and replacing y by loga+logγ+iπ2y+1 and then multiplying both sides by(12)2iπγmeiπm2y+1β2+γ2,we obtain(13)2iikπk+1γme2iπmy+iπmk!β2+γ2−ilogaπ−ilogγπ+2y+1k=−12πi∫C2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(14)2iπk+1eiπmγmΦe2imπ,−k,−iloga−ilogγ+π/2πk!β2+γ2=−12πi∑y=0∞∫C2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw=−12πi∫C∑y=0∞2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw=−12πi∫Cπaww−k−1γm+wcscπm+wβ2+γ2dw,from (1.232.3) in [6], where Imm+w>0 in order for the sum to converge, and cschix=−icscx from (4.5.10) in [8].


### 4.3. Derivation of the Third Contour Integral

In this section, we will derive the Lerch function representation for the third contour integral given by(15)−12πi∫Cπγaww−k−1βm+w−1sec1/2πm+w2β2+γ2dw.Using equation (2) and replacing y by loga+logβ+1/2iπ2y+1 and then multiplying both sides by(16)πγ−1yβm−1e1/2iπm2y+1β2+γ2,we obtain(17)γi/2kπk+1−1yβm−1eiπmy+iπm/2k!β2+γ2−2ilogaπ−2ilogβπ+2y+1k=12πi∫Cπγ−1yw−k−1βm−1β2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(18)γπk+1e1/2iπk+mβm−1k!β2+γ2Φ−eimπ,−k,−2iloga−2ilogβ+π/2π=12πi∑y=0∞∫Cπγ−1yw−k−1βm−1β2+γ2expwloga+logβ+1/2iπ2y+1+12iπm2y+1dw=12πi∫C∑y=0∞πγ−1yw−k−1βm−1β2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=12πi∫Cπγaww−k−1βm+w−1sec1/2πm+w2β2+γ2dw,from (1.232.2) in [6], where Imm+w>0 in order for the sum to converge, and sechix=isecx from (4.5.11) in [8].


## 4.1. Derivation of the First Contour Integral

In this section, we will derive the Lerch function representation for the first contour integral given by(7)12πi∫Cπaww−k−1βm+wcsc1/2πm+w2β2+γ2dw.Using equation (2) and replacing y by loga+logβ+1/2iπ2y+1 and then multiplying both sides by(8)−iπβme1/2iπm2y+1β2+γ2,we obtain(9)−ii/2kπk+1βmeiπmy+iπm/2k!β2+γ2−2ilogaπ−2ilogβπ+2y+1k=−12πi∫Ciπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(10)−iπk+1eiπm/2βmΦeimπ,−k,−2iloga−2ilogβ+π/2πk!β2+γ2=−12πi∑y=0∞∫Ciπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=−12πi∫C∑y=0∞iπw−k−1βmβ2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=12πi∫Cπaww−k−1βm+wcsc1/2πm+w2β2+γ2dw,from (1.232.3) in [6], where Imm+w>0 in order for the sum to converge, and cschix=−icscx from (4.5.10) in [8].


## 4.2. Derivation of the Second Contour Integral

In this section, we will derive the Lerch function representation for the second contour integral given by(11)−12πi∫Cπaww−k−1γm+wcscπm+wβ2+γ2dw.Using equation (2) and replacing y by loga+logγ+iπ2y+1 and then multiplying both sides by(12)2iπγmeiπm2y+1β2+γ2,we obtain(13)2iikπk+1γme2iπmy+iπmk!β2+γ2−ilogaπ−ilogγπ+2y+1k=−12πi∫C2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(14)2iπk+1eiπmγmΦe2imπ,−k,−iloga−ilogγ+π/2πk!β2+γ2=−12πi∑y=0∞∫C2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw=−12πi∫C∑y=0∞2iπw−k−1γmβ2+γ2expwloga+logγ+iπ2y+1+iπm2y+1dw=−12πi∫Cπaww−k−1γm+wcscπm+wβ2+γ2dw,from (1.232.3) in [6], where Imm+w>0 in order for the sum to converge, and cschix=−icscx from (4.5.10) in [8].


