EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 3, 2021, 723-736
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Note on a Stieltjes transform in terms of the Lerch
function

Robert Reynolds1,∗, Allan Stauffer1

1 Department of Mathematics and Statistics, Faculty of Science, York University, Toronto,
Ontario, Canada

Abstract. In this work the authors derive the Stieltjes transform of the logarithmic function
in terms of the Lerch function. This transform is used to derive closed form solutions involving
fundamental constants and special functions. Specifically we derive the definite integral given by∫ ∞

0

(1− bx)m logk(c(1− bx)) + (bx+ 1)m logk(c(bx+ 1))

a+ x2
dx

where a, b, c,m and k are general complex numbers subject to the restrictions given in connection
with the formulas.

2020 Mathematics Subject Classifications: 01A55, 11M06, 11M35, 30-02, 30D10, 30D30,
30E20

Key Words and Phrases: Stieltjes transform | Lerch | Definite integral | entries in Gradshteyn
and Rhyzik

1. Significance Statement

The Stieltjes transform and Lerch function were both developed between 1856-1922,
by famous mathematicians Thomas Joannes Stieltjes and Mathias Lerch respectively.
The two functions are not well documented in current literature. The Stieltjes trans-
form has many real world applications such as in Cognitive Radio Communication and
Networking and the Lerch function is closely related to Poisson’s summation formula.
In this article we generate a new table of definite integral formulas which can be used
as reference similar to current Table of definite integrals.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.3991

Email addresses: milver@my.yorku.ca (R. Reynolds), stauffer@yorku.ca (A. Stauffer)

http://www.ejpam.com 723 © 2021 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 724

2. Introduction

n the late 1800’s Thomas Joannes Stieltjes, a Dutch mathematician found the Stielt-
jes transform. In this work we will derive this formula as shown in the abstract which
does not exist in current literature. We will use this definite integral to produce formal
derivations of known integral formulas in [1] and [2]. We will also derive new formula
which can be considered an extension to Erdeyli’s extensive table of transforms. The
derivations follow the method used by us in [5], [4], [6], [7], [9] and [8]. This method
involves using a form of the generalized Cauchy’s integral formula given by

(1)
yk

k!
=

1

2πi

∫
C

ewy

wk+1
dy.

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. Then multiply both
sides by a function, then takes 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.

3. Definite integral of the contour integral

We use the method in [9]. 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. We replace
y in 1 by logk(c(1 − bx)) and multiply by (1 − bx)m/(a2 + x2) in the first case and by
logk(c(1 + bx)) and multiply by (1 + bx)m/(a2 + x2) in the second case and add them.
The integration is over x now instead of y.

(2)

1

k!

∫ ∞
0

(1− bx)m logk(c(1− bx)) + (bx+ 1)m logk(c(bx+ 1))

a+ x2
dx

=
1

2πi

∫ ∞
0

∫
C

cww−k−1 ((1− bx)m+w + (bx+ 1)m+w)

a+ x2
dwdx

=
1

2πi

∫
C

∫ ∞
0

cww−k−1 ((1− bx)m+w + (bx+ 1)m+w)

a+ x2
dxdw

=
1

2πi

∫
C
πcww−k−1a

1
2
(m+w−1)(−b)m+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
π − 2 cot−1

(√
ab
)))

+ πcww−k−1a
1
2
(m+w−1)bm+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
2 cot−1

(√
ab
)

+ π
))

dw



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 725

from equations (3.227.1) and (3.227.2) in [3], using partial fractions, where Re(a) ≥ 0
and Re(m+ w) < 1.

4. The Lerch function

The Lerch function has a series representation given by

(3)Φ(z, s, v) =
∞∑
n=0

(v + n)−szn

where |z|< 1, v 6= 0,−1, .. and is continued analytically by its integral representation
given by

(4)
Φ(z, s, v) =

1

Γ(s)

∫ ∞
0

ts−1e−vt

1− ze−t
dt

=
1

Γ(s)

∫ ∞
0

ts−1e−(v−1)t

et − z
dt

where Re(v) > 0, or |z|≤ 1, z 6= 1, Re(s) > 0, or z = 1, Re(s) > 1.

