EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 1, 2022, 229-237
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Definite Integrals involving Logarithmic Powers,
Binomials and Polynomials expressed in terms of the

Lerch Function

Robert Reynolds1,∗, Allan Stauffer1

1 Department of Mathematics and Statistics, Faculty of Science, York University, Toronto,
Ontario, Canada, M3J1P3

Abstract. Closed expressions using the Lerch function for a definite integral are derived and
evaluated. Some of these closed expressions are given in Gradshteyn and Ryzhik. Some special
cases of the integral are derived and discussed. The majority of the results in this work are new.

2020 Mathematics Subject Classifications: 30E20, 33-01, 33-03, 33-04, 33-33B

Key Words and Phrases: Entries of Gradshteyn and Ryzhik; Lerch function; analytic continu-
ation

1. Introduction

In this manuscript we focus on the derivation of the definite integral given by∫ 1

0

xm logk(ax)− x−m logk
(
a
x

)
x2 − 1

dx, (1)

which has a closed form solution in terms of the Lerch function. In our case the parameters
in the formula are general complex numbers subject to the restrictions given below. This
integral and its closed form solution are important because it allows us to provide deriva-
tions for integrals in the books of Gradshteyn and Rhyzik [6] and Birens de haan [8]. We
also derive new forms of definite integrals such as tan−1(log(x)) not available in current
literature. Since equation (1) is expressed in terms of the Lerch function, all solutions
of the integrals are analytically continued which widens the range of computation. The
derivations follow the method used by us in [10]. This method involves using a form of
the generalized Cauchy’s integral formula given by

yk

Γ(k + 1)
=

1

2πi

∫
C

ewy

wk+1
dw. (2)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i1.4153

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

http://www.ejpam.com 229 © 2022 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 230

where C is in general an open contour in the complex plane where the bilinear concomitant
[10] has the same value at the end points of the contour. Then we multiply both sides by a
function and 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 (2) by another function
and take the infinite sum of both sides such that the contour integral of both equations
are the same.

2. Definite integral of the contour integral

We use the method in [10]. Here we use the contour of Figure 2 in [10] but for the
z-plane where z = m+w except we replace the vertical lines ±0 by ±ℜ(a). Note Figure 2
represents a Hankel contour which is in the z-plane, with the cut along the positive y-axis
and the contour on opposite sides of the cut but along the y-axis. Using a generalization
of Cauchy’s integral formula we first replace y by log(ax) then y by log(a/x) takig their
difference followed by multiplying both sides by 1

x2−1
then taking the definite integral with

respect x ∈ [0, 1] to get

(3)

∫ 1

0

xm logk(ax)− x−m logk
(
a
x

)
x2 − 1

dx =
1

2πi

∫ 1

0

∫
C

aww−k−1 (xm+w − x−m−w)

x2 − 1
dwdx

=
1

2πi

∫
C

∫ 1

0

aww−k−1 (xm+w − x−m−w)

x2 − 1
dxdw

=
1

2πi

∫
C

1

2
πaww−k−1 tan

(
1

2
π(m+ w)

)
dw,

from (3.269.3) in [6] where the digamma function ψ0(x) can be written out using equation
(44:5:3) in [9] and −1 < ℜ(m+ w) < 1. The logarithmic function is given for example in
section (4.1) in [1]. We are able to switch the order of integration over z = w +m and x
using Fubini’s theorem since the integrand is of bounded measure over the space C× [0, 1].

3. The Lerch function

The Lerch function [3] has a series representation given by

Φ(z, s, v) =
∞∑
n=0

(v + n)−szn, (4)

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

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 (5)

where ℜ(v) > 0, and either |z|≤ 1, z ̸= 1,ℜ(s) > 0, or z = 1,ℜ(s) > 1.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 231

4. Infinite sum of the contour integral

Using equation (2) and replace y by log(a) + iπ(y + 1) then multiply both sides by
−iπ(−1)yeiπm(y+1) followed by taking the infinite sum over y ∈ [0,∞), simplify to get

(6)

−
(iπ)k+1eiπmΦ

(
−eimπ,−k, 1− i log(a)

π

)
Γ(k + 1)

= − 1

2πi

∞∑
y=0

∫
C
iπ(−1)yw−k−1 exp(w(log(a) + iπ(y + 1)) + iπm(y + 1))dw

= − 1

2πi

∫
C

∞∑
y=0

iπ(−1)yw−k−1 exp(w(log(a) + iπ(y + 1)) + iπm(y + 1))dw

=
1

2πi

∫
C

(
1

2
πaww−k−1 tan

(
1

2
π(m+ w)

)
− 1

2
iπaww−k−1

)
dw.

from equation (1.232.1) in [6], where ℑ(m+ w) > 0 in order for the sum to converge.