## 4.3. Derivation of the Third Contour Integral

In this section, we will derive the Lerch function representation for the third contour integral given by(15)−12πi∫Cπγaww−k−1βm+w−1sec1/2πm+w2β2+γ2dw.Using equation (2) and replacing y by loga+logβ+1/2iπ2y+1 and then multiplying both sides by(16)πγ−1yβm−1e1/2iπm2y+1β2+γ2,we obtain(17)γi/2kπk+1−1yβm−1eiπmy+iπm/2k!β2+γ2−2ilogaπ−2ilogβπ+2y+1k=12πi∫Cπγ−1yw−k−1βm−1β2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw.We then take the infinite sum overy∈0,∞ simplifying in terms of the Lerch function to obtain(18)γπk+1e1/2iπk+mβm−1k!β2+γ2Φ−eimπ,−k,−2iloga−2ilogβ+π/2π=12πi∑y=0∞∫Cπγ−1yw−k−1βm−1β2+γ2expwloga+logβ+1/2iπ2y+1+12iπm2y+1dw=12πi∫C∑y=0∞πγ−1yw−k−1βm−1β2+γ2expwloga+logβ+12iπ2y+1+12iπm2y+1dw=12πi∫Cπγaww−k−1βm+w−1sec1/2πm+w2β2+γ2dw,from (1.232.2) in [6], where Imm+w>0 in order for the sum to converge, and sechix=isecx from (4.5.11) in [8].


## 5. Definite Integral in terms of the Lerch Function

Theorem 1.
Fork,a,m∈ℂ,Reβ>0,argγ<π,(19)∫0∞xmlogkaxβ2+x2γ+xdx=γπk+1e1/2iπk+mβm−1Φ−eimπ,−k,−2iloga−2ilogβ+π/2πβ2+γ2−iπk+1eiπm/2βmΦeimπ,−k,−2iloga−2ilogβ+π/2πβ2+γ2+2iπk+1eiπmγmΦe2imπ,−k,−iloga−ilogγ+π/2πβ2+γ2.Proof.
Since the right-hand side of equation (4) is equal to the sum of equations (10), (13), and (18), we may equate the left-hand sides to get the stated result.Note whenm is replaced by m−1, we get the Mellin transform representation.


## 6. Derivation of an Entry in Gradshteyn and Ryzhik

In this section, we will derive equation (3.263) in [6]. Using equation (19) and setting k=0 and simplifying, we obtain(20)∫0∞xmβ2+x2γ+xdx=πcscπmβmβcosπm/2+γsinπm/2−βγmββ2+γ2,from entry (2) below Table (64 : 12 : 7) in [9].


## 7. Definite Integral in terms of the Hypergeometric Function

In this section, we will look at derivations in terms of the Hypergeometric function which can be derived from the Lerch function and apply it to get special cases. We will make use of equation (9.559) in [6] and Table 18-1 in [9] for evaluation of specific values of the parameters.Using equation (19) and setting k=1 and simplifying, we obtain(21)∫0∞xmβ2+x2γ+xlogaxdx=−iγeiπm/2βm−1Φ−eimπ,1,−2iloga−2ilogβ+π/2πβ2+γ2−eiπm/2βmΦeimπ,1,−2iloga−2ilogβ+π/2πβ2+γ2+eiπmγmΦe2imπ,1,−iloga−ilogγ+π/2πβ2+γ2.


### 7.1. Entry H1

Using (21) and setting a=−1, β=1, γ=1, and m=1/2 and simplifying, we obtain(22)∫0∞xx+1x2+1log−xdx=12ilog6+42−i2.Next, we rationalize the denominator on the left-hand side and simplify to obtain(23)∫0∞πxx+1x2+1log2x+π2dx=2−12log6+42,∫0∞xlogxx+1x2+1log2x+π2dx=0.


### 7.2. Entry H2

Using equation (21), setting β=γ=1, formulating a second equation by replacing m by p, and taking the difference and simplifying, we obtain(24)∫0∞xm−xpx+1x2+1logxdx=12logtanπm/4+12cos2πp/4sinπp/2+1.