5. Infinite sum of the contour integral

In this section we will again use Cauchy’s integral formula (1) and taking the infinite
sum to derive equivalent sum representations for the contour integrals.

5.1. Derivation of the general sine contour integral

Use equation (1) and replace y by y + it and multiply by emit for the first equation.
Next we form the second equation by replacing t by −t and taking their difference to
get

(5)
ie−imt

(
(y − it)k − e2imt(y + it)k

)
2k!

=
1

2πi

∫
C
w−k−1ewy sin(t(m+ w))dw

5.2. Derivation of the contour integral

In this section we will derive the contour integral given by∫
C
πcww−k−1a

1
2
(m+w−1)(−b)m+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
π − 2 cot−1

(√
ab
)))

dw

Using equation (5) and substituting y by 1
2 log

(
1
ab2

+ 1
)

+ log(a)
2 + log(−b) + log(c) +

iπ(2y + 1) and multiplying both sides by e2iπmy+iπm, simplifying we get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 726

ie−imt+2iπmy+iπm

2k!

(2iπ)k

(
−
i log

(
1
ab2

+ 1
)

4π
− i log(a)

4π

− i log(−b)
2π

− i log(c)

2π
− t

2π
+

1

2
(2y + 1)

)k

− (2iπ)ke2imt

(
−
i log

(
1
ab2

+ 1
)

4π
− i log(a)

4π
− i log(−b)

2π
− i log(c)

2π
+

t

2π
+

1

2
(2y+ 1)

)k
=

1

2πi

∫
C
w−k−1 sin(t(m+w)) exp

(
w

(
1

2
log

(
1

ab2
+ 1

)
+

log(a)

2
+ log(−b) + log(c)

+ iπ(2y + 1)

)
+ 2iπmy + iπm

)
dw

(6)

Next we take the infinite sum over y ∈ [0,∞) and replace t by 1
2

(
π − 2 cot−1 (

√
ab)
)

and multiply both sides by −2iπa
m−1

2 (−b)m
(

1
ab2

+ 1
)m/2

simplifying to get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 727

(7)
2ke

iπk
2 πk+1a

m−1
2 (−b)m

k!

(
1

ab2
+ 1

)m/2(
e

1
2
im(2 cot−1(

√
ab)+π)

Φ

(
e2imπ,−k,−

i
(
2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(−b) + 2 log(c) + iπ

)
4π

)
− e

1
2
im(3π−2 cot−1(

√
ab))

Φ

(
e2imπ,−k,−

i
(
−2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(−b) + 2 log(c) + 3iπ

)
4π

))

=
1

2πi

∞∑
y=0

∫
C
w−k−1 sin(t(m+ w))

exp

(
w

(
1

2
log

(
1

ab2
+ 1

)
+

log(a)

2
+ log(−b) + log(c) + iπ(2y + 1)

)
+ 2iπmy + iπm

)
dw

=
1

2πi

∫
C

∞∑
y=0

w−k−1 sin(t(m+ w))

exp

(
w

(
1

2
log

(
1

ab2
+ 1

)
+

log(a)

2
+ log(−b) + log(c) + iπ(2y + 1)

)
+ 2iπmy + iπm

)
dw

=
1

2πi

∫
C
πcww−k−1a

1
2
(m+w−1)(−b)m+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
π − 2 cot−1

(√
ab
)))

dw

from equation (1.232.3) in [3] where Im(m+w) > 0 in order for the sum to converge.

5.3. Derivation of the second contour integral

In this section we derive the second contour integral given by

(8)

1

2πi

∫
C
πcww−k−1a

1
2
(m+w−1)bm+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
2 cot−1

(√
ab
)

+ π
))

dw

In this derivation we proceed as above but multiply by −2iπa
m−1

2 bm
(

1
ab2

+ 1
)m/2

and replace t by 1
2

(
2 cot−1 (

√
ab) + π

)
and simplifying to get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 728

2ke
iπk
2 πk+1a

m−1
2 bm

k!