5. The additional contour integral

Using equation (2) and replace y by log(a) followed by multiplying both sides by π
2i to

get

− iπ log
k(a)

2Γ(k + 1)
= − 1

2πi

∫
C

1

2
iπaww−k−1dw, (7)

6. A Note on the Hypergeometric function

In this manuscript we will derive definite integrals in terms of the Lerch function which
simplify to the Hypergeometric function by equation (1.11.10) in [4].

Φ(z, 1, v) =

∞∑
n=0

zn

n+ v
= v−1

2F1(1, v, 1 + v; z). (8)

7. The definite integral in terms of the Lerch function

Since the right-hand sides of equations (3), (6) and (7) are equal we may equate the
left hand sides simplify to get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 232∫ 1

0

xm logk(ax)− x−m logk
(
a
x

)
x2 − 1

dx =
1

2
iπ

(
logk(a)− 2(iπ)keiπmΦ

(
−eimπ,−k, 1− i log(a)

π

))
,

(9)

where −1 < ℜ(m) < 1.

8. Derivation of entry 4.282.13 in [6]

Using equation (9) first replacing a by eqi then setting k = −1, we then replace m
by p and −p to get a second equation, then taking the difference of these two equations
simplify we get

∫ 1

0

(xp − x−p)

(x2 − 1)
(
q2 + log2(x)

)dx =
ie−iπp

2q

(
Φ

(
−e−ipπ, 1,

q + π

π

)
− e2iπpΦ

(
−eipπ, 1, q + π

π

))

=
iπe−iπp

2q(q + π)

(
2F1

(
1,
q + π

π
;
q

π
+ 2;−e−ipπ

)
− e2iπp 2F1

(
1,
q + π

π
;
q

π
+ 2;−eipπ

))
.

(10)

This solution represents the analytic continuation of the integral in [6]. The solution listed
in [6] is slowly convergent and limited in the variable domain of evaluation.

9. Derivation of entry 4.282.4 in [6]

In this section we will use the formula 2F1(1, 2; 3; z) = −2(z+log(1−z))
z2

where z = −1,
which is derived from section (15.2) (relations between contiguous functions) in [1]. Using
equation (10) then taking the first partial derivative with respect to p followed by setting
q = π and p = 0 we get∫ 1

0

log(x)

(x2 − 1)
(
log2(x) + π2

)dx =
1

4
(log(4)− 1). (11)

10. Derivation of entry 4.282.8 in [6]

In this section we will use the formula 2F1

(
1, 32 ;

5
2 ; z
)
= −3

z+
3 tanh−1(

√
z)

z3/2
where z = −1,

which is derived from section (15.2) (relations between contiguous functions) in [1]. Using
equation (10) then taking the first partial derivative with respect to p followed by setting
q = π/2 and p = 0 we get∫ 1

0

log(x)

(x2 − 1)
(
4 log2(x) + π2

)dx =
1

16
(π − 2). (12)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 233

11. Derivation of entry 4.282.10 in [6]

In this section we will use the formula 2F1

(
1, 54 ;

7
4 ; z
)
=

3Bz( 3
4
, 1
2)

4
√
1−zz3/4

where z = −1,

which is derived from section (15.2) (relations between contiguous functions) in [1]. Using
equation (10) then taking the first partial derivative with respect to p followed by setting
q = π/4 and p = 0 we get

∫ 1

0

log(x)

(x2 − 1)
(
16 log2(x) + π2

)dx =
1

64

(
−4 +

√
2π + 2

√
2 log

(
cot
(π
8

)))
. (13)

12. A special case in terms of the Hypergeometric function

Using equation (9) and first replacing a and eai then setting k = −1, and replacing m
by −m to form a second equation and adding both simplify to get

∫ 1

0

x−p
(
x2p + 1

)
log(x)

(x2 − 1)
(
q2 + log2(x)

)dx
=

1

2
π

(
1

q
− e−iπp

q + π

(
2F1

(
1,
q + π

π
;
q

π
+ 2;−e−ipπ

)
+ e2iπp 2F1

(
1,
q + π

π
;
q

π
+ 2;−eipπ

)))
.

(14)

13. A special case in terms of the Polylogarithm function

Using equation (9) and first setting a = 1 and replacing m by −m to form a second
equation and subtracting both simplify to get

∫ 1

0

x−m
(
x2m − 1

)
logk(x)

x2 − 1
dx = −1

2
ieiπkπk+1 sec

(
πk

2

)(
Li−k

(
−e−imπ

)
− Li−k

(
−eimπ

))
,

(15)
from equation (6) in [7].