### 7.3. Entry H3

Using equation (21), taking the first partial derivative with respect to m, and setting β=γ=1, a=eπi, and m=−1/2 simplifying, we obtain(25)∫0∞logxxx+1x2+1log2x+π2dx=2π−π2/2π,∫0∞log2xxx+1x2+1log2x+π2dx=π2−12πlog2,after rationalizing the real and imaginary parts.


### 7.4. Entry H4

Using equation (21), taking the first partial derivative with respect to m, and setting β=γ=1, a=eπi/2, and m=−1/2 simplifying, we obtain(26)∫0∞log2xxx+1x2+14log2x+π2dx=164π8−22π+2log17+122,∫0∞logxxx+1x2+14log2x+π2dx=log4−π162,after rationalizing the real and imaginary parts.


## 7.1. Entry H1

Using (21) and setting a=−1, β=1, γ=1, and m=1/2 and simplifying, we obtain(22)∫0∞xx+1x2+1log−xdx=12ilog6+42−i2.Next, we rationalize the denominator on the left-hand side and simplify to obtain(23)∫0∞πxx+1x2+1log2x+π2dx=2−12log6+42,∫0∞xlogxx+1x2+1log2x+π2dx=0.


## 7.2. Entry H2

Using equation (21), setting β=γ=1, formulating a second equation by replacing m by p, and taking the difference and simplifying, we obtain(24)∫0∞xm−xpx+1x2+1logxdx=12logtanπm/4+12cos2πp/4sinπp/2+1.


## 7.3. Entry H3

Using equation (21), taking the first partial derivative with respect to m, and setting β=γ=1, a=eπi, and m=−1/2 simplifying, we obtain(25)∫0∞logxxx+1x2+1log2x+π2dx=2π−π2/2π,∫0∞log2xxx+1x2+1log2x+π2dx=π2−12πlog2,after rationalizing the real and imaginary parts.


## 7.4. Entry H4

Using equation (21), taking the first partial derivative with respect to m, and setting β=γ=1, a=eπi/2, and m=−1/2 simplifying, we obtain(26)∫0∞log2xxx+1x2+14log2x+π2dx=164π8−22π+2log17+122,∫0∞logxxx+1x2+14log2x+π2dx=log4−π162,after rationalizing the real and imaginary parts.


## 8. Derivation of New Malmsten-Type Integrals over the Domainx∈0,∞

### 8.1. Definite Integrals ofloglogx with a Cubic Denominator

In this section, we will use equation (19) to derive a Table of Malmsten’s type loglogx integrals [4] in terms of the Log gamma function logΓx and the fundamental constant π. The forms, which we will derive, are given by(27)∫0∞loglogxβ2+x2γ+xdx,and we use equations (1.10.10) in [10], (64 : 13 : 3) in [9], and (24.14.2) in [11] for simplifying the formula derived.


### 8.2. Entry M1

Using equation (19) and setting m=0 and a=1 followed by taking the first partial derivative with respect to k and then setting k=0 and simplifying, we obtain(28)∫0∞loglogxβ2+x2γ+xdx=πγ2ββ2+γ2logπ+2log1+iΓ3/4−ilogβ/2πΓπ−2ilogβ/4π−2βlog2πlogγ+2iπlog2π4γΓ4π−ilogγ/2π−π−2ilogβΓ−ilogβ/π−1/2+β2logπ+iπlogβ.


### 8.3. Entry M2

Next, using equation (28) and setting β=γ=1 and simplifying, we obtain(29)∫0∞loglogxx+1x2+1dx=18iππ+log4+log2π−2i3Γ−3/4Γ−1/44i.


### 8.4. Entry M3

Next, using equation (28) and setting β=1 and γ=2 and simplifying, we obtain(30)∫0∞loglogxx+2x2+1dx=16−log2log2π+π2logπ+22log1+iΓ3/4Γ1/4−4ilog−Γ1/2−ilog2/4π28π.