(
1

ab2
+ 1

)m/2(
e

1
2
im(π−2 cot−1(

√
ab))

Φ

(
e2imπ,−k,−

i
(
−2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(b) + 2 log(c) + iπ

)
4π

)
− e

1
2
im(2 cot−1(

√
ab)+3π)

Φ

(
e2imπ,−k,−

i
(
2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(b) + 2 log(c) + 3iπ

)
4π

))

=
1

2πi

∫
C
πcww−k−1a

1
2
(m+w−1)bm+w csc(π(m+ w))

(
1

ab2

+ 1

)m+w
2

sin

(
1

2
(m+ w)

(
2 cot−1

(√
ab
)

+ π
))

dw

(9)

from equation (1.232.3) in [3] where Im(m+w) > 0 in order for the sum to converge.

6. The Stieltjes transform in terms of the Lerch function

Since the right-hand side of equation (2) is equal to the sum of the right-hand sides
of equations (8) and (9) we can equate the left-hand sides and simplifying the factorial
to get

(10)

∫ ∞
0

(1− bx)m logk(c(1− bx)) + (bx+ 1)m logk(c(bx+ 1))

a2 + x2
dx

= 2ke
iπk
2 πk+1am−1(−b)m

(
1

a2b2
+ 1

)m/2(
e

1
2
im(2 cot−1(ab)+π)

Φ

(
e2imπ,−k,−

i
(
2i cot−1(ab) + log

(
a2
)

+ log
(
1 + 1

a2b2

)
+ 2 log(−b) + 2 log(c) + iπ

)
4π

)
− e

1
2
im(3π−2 cot−1(ab))

Φ

(
e2imπ,−k,−

i
(
−2i cot−1(ab) + log

(
a2
)

+ log
(
1 + 1

a2b2

)
+ 2 log(−b) + 2 log(c) + 3iπ

)
4π

))

+ 2ke
iπk
2 πk+1am−1bm

(
1

a2b2
+ 1

)m/2(
e

1
2
im(π−2 cot−1(ab))

Φ

(
e2imπ,−k,−

i
(
−2i cot−1(ab) + log

(
a2
)

+ log
(
1 + 1

a2b2

)
+ 2 log(b) + 2 log(c) + iπ

)
4π

)
− e

1
2
im(2 cot−1(ab)+3π)

Φ

(
e2imπ,−k,−

i
(
2i cot−1(ab) + log

(
a2
)

+ log
(
1 + 1

a2b2

)
+ 2 log(b) + 2 log(c) + 3iπ

)
4π

))



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 729

7. Definite integrals in terms of the Hurwitz zeta function

Using equation (10) replacing b by β, a by α and setting c = 1 followed by taking
the sum over m ∈ [0,∞) and simplifying in terms of the Hurwitz zeta function we get∫ ∞
0

logk(1− βx)− logk(βx+ 1)

x (α2 + x2)
dx = − i2

ke
iπk
2 πk+1

α2

(
ζ

(
−k, i(−2 log(i− αβ)− 3iπ)

4π

)
− ζ

(
−k,− i(2 log(αβ − i) + iπ)

4π

))
(11)

Using equation (10) and setting m = 0 and replacing a by
√
a and simplifying in

terms of the Hurwitz zeta function ζ(s, v) we get∫ ∞
0

logk(c− bcx) + logk(bcx+ c)

a+ x2
dx

=
(2i)kπk+1

√
a

(
ζ

(
−k,

−
i
(
−2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(b) + 2 log(c) + iπ

)
4π

)
− ζ

(
−k,

−
i
(
−2i cot−1 (

√
ab) + log(a) + log

(
1 + 1

b2a

)
+ 2 log(b) + 2 log(c) + 5iπ

)
4π

))
(12)

8. Derivation of entry 4.535.7 in [3]

Using (11) and setting k = α = 1 and replacing β by −ip and simplifying we get

(13)

∫ ∞
0

tan−1(px)

x3 + x
dx =

1

2
π log(p+ 1)

9. Derivation of entry 4.535.8 in [3]

Using (11) and setting k = 1, β = −p, α = i and simplifying we get

(14)

∫ ∞
0

tan−1(px)

x− x3
dx =

1

4
π(2 log(p+ i)− iπ)

