EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 4, 2021, 1132-1147
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Definite integral of a hyperbolic quotient function
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. In applied sciences it is always useful to improve the catalogue of definite integrals
available in tables. This present paper is a compendium of definite integrals involving a hyperbolic
quotient function expressed in terms of the Lerch function. A substantial portion of the results are
new.

2020 Mathematics Subject Classifications: 30E20, 33-01, 33-03, 33-04, 33-33B, 33E20, 33E33

Key Words and Phrases: Entries in Gradshteyn and Ryzhik, Mellin transform, Hyperbolic
function, Definite integral, Catalan’s constant

1. Significance Statement

The works of Bonderson and Gorda [2, 4] are concerned with some very interesting
topics in particle and plasma physics. Within these articles the authors used some math-
ematical formula from the book of Gradshteyn and Ryzhik [8]. In this paper the authors
provided a formal derivation for the formulae used in [2, 4] along with deriving generalized
forms for some known and new integrals. The definite integrals derived in this work are
useful in applications, in particular in perturbation analysis of solitons and plasma physics
[1, 3, 5, 6, 9, 14].

The derived integral formula in this present work is expressed in terms of Lerch function.
The Lerch function being a special function has the fundamental property of analytic con-
tinuation, which enables us to widen the range of evaluation for the parameters involved.
We provide formal derivations of some formula in the books of [8] and [12] not previously
published to the best of our knowledge.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4029

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

http://www.ejpam.com 1132 © 2021 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1133

2. Introduction

The definite integral derived in this manuscript is given by∫ ∞

0
csch2(cx)

(
e−2mx(log(a)− 2x)k + e2mx(log(a) + 2x)k − 2 logk(a)

)
dx (1)

where the parameters k, a are general complex numbers and |Re(c)|> m. The derivation of
the definite integral follows the method used by us in [13] which involves Cauchy’s integral
formula. The generalized Cauchy’s integral formula is given by

yk

Γ(k + 1)
=

1

2πi

∫
C

ewy

wk+1
dw. (2)

where C is in general an open contour in the complex plane where the theorem in [13] gives
the result. This method involves using a form of equation (2) then multiply both sides
by a different function, then take 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 (2) by a function and performing some substitutions so that the contour integrals
are the same.

3. Definite integral of the contour integral

We use the method in [13]. We will derive three contour integral representations and
add them such that we get an equivalent form for the infinite sum. Deriving the first
contour we use equation (2) and replace y by log(a) and multiply by −1

2csch
2(cx) to get

− logk(a)csch2(cx)

2Γ(k + 1)
= − 1

4πi

∫
C
aww−k−1csch2(cx)dw (3)

Deriving the second contour we use equation (2) and replace y by log(a)−2x and multiply
by 1

8e
−2mxcsch2(cx) to get

e−2mxcsch2(cx)(log(a)− 2x)k

8Γ(k + 1)
=

1

16πi

∫
C
aww−k−1csch2(cx)e−2x(m+w)dw (4)

Deriving the third contour we use equation (2) and replace y by log(a) + 2x and multiply
by 1

4e
2mxcsch2(cx) to get

e2mxcsch2(cx)(log(a) + 2x)k

4Γ(k + 1)
=

1

8πi

∫
C
aww−k−1csch2(cx)e2x(m+w)dw (5)

Next we add equations (3), (4) and (5) then take the infinite integral over x ∈ [0,∞) to
get

1

Γ(k + 1)

∫ ∞

0
csch2(cx)

(
e−2mx(log(a)− 2x)k + e2mx(log(a) + 2x)k − 2 logk(a)

)
dx



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1134

=
1

2πi

∫ ∞

0

∫
C
aww−k−1csch2(cx) sinh2(x(m+ w))dwdx

=
1

2πi

∫
C

∫ ∞

0
aww−k−1csch2(cx) sinh2(x(m+ w))dxdw

=
1

2πi

∫
C

aww−k−1
(
c− π(m+ w) cot

(
π(m+w)

c

))
2c2

dw (6)

from equation (2.4.4.2) in [12] where z = m+w
c , −1 < Re(z) < 1 and Im(z) > 0,

|Re(c)|> m. 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,∞).