### 8.5. Entry M4

Next, using equation (28) and setting β=1 and γ=3 and simplifying, we obtain(31)∫0∞loglogxx+3x2+1dx=18−log3log2π+π3logπ+23log1+iΓ3/4Γ1/4−4ilog−38Γ1/2−ilog3/4π24π.


### 8.6. Entry M5

Next, using equation (28) and setting β=21/4 and γ=3 and simplifying, we obtain(32)∫0∞loglogxx+3x2+2dx=183+2log4logπ−4log3log2π+π83+2ilog2+223/43−2ilogπ+423/43log1+iΓ3/4−ilog2/8πΓ1/4−ilog2/8π−2ilog192Γ1/2−ilog3/4π8log2−2iπ4Γ−1/2−ilog2/4π4.


### 8.7. Entry M6

Using equation (28) and setting β=i and simplifying, we obtain(33)∫0∞loglogxx2−1γ+xdx=1γ2−1−log2πlogγ+2iπlogπ21/4−γ/4e−1/8iπγ−1γ4Γπ−ilogγ/2π.The singularity is atx=1 which includes the loglogx term, so we have to use a principle value integral here.


### 8.8. Entry M7

Using equation (33) and applying L’Hopital’s rule as γ⟶1 and simplifying, we obtain(34)∫0∞loglogxx−1x+12dx=184γ+π2−2iπ1+log2+log16π4,where γ is Euler’s constant.


### 8.9. Entry M8

Using equation (33) and applying L’Hopital’s rule to the right-hand side as γ⟶−1 and simplifying, we obtain(35)∫0∞loglogxx−12x+1dx=184γ−π2+log2−4+2iπ−4logπ.


### 8.10. Entry M9

Using equation (33) and setting γ=2 and simplifying, we obtain(36)∫0∞loglogxx+2x2−1dx=13−log2log2π+2iπloge−iπ/8π/2Γπ−ilog2/2π.


### 8.11. Entry M10

Using equation (33) and setting γ=−2 and simplifying, we obtain(37)∫0∞loglogxx−2x2−1dx=112−4log2log2π−ππ+8ilogΓ1−ilog22π.


### 8.12. Entry M11

Using equation (33), then taking the first partial derivative with respect to γ, and setting γ=1 and simplifying, we obtain(38)∫0∞loglogxx−1x+13dx=116π2−iπ5+log4+8γ+log2π.


### 8.13. Entry M12

Using equation (33), then taking the first partial derivative with respect to γ, and setting γ=−1 and simplifying, we obtain(39)∫0∞loglogxx−13x+1dx=148−24γ+3ππ+3i+log644−iπ+24logπ.


### 8.14. Entry M13

Using equation (33), then taking the second partial derivative with respect to γ, and setting γ=1 and simplifying, we obtain(40)∫0∞loglogxx−1x+14dx=−28ζ3+3π4−iπ329+log64+40π2γ+log2/π96π2.


### 8.15. Entry M14

Using equation (33), then taking the second partial derivative with respect to γ, and setting γ=−1 and simplifying, we obtain(41)∫0∞loglogxx−14x+1dx=196−4ζ3π2+40γ−3π2+iπlog64−23−40log2π.


### 8.16. Definite Integrals ofloglogx and a Rational Quotient Function

In this section, we again look at Malmsten integrals [4] of the form(42)∫0∞loglogxxn−xmβ2+x2γ+xdx.Using equation (19) and formulating a second equation by replacing m by n, then taking the difference followed by taking the first partial derivative with respect to k, and then setting k=0 and simplifying in terms of the derivative of the Lerch function, we obtain/(43)∫0∞loglogxxn−xmβ2+x2γ+xdx=π4ββ2+γ2−iπ−2ilogπβm+1cscπm2+β2iπ+log16+4logπγmcscπm+2logπ+iπβnβcscπn2+γsecπn2−γβmsecπm2−2iβπ−ilog4π2γncscπn+πγeiπm/2βm−1Φ′−eiπm,0,1/2−ilogβ/πβ2+γ2−iπeiπm/2βmΦ′eiπm,0,1/2−ilogβ/πβ2+γ2+2iπeiπmγmΦ′e2iπm,0,π−ilogγ/2πβ2+γ2−πγeiπn/2βn−1Φ′−eiπn,0,1/2−ilogβ/πβ2+γ2+iπeiπn/2βnΦ′eiπn,0,1/2−ilogβ/πβ2+γ2−2iπeiπnγnΦ′e2iπn,0,π−ilogγ/2πβ2+γ2.