10. Derivation of entry 4.535.9 in [3]

Using (11) and setting k = 1 and replacing β by −iq and α by p and simplifying we
get

(15)

∫ ∞
0

tan−1(qx)

p2x+ x3
dx =

π log(pq + 1)

2p2



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 730

11. Derivation of entry 4.535.10 in [3]

In this evaluation we will derive the definite integrals for both the Arctangent and
Hyperbolic tangent functions. Using (11) and setting k = 1 and replacing β by q and α
by 1/p simplifying we get

(16)

∫ ∞
0

tanh−1(qx)

p2x3 + x
dx =

1

8

(
− log

(
−q
p

+ i

)
+ log

(
q

p
− i
)
− iπ

)(
log

(
−q
p

+ i

)
+ log

(
q

p
− i
))

Next we multiply by −1/i, replace p by −ip and q by −iq simplifying to get

(17)

∫ ∞
0

tan−1(qx)

x− p2x3
dx =

1

2
π log

(
1 +

iq

p

)
The formula given in [3] is in error.

12. Derivation of entry 4.535.11 in [3]

In this evaluation we will derive the definite integrals for both the Arctangent and
Hyperbolic tangent functions. Using (10) and setting m = 0, k = 1 and replacing β by
q and α by 1/p and simplifying we get∫ ∞

0

tanh−1(bx)

ax+ x3
dx

= −
(log(−b)− log(b))

(
log
(

1
ab2

+ 1
)

+ log(a) + log(−b) + log(b)
)

+ 2π cot−1 (
√
ab)

4a
(18)

Next we split the left-hand side to get∫ ∞
0

(
tanh−1(bx)

x
− x tanh−1(bx)

a+ x2

)
dx=

1

4

(
−(log(−b)− log(b))

(
log

(
1

ab2
+1

)
+log(a)

+ log(−b) + log(b)

)
− 2π cot−1

(√
ab
))

(19)

Next we take the first partial derivative with respect to a simplifying to get

(20)

∫ ∞
0

x tanh−1(bx)

(a2 + x2)2
dx =

b
(

π√
a2
− b log(−b) + b log(b)

)
4a2b2 + 4

Next replacing b by ib and simplifying we get

(21)

∫ ∞
0

x tan−1(bx)

(a2 + x2)2
dx =

πb

4a(ab+ 1)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 731

13. Derivation of entry 4.295.1 in [3]

Using (12) and setting k = c = 1 and simplifying we get∫ ∞
0

log
(
1− b2x2

)
a+ x2

dx

=
1
2π
(
log
(

1
ab2

+ 1
)

+ log(a) + log(−b) + log(b)
)

+ (log(b)− log(−b)) cot−1 (
√
ab)

√
a

(22)

Next we replace b by
i
√
µ√
β

and a by γ simplifying to get

(23)

∫ ∞
0

log
(
β + µx2

)
γ + x2

dx =
π log

(√
β +
√
γ
√
µ
)

√
γ

The equation quoted in [3] is not valid for general complex numbers, for example
when Re(γ) < 0.

14. Derivation of entry 4.295.7 in [3]

Using (23) and replacing β by a2, µ by b2 and γ by
(
c
g

)2
and simplifying we get

(24)

∫ ∞
0

log
(
a2 + b2x2

)
c2 + g2x2

dx =
π log

(
ag+bc
g

)
cg

15. Derivation of entry 4.295.8 in [3]

Using (24) and replacing g by ig and simplifying we get

(25)

∫ ∞
0

log
(
a2 + b2x2

)
c2 − g2x2

dx = −
iπ log

(
− i(bc+iag)

g

)
cg

16. Derivation of entry 4.295.9 in [3]

We will form two equations by using (24) and setting β = 1 and replacing µ by p2 for
the first equation and then replacing p by q for the second, subtracting and simplifying
we get

(26)

∫ ∞
0

log
(
p2x2 + 1

)
− log

(
q2x2 + 1

)
γ + x2

dx =
π log

(√
γp+1√
γq+1

)
√
γ

Next we apply L’Hopital’s rule as γ → 0 to the right-hand side simplifying to get

(27)