4. The Lerch function

We use (9.550) and (9.556) in [8] where Φ(z, s, v) is the Lerch function which is a
generalization of the Hurwitz zeta ζ(s, v) and Polylogarithm functions Lin(z). The Lerch
function has a series representation given by

Φ(z, s, v) =

∞∑
n=0

(v + n)−szn (7)

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 (8)

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

4.1. Infinite sum of the first contour integral

In this section we will again use Cauchy’s integral formula (2) and taking the infinite
sum to derive equivalent sum representations for the contour integrals. We proceed using

equation (2) and replace y by log(a) + 2iπ(y+1)
c and multiply both sides by iπm

c2
e

2iπm(y+1)
c

and simplifying to get

πk+1m
(
i
c

)k+1
e

2iπm(y+1)
c

(
− ic log(a)

π + 2y + 2
)k

cΓ(k + 1)

=
1

2πi

∫
C

iπmaww−k−1e
2iπ(y+1)(m+w)

c

c2
dw (9)

Next we take the infinite sum over y ∈ [0,∞) and simplify using the Lerch function to get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1135

2kπk+1m
(
i
c

)k+1
e

2iπm
c Φ

(
e

2imπ
c ,−k, 1− ic log(a)

2π

)
cΓ(k + 1)

=
1

2πi

∞∑
y=0

∫
C

iπmaww−k−1e
2iπ(y+1)(m+w)

c

c2
dw

=
1

2πi

∫
C

∞∑
y=0

iπmaww−k−1e
2iπ(y+1)(m+w)

c

c2
dw

= − 1

2πi

∫
C

πmaww−k−1
(
cot
(
π(m+w)

c

)
+ i
)

2c2
dw (10)

from (1.232.1) in [8] and Im(w +m) > 0 for convergence of the sum.

4.2. Infinite sum of the second contour integral

The derivation in this section is equivalent to Section (2.1) after multiplication by m
and replacement of k → (k − 1).

2k−1πk
(
i
c

)k
e

2iπm
c Φ

(
e

2imπ
c , 1− k, 1− ic log(a)

2π

)
c(k − 1)!

= − 1

2πi

∫
C

πaww−k
(
cot
(
π(m+w)

c

)
+ i
)

2c2
dw (11)

from (1.232.1) in [8] and Im(w +m) > 0 for convergence of the sum.

4.3. Additional contours

In this section we will derive the additional contours from equations (6), (10) and (11).
We proceed using equation (2) and replace y by log(a) and multiply both sides by 1

2c and
simplifying to get

logk(a)

2cΓ(k + 1)
=

1

2πi

∫
C

aww−k−1

2c
dw (12)

Next using equation (2) and replace y by log(a) and multiply both sides by − iπm
2c2

and
simplifying to get

iπm logk(a)

2c2Γ(k + 1)
= − 1

2πi

∫
C

iπmaww−k−1

2c2
dw (13)

Next using equation (2) and replace y by log(a), k by k − 1 and multiply both sides by
− iπ

2c2
and simplifying to get



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1136

− iπ logk−1(a)

2c2Γ(k)
= − 1

2πi

∫
C

iπaww−k

2c2
dw (14)

Note that the equations (12), (13) and (14) are equivalent to (3) with csch(cx) dropped
and different multipliers.

5. Definite integral in terms of the Lerch function

Theorem 1. For all k, a ∈ C, |Re(c)|> m,∫ ∞

0
csch2(cx)

(
e−2mx(log(a)− 2x)k + e2mx(log(a) + 2x)k − 2 logk(a)

)
dx

=
i2k+2πk+1m

(
i
c

)k
e

2iπm
c Φ

(
e

2imπ
c ,−k, 1− ic log(a)

2π

)
c2

+
2iπm logk(a)

c2

+
2iπk logk−1(a)

c2
+

2k+1kπk
(
i
c

)k
e

2iπm
c Φ

(
e

2imπ
c , 1− k, 1− ic log(a)

2π

)
c

+
2 logk(a)

c
(15)

Proof. Since the right-hand side of equation (6) is equal to the sum of the right-hand
sides of equations (10), (11), (12), (13) and (14), we can equate the left-hand sides to
achieve the stated result.