### 8.17. Enrty MQ1

Using equation (43), we set β=γ=1, m=1/2, and n=1/3 simplifying to obtain(44)∫0∞x3−xloglogxx+1x2+1dx=−πΦ′−1,0,12+124−1πΦ′−i,0,12−12−13/4πΦ′i,0,12−12−16πΦ′−−13,0,12+12−12/3πΦ′−13,0,12−−15/6πΦ′−12/3,0,12+iπ22−iπ222−iπ243+112πlog64+πlogπ−πlogπ2−πlog4π23.


## 8.1. Definite Integrals ofloglogx with a Cubic Denominator

In this section, we will use equation (19) to derive a Table of Malmsten’s type loglogx integrals [4] in terms of the Log gamma function logΓx and the fundamental constant π. The forms, which we will derive, are given by(27)∫0∞loglogxβ2+x2γ+xdx,and we use equations (1.10.10) in [10], (64 : 13 : 3) in [9], and (24.14.2) in [11] for simplifying the formula derived.


## 8.2. Entry M1

Using equation (19) and setting m=0 and a=1 followed by taking the first partial derivative with respect to k and then setting k=0 and simplifying, we obtain(28)∫0∞loglogxβ2+x2γ+xdx=πγ2ββ2+γ2logπ+2log1+iΓ3/4−ilogβ/2πΓπ−2ilogβ/4π−2βlog2πlogγ+2iπlog2π4γΓ4π−ilogγ/2π−π−2ilogβΓ−ilogβ/π−1/2+β2logπ+iπlogβ.


## 8.3. Entry M2

Next, using equation (28) and setting β=γ=1 and simplifying, we obtain(29)∫0∞loglogxx+1x2+1dx=18iππ+log4+log2π−2i3Γ−3/4Γ−1/44i.


## 8.4. Entry M3

Next, using equation (28) and setting β=1 and γ=2 and simplifying, we obtain(30)∫0∞loglogxx+2x2+1dx=16−log2log2π+π2logπ+22log1+iΓ3/4Γ1/4−4ilog−Γ1/2−ilog2/4π28π.


## 8.5. Entry M4

Next, using equation (28) and setting β=1 and γ=3 and simplifying, we obtain(31)∫0∞loglogxx+3x2+1dx=18−log3log2π+π3logπ+23log1+iΓ3/4Γ1/4−4ilog−38Γ1/2−ilog3/4π24π.


## 8.6. Entry M5

Next, using equation (28) and setting β=21/4 and γ=3 and simplifying, we obtain(32)∫0∞loglogxx+3x2+2dx=183+2log4logπ−4log3log2π+π83+2ilog2+223/43−2ilogπ+423/43log1+iΓ3/4−ilog2/8πΓ1/4−ilog2/8π−2ilog192Γ1/2−ilog3/4π8log2−2iπ4Γ−1/2−ilog2/4π4.


## 8.7. Entry M6

Using equation (28) and setting β=i and simplifying, we obtain(33)∫0∞loglogxx2−1γ+xdx=1γ2−1−log2πlogγ+2iπlogπ21/4−γ/4e−1/8iπγ−1γ4Γπ−ilogγ/2π.The singularity is atx=1 which includes the loglogx term, so we have to use a principle value integral here.


## 8.8. Entry M7

Using equation (33) and applying L’Hopital’s rule as γ⟶1 and simplifying, we obtain(34)∫0∞loglogxx−1x+12dx=184γ+π2−2iπ1+log2+log16π4,where γ is Euler’s constant.


## 8.9. Entry M8

Using equation (33) and applying L’Hopital’s rule to the right-hand side as γ⟶−1 and simplifying, we obtain(35)∫0∞loglogxx−12x+1dx=184γ−π2+log2−4+2iπ−4logπ.