∫ ∞
0

log
(
p2x2 + 1

)
− log

(
q2x2 + 1

)
x2

dx = π(p− q)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 732

17. Derivation of entry 4.295.22 in [3]

Using (24) setting a = 1 and replacing c by r, g by q, b by p and simplifying we get

(28)

∫ ∞
0

log
(
p2x2 + 1

)
q2x2 + r2

dx =
π log

(
pr+q
q

)
qr

18. Derivation of entry 4.295.25 in [3]

Using (24) and taking the first partial derivative with respect to c and simplifying
we get

(29)

∫ ∞
0

log
(
a2 + b2x2

)
(c2 + g2x2)2

dx =
π log

(
ag+bc
g

)
2c3g

− πb

2c2g(ag + bc)

19. Derivation of entry 4.295.26 in [3]

Using (24) and taking the first partial derivative with respect to g and simplifying
we get

(30)

∫ ∞
0

x2 log
(
a2 + b2x2

)
(c2 + g2x2)2

dx =
πb

2g3(ag + bc)
+
π log

(
ag+bc
g

)
2cg3

20. Definite logarithmic integral in terms π

Using (12) and setting c = 1, b = 1 and a = 1 simplifying we get

(31)

∫ ∞
0

logk(1− x) + logk(x+ 1)

x2 + 1
dx = (2i)kπk+1

(
ζ

(
−k,−

−7π
2 + i log(2)

4π

)

− ζ

(
−k,

9π
2 − i log(2)

4π

))

+ (2i)kπk+1

(
ζ

(
−k,−

−π
2 + i log(2)

4π

)

− ζ

(
−k,

7π
2 − i log(2)

4π

))

Next we apply L’Hopital’s rule as k → −1 and simplifying to get

(32)

∫ ∞
0

log
(
1− x2

)
(x2 + 1) log(1− x) log(x+ 1)

dx =
4π

log(4) + iπ



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 733

21. Definite nested logarithmic integral in terms π

Using (12) and taking the first partial derivative with respect to k setting k = 0,
c = 1, b = 1 and a = 1 and simplifying we get

(33)

∫ ∞
0

log(log(1− x)) + log(log(x+ 1))

x2 + 1
dx =

1

2
π

(
2logΓ

(
π − 2i log(2)

8π

)
− 2logΓ

(
−7

8
− i log(2)

4π

)
+ 3iπ

+ log

(
4π2(π − 2i log(2))2

(7π + 2i log(2))2

))
Next simplifying the right-hand side we get

(34)

∫ ∞
0

log(log(1− x) log(x+ 1))

x2 + 1
dx = π log

(
1

4
i(π − 2i log(2))

)
22. Definite integral of the hyperbolic tangent and logarithmic

functions

Using (11) and setting k = 2 and simplifying we get∫ ∞
0

log
(
1− β2x2

)
tanh−1(βx)

x3 + α2x
dx

= −4 log3(−αβ + i) + 6iπ log2(−αβ + i)− 4 log3(αβ − i) + 6iπ log2(αβ − i) + 4iπ3

48α2

(35)

Next we apply L’Hopital’s rule to the right-hand side as α→ 0 and upon inspection
of this closed form solution we are able to write down the conditional form given by

∫ ∞
0

log
(
1− β2x2

)
tanh−1(βx)

x3
dx =


1
2 iπβ

2 if Im(β) < 0,
1
2 iπβ|β| if Im(β) = 0,

−1
2 iπβ|β| if Im(β) > 0.

(36)

We can also expand the left-hand side of equation (35) to get

(37)

∫ ∞
0

(
log
(
1− β2x2

)
tanh−1(βx)

x
−
x log

(
1− β2x2

)
tanh−1(βx)

α2 + x2

)
dx

=
1

48

(
−4 log3(−αβ + i)− 6iπ log2(−αβ + i)− π2 log(−αβ + i)

− (π + i log(αβ − i))
(
2(π + 2i log(αβ − i)) log(αβ − i) + 3iπ2

))
and take the first partial derivative with respect to α, replacing β by −iβ simplifying

to get

(38)

∫ ∞
0

x log
(
β2x2 + 1

)
tan−1(βx)