6. Derivation of entry (2.4.4.2) in [12]

Corollary 1. For |Re(c)|> m,∫ ∞

0
csch2(cx) sinh2(mx)dx =

c− πm cot
(
πm
c

)
2c2

(16)

Proof. Use equation (15) and set k = 0 and simplify using entry (2) in Table below
(64:12:7) in [11].

7. Derivation of new entry for Table 2.4.4 in [12]

Corollary 2. For all Re(β) > |Re(α)|/2,

∫ ∞

0
x sinh(αx) csch2(βx)dx =

π
(
πα− β sin

(
πα
β

))
csc2

(
πα
2β

)
4β3

(17)

Proof. Use equation (15) and set k = 1, a = 1,m = α/2, c = β and simplify using
entry (1) in Table below (64:12:7) in [11].



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1137

8. Derivation of the Mellin transform

In this section we will derive a generalize Mellin transform of equations (2.3.1.19) in
[7], (3.527.8) and (3.527.9) in [8].

Theorem 2. For all Re(s) > 0, Re(β) =
√
|Re(α)|,∫ ∞

0
xs−1 cosh(αx) csch2(βx)dx

= −1

2
πs−1

(
1

β

)s

csc
(πs
2

)
Li2−s

(
e
− iπα

β

)
+

1

2
πs−1s

(
1

β

)s

csc
(πs
2

)
Li2−s

(
e
− iπα

β

)
− 1

2
πs−1

(
1

β

)s

csc
(πs
2

)
Li2−s

(
e

iπα
β

)
+

1

2
πs−1s

(
1

β

)s

csc
(πs
2

)
Li2−s

(
e

iπα
β

)
− 1

2
iαπs

(
1

β

)s+1

csc
(πs
2

)
Li1−s

(
e
− iπα

β

)
+

1

2
iαπs

(
1

β

)s+1

csc
(πs
2

)
Li1−s

(
e

iπα
β

)
(18)

Proof. Use equation (15) set m = m/2, a = e2a and replace m by −m to form a second
equation and take their difference to get

∫ ∞

0

(
(a− x)k − (a+ x)k

)
csch2(cx) sinh(mx)dx

= − iπmak

c2
−

iπk+1m
(
i
c

)k
e−

iπm
c Φ

(
e−

imπ
c ,−k, 1− iac

π

)
c2

−
iπk+1m

(
i
c

)k
e

iπm
c Φ

(
e

imπ
c ,−k, 1− iac

π

)
c2

+
kπk

(
i
c

)k
e−

iπm
c Φ

(
e−

imπ
c , 1− k, 1− iac

π

)
c

−
kπk

(
i
c

)k
e

iπm
c Φ

(
e

imπ
c , 1− k, 1− iac

π

)
c

(19)

Next set a = 0,m = α, c = β, k = s− 1 and simplify using equation (64:12:2) in [11].

9. Derivation of (3.527.1) in [8]

Theorem 3. For all Re(a) > 0, Re(µ) > 2,∫ ∞

0
xµ−1 csch2(ax)dx = 22−µ

(
1

a

)µ

Γ(µ)ζ(µ− 1) (20)

Proof. Use equation (18) and set α = 0, s = µ, β = a and simplify using equation
(25:12:5) in [11].



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1138

10. Derivation of entry (3.527.2) in [8]

Theorem 4. For all Re(β) > 0,∫ ∞

0
x2m csch2(βx)dx = π2m

(
1

β

)2m+1

|B2m| (21)

Proof. Use equation (20) and set µ = 2m + 1 and simplify using equation (3:13:1) in
[11].

11. Derivation of entry (3.527.9) in [8]

Theorem 5. For all Re(m) > 0, Re(a) > 0,∫ ∞

0
x2m+1 coth(ax) csch(ax)dx

=
(
1− 22m+1

)
(2m+ 1)π2m+1

(
1

a

)2m+2

ζ(−2m) csc(πm) (22)

Proof. Use equation (18) and set s = 2m+ 2, α = β = a and simplify using entry (4)
in Table below (25:12:5) in [11].