## 8.10. Entry M9

Using equation (33) and setting γ=2 and simplifying, we obtain(36)∫0∞loglogxx+2x2−1dx=13−log2log2π+2iπloge−iπ/8π/2Γπ−ilog2/2π.


## 8.11. Entry M10

Using equation (33) and setting γ=−2 and simplifying, we obtain(37)∫0∞loglogxx−2x2−1dx=112−4log2log2π−ππ+8ilogΓ1−ilog22π.


## 8.12. Entry M11

Using equation (33), then taking the first partial derivative with respect to γ, and setting γ=1 and simplifying, we obtain(38)∫0∞loglogxx−1x+13dx=116π2−iπ5+log4+8γ+log2π.


## 8.13. Entry M12

Using equation (33), then taking the first partial derivative with respect to γ, and setting γ=−1 and simplifying, we obtain(39)∫0∞loglogxx−13x+1dx=148−24γ+3ππ+3i+log644−iπ+24logπ.


## 8.14. Entry M13

Using equation (33), then taking the second partial derivative with respect to γ, and setting γ=1 and simplifying, we obtain(40)∫0∞loglogxx−1x+14dx=−28ζ3+3π4−iπ329+log64+40π2γ+log2/π96π2.


## 8.15. Entry M14

Using equation (33), then taking the second partial derivative with respect to γ, and setting γ=−1 and simplifying, we obtain(41)∫0∞loglogxx−14x+1dx=196−4ζ3π2+40γ−3π2+iπlog64−23−40log2π.


## 8.16. Definite Integrals ofloglogx and a Rational Quotient Function

In this section, we again look at Malmsten integrals [4] of the form(42)∫0∞loglogxxn−xmβ2+x2γ+xdx.Using equation (19) and formulating a second equation by replacing m by n, then taking the difference followed by taking the first partial derivative with respect to k, and then setting k=0 and simplifying in terms of the derivative of the Lerch function, we obtain/(43)∫0∞loglogxxn−xmβ2+x2γ+xdx=π4ββ2+γ2−iπ−2ilogπβm+1cscπm2+β2iπ+log16+4logπγmcscπm+2logπ+iπβnβcscπn2+γsecπn2−γβmsecπm2−2iβπ−ilog4π2γncscπn+πγeiπm/2βm−1Φ′−eiπm,0,1/2−ilogβ/πβ2+γ2−iπeiπm/2βmΦ′eiπm,0,1/2−ilogβ/πβ2+γ2+2iπeiπmγmΦ′e2iπm,0,π−ilogγ/2πβ2+γ2−πγeiπn/2βn−1Φ′−eiπn,0,1/2−ilogβ/πβ2+γ2+iπeiπn/2βnΦ′eiπn,0,1/2−ilogβ/πβ2+γ2−2iπeiπnγnΦ′e2iπn,0,π−ilogγ/2πβ2+γ2.


## 8.17. Enrty MQ1

Using equation (43), we set β=γ=1, m=1/2, and n=1/3 simplifying to obtain(44)∫0∞x3−xloglogxx+1x2+1dx=−πΦ′−1,0,12+124−1πΦ′−i,0,12−12−13/4πΦ′i,0,12−12−16πΦ′−−13,0,12+12−12/3πΦ′−13,0,12−−15/6πΦ′−12/3,0,12+iπ22−iπ222−iπ243+112πlog64+πlogπ−πlogπ2−πlog4π23.


## 9. Definite Integrals of Bierens De Haan Type

In this section, we will derive definite integrals similar to works published in [12] given by(45)∫0∞loglogxx+γx2+β2logxdx,∫0∞logxloglogxx+γx2+β2dx,and use equation (64 : 13 : 3) in [9] and (25.14.2) in [11] for simplifying the derived formula.


### 9.1. Entry B1

Using equation (19) and setting a=1 and m=0 and simplifying, we obtain(46)∫0∞logkxβ2+x2γ+xdx=eiπk/2πk+1ββ2+γ22kγζ−k,π−2ilogβ4π−γζ−k,34−ilogβ2π+2iβζ−k,π−ilogγ2π−iβζ−k,12−ilogβπ.