(α2 + x2)2
dx =

πβ log
(
(αβ + 1)2

)
4α(αβ + 1)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (3) (2021), 723-736 734

23. Definite integral representation for π

Using (10) and setting b = c = k = 1 and simplifying we get

∫ ∞
0

(1− x)m log(1− x) + (x+ 1)m log(x+ 1)

a+ x2
dx =

πe
iπm
2 (
√
a− i)m log

(
− (
√
a− i)2

)
2
√
a

(39)

Next we apply L’Hopital’s rule as a→ 0 to the right-hand side to get

(40)

∫ ∞
0

(1− x)m log(1− x) + (x+ 1)m log(x+ 1)

x2
dx = iπ

Next we take the definite integral over m ∈ [0,m] to get

(41)

∫ ∞
0

(1− x)m + (x+ 1)m − 2

x2
dx = iπm

where −1 < Re(m) < 1.



REFERENCES 735

24. Table of integrals

f(x)
∫∞
0 f(x)dx

tan−1(px)
x3+x

1
2π log(p+ 1)

tan−1(px)
x−x3

1
4π(2 log(p+ i)− iπ)

tan−1(qx)
p2x+x3

π log(pq+1)
2p2

tan−1(qx)
x−p2x3

1
4π log

(
p2+q2

p2

)
x tan−1(bx)

(a2+x2)2
πb

4a(ab+1)

log(β+µx2)
γ+x2

π log(
√
β+
√
γ
√
µ)√

γ

log(a2+b2x2)
c2+g2x2

π log
(
ag+bc
g

)
cg

log(a2+b2x2)
c2−g2x2 −

iπ log
(
− i(bc+iag)

g

)
cg

log(p2x2+1)−log(q2x2+1)
x2

π(p− q)
log(p2x2+1)
q2x2+r2

π log
(
pr+q
q

)
qr

log(a2+b2x2)
(c2+g2x2)2

π log
(
ag+bc
g

)
2c3g

− πb
2c2g(ag+bc)

x2 log(a2+b2x2)
(c2+g2x2)2

πb
2g3(ag+bc)

+
π log

(
ag+bc
g

)
2cg3

log(1−x2)
(x2+1) log(1−x) log(x+1)

4π
log(4)+iπ

log(log(1−x) log(x+1))
x2+1

π log
(
1
4 i(π − 2i log(2))

)
(1−x)m log(1−x)+(x+1)m log(x+1)

x2
iπ

Table 1: Table of definite integrals

Acknowledgements

Supported by The Natural Sciences and Engineering Research Council of Canada
(NSERC), Grant number 504070.

References

[1] D. Bierens de Haan. Nouvelles tables d’intégrales définies. P. Engels, Leiden, 1867.

[2] A. Erdélyi. Tables of Integral Transforms. McGraw-Hill, New York, 1954.



REFERENCES 736

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vier/Academic Press, Amsterdam, seventh edition, 2007. Translated from the Rus-
sian, Translation edited and with a preface by Alan Jeffrey and Daniel Zwillinger,
With one CD-ROM (Windows, Macintosh and UNIX).

[4] A Reynolds, R.; Stauffer. A definite integral involving the logarithmic function in
terms of the lerch function. Mathematics, 1148(7), 2019.

[5] A Reynolds, R.; Stauffer. Definite integral of arctangent and polylogarithmic func-
tions expressed as a series. Mathematics, 1099(7), 2019.

[6] A Reynolds, R.; Stauffer. A definite integral involving the logarithmic function in
terms of the lerch function. Mathematics, 687(7), 2020.

[7] A Reynolds, R.; Stauffer. Definite integrals involving product of logarithmic functions
and logarithm of square root functions expressed in terms of special functions. AIMS
Mathematics, (5), 2020.

[8] A Reynolds, R.; Stauffer. Integrals in gradshteyn and ryzhik: hyperbolic and alge-
braic functions. International Mathematical Forum, 15(6):255–263, 2020.

[9] A Reynolds, R.; Stauffer. A method for evaluating definite integrals in terms of
special functions with examples. International Mathematical Forum, 15(5):235–244,
2020.