12. Derivation of entry (3.527.10) in [8]

Corollary 3. For all Re(a) > 0, Re(m) > 1/2,∫ ∞

0
x2m coth(ax) csch(ax)dx = 41−m (4m − 1)m

(
1

a

)2m+1

ζ(2m)Γ(2m) (23)

Proof. Use equation (18) and set s = 2m + 1, α = a, β = a and simplify using entry
(4) in Table below (3:13:1) in [11].

13. Derivation of entry (3.527.12) in [8]

Corollary 4. ∫ ∞

0
x2 csch2(x)dx =

π2

6
(24)

Proof. Use equation (21) and set m = β = 1 and simplify.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1139

14. Derivation of entry (3.527.13) in [8]

Corollary 5. For all Re(a) > 0∫ ∞

0
x2 coth(ax) csch(ax)dx =

π2

2a3
(25)

Proof. Use equation (18) and set s = 3, α = β = a and simplify.

15. Derivation of entry (3.527.16) in [8] and (2.4.5.12) in [12]

Theorem 6. For all Re(a) > 0, Re(µ) > 2∫ ∞

0
xµ−1 coth(ax) csch(ax)dx = 21−µ (2µ − 2)

(
1

a

)µ

Γ(µ)ζ(µ− 1) (26)

Proof. Use equation (18) and set s = µ, α = β = a and simplify using entry (4) in
Table below (25:12:5) in [11].

16. Derivation of entry (2.3.1.9) in Brychkov, (3.523.1) in [8]

Theorem 7. For all Re(s) > 1, Re(α) > 0∫ ∞

0
xs−1 csch(αx)dx = 21−s (2s − 1)

(
1

α

)s

ζ(s)Γ(s) (27)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β, s = s− 1 and simplify using entry (4) in Table below (25:12:5) in [11].

17. Derivation of entry (3.523.2) in [8]

Theorem 8. For all Re(a) > 0, n = 1, 2, ..∫ ∞

0
x2n−1 csch(αx)dx =

(4n − 1)π2n
(
1
α

)2n |B2n|
2n

(28)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β, s = 2n and simplify using equation (3:13:1) in [11].

18. Derivation of entry (3.523.6) in [8]

Corollary 6. ∫ ∞

0
x3 csch(x)dx =

π4

8
(29)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β = 1, s = 3 and simplify using entry (2) in Table below (25:12:5) in [11].



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1140

19. Derivation of entry (3.523.8) in [8]

Corollary 7. ∫ ∞

0
x5 csch(x)dx =

π6

4
(30)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β = 1, s = 5 and simplify using entry (2) in Table below (25:12:5) in [11].

20. Derivation of entry (3.523.10) in [8]

Corollary 8. ∫ ∞

0
x7 csch(x)dx =

17π8

16
(31)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β = 1, s = 7 and simplify using entry (2) in Table below (25:12:5) in [11].

21. Derivation of entry (3.521.1) in [8]

Theorem 9. For α ∈ C,

∫ ∞

0
x csch(αx)dx =


π2

4α2
, for Re(α) > 0

− π2

4α2
, for Re(α) < 0

(32)

Proof. Use equation (18) and take the first partial derivative with respect to α then
set α = β, s = 1 and simplify using entry (2) in Table below (25:12:5) in [11].