### 9.2. Entry B2

Next, we take the first partial derivative with respect tok and set k=−1/2 and β=γ=1 and simplify to obtain(47)∫0∞loglogxx+1x2+1logxdx=18−14π2i2ζ′12,14−2i2ζ12,34+2ζ12,14π−2ilog2π−2ζ12,34π−2ilog2π+22−3ζ122γ+1−2iπ+log64π2,where γ is Euler’s constant given by equation (9.73) in [6].


### 9.3. Entry B3

Another example is by once again using equation (19) taking the first partial derivative with respect to k setting k=1/2 and β=γ=1 and simplifying to obtain(48)∫0∞logxloglogxx+1x2+1dx=132−1π3/24−162ζ′−12,14+162ζ′−12,34−48iζ′−12+40i2ζ′−12+82ζ−12,34log14π2+82ζ−12,14log4π2+6ζ32−52ζ32−−13/4π5/2ζ−1/2,3/4−ζ−1/2,1/422+116−13/4πζ322log8−6logπ+52logπ.


## 9.1. Entry B1

Using equation (19) and setting a=1 and m=0 and simplifying, we obtain(46)∫0∞logkxβ2+x2γ+xdx=eiπk/2πk+1ββ2+γ22kγζ−k,π−2ilogβ4π−γζ−k,34−ilogβ2π+2iβζ−k,π−ilogγ2π−iβζ−k,12−ilogβπ.


## 9.2. Entry B2

Next, we take the first partial derivative with respect tok and set k=−1/2 and β=γ=1 and simplify to obtain(47)∫0∞loglogxx+1x2+1logxdx=18−14π2i2ζ′12,14−2i2ζ12,34+2ζ12,14π−2ilog2π−2ζ12,34π−2ilog2π+22−3ζ122γ+1−2iπ+log64π2,where γ is Euler’s constant given by equation (9.73) in [6].


## 9.3. Entry B3

Another example is by once again using equation (19) taking the first partial derivative with respect to k setting k=1/2 and β=γ=1 and simplifying to obtain(48)∫0∞logxloglogxx+1x2+1dx=132−1π3/24−162ζ′−12,14+162ζ′−12,34−48iζ′−12+40i2ζ′−12+82ζ−12,34log14π2+82ζ−12,14log4π2+6ζ32−52ζ32−−13/4π5/2ζ−1/2,3/4−ζ−1/2,1/422+116−13/4πζ322log8−6logπ+52logπ.


## 10. Table of Mellin Transforms

In this section, we will use equation (19) to derive a table of Mellin transforms not published in current literature.


### 10.1. Entry MT1

We take the first partial derivative with respect toγ and set β=γ=a=k=1 and replace m by s−1 and simplify to obtain(49)∫0∞xs−1logxx+12x2+1dx=−12πcscπs−πs−2cotπs+πsin3πs2cscπs+1,from entry (1) below Table (64 : 12 : 7) in [9].


### 10.2. Entry MT2

We take the first partial derivative with respect toβ and set β=γ=a=k=1 and replace m by s−1 and simplify to obtain(50)∫0∞xs−1logxx+1x2+12dx=116πcsc2πs3πs−2−2sinπs2−πs−2+2sin3πs2+3πs−3+2cosπs2−4πcosπs+πs−3−2cos3πs2,from entry (1) below Table (64 : 12 : 7) in [9].


### 10.3. Entry MT3

We take the first partial derivative with respect toγ and set β=γ=a=1, k=1/2 and replace m by s−1 and simplify to obtain(51)∫0∞xs−1logxx+12x2+1dx=14+i4π2eiπs/22eiπs/24iπs−2Φe2iπs,−12,12+Φe2iπs,12,12−2πΦ−eiπs,−12,12.


### 10.4. Entry MT4

We take the first partial derivative with respect toa and set β=γ=a=1 and k=1/3 and replace m by s−1 and simplify to obtain(52)∫0∞xs−1x+1x2+1log2/3xdx=12−12/3π3eiπs/2Φ−eiπs,23,12+iΦeiπs,23,12−−16π3eiπsΦe2iπs,2/3,1/222/3.