14. A special case in terms of the Lerch function

Using equation (9) and first setting k = −2 and replacing a by eqi then replacing m
by −m to form a second equation and subtracting both simplify to get

∫ 1

0

x−m
(
x2m − 1

) (
q2 − log2(x)

)
(x2 − 1)

(
q2 + log2(x)

)2 dx



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 234

=
ie−iπm

2π

(
Φ

(
−e−imπ, 2,

q + π

π

)
− e2iπmΦ

(
−eimπ, 2,

q + π

π

))
. (16)

15. Definite integral of nested logarithm function in terms of the
derivative of the Polylogarithm function

Using equation (9) and first setting k = −2 and replacing a by eqi then replacing m
by −m to form a second equation and subtracting both simplify to get

(17)

∫ 1

0

1
√
x(x+ 1)

(
log2(x) + π2

)dx =
log(2)

2π

and∫ 1

0

iπ log
(
log2(x) + π2

)
+ log(x) log

(
π+i log(x)
π−i log(x)

)
√
x(x+ 1)

(
log2(x) + π2

) dx = iLi′1(−i) + iLi′1(i) + i log(2) log(π),

(18)

from equation (27) in [2].

16. Derivation of entry BI(131)(3) in [8]

In this section we will use the formula 2F1(1, 2; 3; z) = −2(z+log(1−z))
z2

where z = −1,
which is derived from section (15.2) (relations between contiguous functions) in [1]. Using
equation (14) and setting q = π simplify we get

∫ 1

0

x−p
(
x2p + 1

)
log(x)

(x2 − 1)
(
log2(x) + π2

)dx =
1

2
(πp sin(πp) + cos(πp) log(2(cos(πp) + 1))− 1). (19)

17. Derivation of entry BI(131)(4) in [8]

In this section we will use the formula 2F1(1, 1; 2; z) = − log(1−z)
z which is derived from

section (15.2) (relations between contiguous functions) in [1]. Using equation (10) and
setting q = π simplify we get

(20)

∫ 1

0

x−p
(
x2p − 1

)
(x2 − 1)

(
log2(x) + π2

)dx =
i
(
e−iπp log

(
1 + eiπp

)
− eiπp log

(
1 + e−iπp

))
2π

=
sin(πp) log(2(cos(πp) + 1))− πp cos(πp)

2π
.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 229-237 235

18. Derivation of arctangent logarithmic integrals

In this section we will look at deriving definite integrals of the arctangent of the
logarithmic function. We will also derive integrals in terms of π and the loggmma function.

Using (9) and setting m = 0 simplifying we get

∫ 1

0

logk(ax)− logk
(
a
x

)
x2 − 1

dx

=
1

2

(
(2iπ)k+1

(
ζ

(
−k, 1− i log(a)

2π

)
− ζ

(
−k, π − i log(a)

2π

))
+ iπ logk(a)

)
, (21)

from equations (64:5:3) in [9] and (25.14.2) in [5]. Then we take the first partial derivative
with respect to k then set k = 0 and replace a = ea simplifying to get

∫ 1

0

tanh−1
(
log(x)

a

)
x2 − 1

dx =
1

8
π

(
−4ilogΓ

(
− ia

2π

)
+ 4ilogΓ

(
− ia+ π

2π

)
−4i log(−ia) + 2i log(a) + 4i log(−π − ia) + π − 2i log(2π)) (22)

Next we replace a by − 1
ai simplifying to get

∫ 1

0

tan−1(a log(x))

x2 − 1
dx =

1

8
iπ

π + 4i log


√

i
aΓ
(
π+ 1

a
2π

)
√
2πΓ

(
1 + 1

2πa

)

 (23)

where Re(a) > 0.

18.1. Example 1

Using equation (23) and setting a = 1 simplifying we get∫ 1

0

tan−1(log(x))

x2 − 1
dx =

1

4
π log

(
2πΓ

(
1 + 1

2π

)2
Γ
(
1+π
2π

)2
)

(24)

18.2. Example 2

Using equation (23) and setting a = 1/π simplifying we get

∫ 1

0

tan−1
(
log(x)

π

)
x2 − 1

dx =
1

4
π log

(π
2

)
(25)



REFERENCES 236

18.3. Example 3

Using equation (23) and setting a = 1/(2π) simplifying we get

∫ 1

0

cot−1
(

2π
log(x)

)
x2 − 1

dx =
1

4
π log

(
4

π

)
. (26)

19. Discussion

In comparing our results with Table 4.282 in [6], our formulae have a wider range of
the parameters than are listed in the Gradshteyn and Ryzhik book [6] due to the use of
the Lerch function in the derivation of these integrals. We also provided correct formula
for an integral supplied by Bierens de Haan. We will be looking at other integrals using
this contour integral method for future work.

20. Conclusion

In this paper, we have presented a novel method for deriving some interesting definite
integrals not previously published in literature using contour integration. The results
presented were numerically verified for both real and imaginary and complex values of the
parameters in the integrals using Mathematica by Wolfram.

Acknowledgements

This research is supported by NSERC Canada under grant 504070.

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