22. Derivations in terms of Catalan’s constant C and π

Corollary 9.∫ ∞

0

csch(x)
(
−4x2 csch(x) + π2 coth(x)− π2 csch(x)

)
4x2 + π2

dx = −2(C − 1) (33)

Corollary 10. ∫ ∞

0

x csch(x)

4x2 + π2
dx =

1

8
(π − 2) (34)

Proof. Use equation (15) and set k = −1, a = −1, c = 1,m = 1/2 and simplify in
terms of Catalan’s constant C, using entries (1) and (4) in Tables below (64:12:7) in [11]
and equations (2.3) and (2.7) in [10] and equation (9.73) in [8] and simplify in terms of
the real and imaginary parts.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1141

Corollary 11. ∫ ∞

0

x csch(x)

x2 + π2
dx = log(2)− 1

2
(35)

Corollary 12. ∫ ∞

0

x sinh(x) csch2(2x)

x2 + π2
dx =

8C − 8 + π(log(4)− 1)

8π
(36)

Proof. Use equation (19) and set k = −1, a = πi,m = 1, c = 1 and k = −1, a = πi,m =
1, c = 2 respectively and simplify using entries (1) and (4) in Tables below (64:12:7) in
[11] and equations (2.3) and (2.7) in [10].

23. Derivation of a new entry for Table 2.4.5 in [12]

Theorem 10. For all Re(c) > 0, Re(m) > 0, z ∈ C,∫ ∞

0

x csch2(cx) sinh(mx)

x2 + z2
dx

=
πm

2c2z
+

me−
iπm
c Φ

(
e−

imπ
c , 1, czπ + 1

)
2c

+
me

iπm
c Φ

(
e

imπ
c , 1, czπ + 1

)
2c

−
ie−

iπm
c Φ

(
e−

imπ
c , 2, czπ + 1

)
2π

+
ie

iπm
c Φ

(
e

imπ
c , 2, czπ + 1

)
2π

(37)

Proof. Use equation (19) and set k = −1, a = zi and simplify.

24. Derivations in terms of Catalan’s constant C and π

Theorem 11. For all k, a ∈ C, Re(c) > 0,∫ ∞

0
2 csch2(cx)

(
cosh(2mx)

(
(log(a)− 2x)k + (log(a) + 2x)k

)
− 2 logk(a)

)
dx

= −
i2k+2πk+1m

(
i
c

)k
e−

2iπm
c Φ

(
e−

2imπ
c ,−k, 1− ic log(a)

2π

)
c2

+
i2k+2πk+1m

(
i
c

)k
e

2iπm
c Φ

(
e

2imπ
c ,−k, 1− ic log(a)

2π

)
c2

+
4iπk logk−1(a)

c2
+

2k+1kπk
(
i
c

)k
e−

2iπm
c Φ

(
e−

2imπ
c , 1− k, 1− ic log(a)

2π

)
c

+
2k+1kπk

(
i
c

)k
e

2iπm
c Φ

(
e

2imπ
c , 1− k, 1− ic log(a)

2π

)
c

+
4 logk(a)

c
(38)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1142

Proof. Use equation (18) and form a second equation by replacing m by −m and
adding both and simplify.

Corollary 13. ∫ ∞

0

(
−4x2 + π2 cosh(x)− π2

)
csch2(x)

2 (4x2 + π2)
dx = 1− C (39)

Proof. Use equation (38) and set k = −1, a = −1,m = 1/2, c = 1and simplify using
entries (1) and (4) in Tables below (64:12:7) in [11] and equations (2.3) and (2.7) in [10]
and equation (9.73) in [8].

Corollary 14.∫ ∞

0

(
−4x2 + π2 cosh(x)− π2

)
csch2(2x)

4x2 + π2
dx =

1

8
(−4C + 6− π log(2)) (40)

Proof. Use equation (38) and set k = −1, a = −1,m = 1/2, c = 2and simplify using
entries (1) and (4) in Tables below (64:12:7) in [11] and equations (2.3) and (2.7) in [10]
and equation (9.73) in [8].

25. Definite integral involving the arctangent function in terms of the
log-gamma and Harmonic number functions

Theorem 12. For all a ∈ C, Re(c) > 0,

(41 )

∫ ∞

0
x tanh−1

(x
a

)
coth(cx) csch(cx)dx

=

ac
(
H− iac

2π
−H− iac+π

2π

)
+ iπ

(
−1 + 2 log

(
i(−π−iac)Γ(− iac+π

2π )
√
2π

√
a
√

i
c
cΓ(− iac

2π )

))
2c2

Proof. Use equation (19) and take the first partial derivative with respect to m. Next
set m = c, followed by taking the first partial derivative with respect to k then applying
L’Hopital’s rule as k → 0 and simplify using equations (64:10:2), (64:4:1), (44:1:1) and
entry (4) in Table below (64:12:7) in [11].