### 10.5. Entry MT5

We take the first partial derivatives with respect toβ then γ and set β=γ=a=1 and k=1/2 and replace m by s−1 and simplify to obtain(53)∫0∞xs−1logxx+12x2+12dx=18+i8π2eiπs/22πsΦ−eiπs,−12,12−6πΦ−eiπs,−12,12−iΦ−eiπs,12,12−2iπΦeiπs,−12,12+18+i8πeiπs4iπsΦe2iπs,−12,12−12iπΦe2iπs,−12,12+Φe2iπs,12,12.


### 10.6. Entry MT6

Here, we derive the Mellin transform in terms of the Polylogarithmic function. We setβ=i/a and a=1 and replace m by s−1 and simplify to obtain(54)∫0∞xs−1logkxx2−1γ+xdx=−e1/2iπk+1πk+1γγ2−12k+1eiπsγsΦe2iπs,−k,π−ilogγ2π+γLi−k−eiπs−γLi−keiπs,from equation (1.11.14) in [10].


## 10.1. Entry MT1

We take the first partial derivative with respect toγ and set β=γ=a=k=1 and replace m by s−1 and simplify to obtain(49)∫0∞xs−1logxx+12x2+1dx=−12πcscπs−πs−2cotπs+πsin3πs2cscπs+1,from entry (1) below Table (64 : 12 : 7) in [9].


## 10.2. Entry MT2

We take the first partial derivative with respect toβ and set β=γ=a=k=1 and replace m by s−1 and simplify to obtain(50)∫0∞xs−1logxx+1x2+12dx=116πcsc2πs3πs−2−2sinπs2−πs−2+2sin3πs2+3πs−3+2cosπs2−4πcosπs+πs−3−2cos3πs2,from entry (1) below Table (64 : 12 : 7) in [9].


## 10.3. Entry MT3

We take the first partial derivative with respect toγ and set β=γ=a=1, k=1/2 and replace m by s−1 and simplify to obtain(51)∫0∞xs−1logxx+12x2+1dx=14+i4π2eiπs/22eiπs/24iπs−2Φe2iπs,−12,12+Φe2iπs,12,12−2πΦ−eiπs,−12,12.


## 10.4. Entry MT4

We take the first partial derivative with respect toa and set β=γ=a=1 and k=1/3 and replace m by s−1 and simplify to obtain(52)∫0∞xs−1x+1x2+1log2/3xdx=12−12/3π3eiπs/2Φ−eiπs,23,12+iΦeiπs,23,12−−16π3eiπsΦe2iπs,2/3,1/222/3.


## 10.5. Entry MT5

We take the first partial derivatives with respect toβ then γ and set β=γ=a=1 and k=1/2 and replace m by s−1 and simplify to obtain(53)∫0∞xs−1logxx+12x2+12dx=18+i8π2eiπs/22πsΦ−eiπs,−12,12−6πΦ−eiπs,−12,12−iΦ−eiπs,12,12−2iπΦeiπs,−12,12+18+i8πeiπs4iπsΦe2iπs,−12,12−12iπΦe2iπs,−12,12+Φe2iπs,12,12.


## 10.6. Entry MT6

Here, we derive the Mellin transform in terms of the Polylogarithmic function. We setβ=i/a and a=1 and replace m by s−1 and simplify to obtain(54)∫0∞xs−1logkxx2−1γ+xdx=−e1/2iπk+1πk+1γγ2−12k+1eiπsγsΦe2iπs,−k,π−ilogγ2π+γLi−k−eiπs−γLi−keiπs,from equation (1.11.14) in [10].


## 11. Conclusion

In this article, we derived a table of definite integrals which can be included in current books such as [1, 2, 6]. We found that using our simultaneous contour integral method, we were able to generate very interesting new integral formulae. We also looked at analytic continuity and principle value methods to evaluate integrals when singularities were present. One of the consequences of this work is the production of a table of definite integrals which researchers may use to advance their work if they find any of these formulae applicable. 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.



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