Corollary 15.∫ ∞

0
x tanh−1(x) coth(x) csch(x)dx

=
1

2
H− i

2π
− 1

2
H− i+π

2π
− iπ

2
+

3π2

4
− 1

2
iπ log

(
2π

(π + i)2

)
− iπ log

(
Γ

(
− i

2π

))
+ iπ log

(
Γ

(
− i+ π

2π

))
(42)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1143

Proof. Use equation (41) and set a = c = 1 and simplify. Note: There exists a
singularity at x = 1.

Corollary 16. For a ∈ C,

∫ ∞

0
x cot

(πx
2a

)
csc
(πx
2a

)
tanh−1

(x
a

)
dx =


− ia2

π

(
−2 + π + log

(
16Γ

(
1
4

)4
Γ
(
−1

4

)4
))

, for Im(a) > 0

ia2

π

(
−2 + π + log

(
16Γ

(
1
4

)4
Γ
(
−1

4

)4
))

, for Im(a) < 0

(43)

Proof. Use equation (41) and set c = iπ
2a and simplify.

Corollary 17.

∫ ∞

0
x tan−1(x) coth

(πx
2

)
csch

(πx
2

)
dx =

−2 + π + log

(
16Γ( 1

4)
4

Γ(− 1
4)

4

)
π

(44)

Proof. Use equation (43) and set a = i.

Corollary 18.∫ ∞

0
x tan−1

(
x√
π

)
coth

(√
πx

2

)
csch

(√
πx

2

)
dx = −2 + π + log

(
16Γ

(
1
4

)4
Γ
(
−1

4

)4
)

(45)

Proof. Use equation (43) and set a =
√
2i.

Corollary 19. For a ∈ C,

(46 )

∫ ∞

0
x cot

(πx
a

)
csc
(πx

a

)
tanh−1

(x
a

)
dx =

{
− ia2

2π (−1 + log(2) + log(π)), for Im(a) > 0
ia2

2π (−1 + log(2) + log(π)), for Im(a) < 0

Proof. Use equation (41) and apply L’Hopital’s rule as c → iπ
a and simplify.

Corollary 20. ∫ ∞

0
x tan−1(x) coth(πx) csch(πx)dx =

log(2π)− 1

2π
(47)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1144

Proof. Use equation (46) and set a = i.

Corollary 21.∫ ∞

0
x tan−1

(
x√
2π

)
coth

(√
π

2
x

)
csch

(√
π

2
x

)
dx = log(2π)− 1 (48)

Proof. Use equation (46) and set a =
√
2πi.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1145

26. Tables of results

Table 1: Table of definite integrals

f(x)
∫∞
0 f(x)dx

csch2(cx) sinh2(mx)
c−πm cot(πm

c )
2c2

x sinh(αx)csch2(βx)
π
(
πα−β sin

(
πα
β

))
csc2

(
πα
2β

)
4β3

xµ−1csch2(ax) 22−µ
(
1
a

)µ
Γ(µ)ζ(µ− 1)

x2mcsch2(βx) π2m
(

1
β

)2m+1
|B2m|

x2m coth(ax)csch(ax) 41−m (4m − 1)m
(
1
a

)2m+1
ζ(2m)Γ(2m)

x2csch2(x) π2

6

x2 coth(ax)csch(ax) π2

2a3

xµ−1 coth(ax)csch(ax) 21−µ (2µ − 2)
(
1
a

)µ
Γ(µ)ζ(µ− 1)

xs−1csch(αx) 21−s (2s − 1)
(
1
α

)s
ζ(s)Γ(s)

x2n−1csch(αx)
(4n−1)π2n( 1

α)
2n|B2n|

2n

x3csch(x) π4

8

x5csch(x) π6

4

x7csch(x) 17π8

16

xcsch(αx) π2

4α2

csch(x)(−4x2csch(x)+π2 coth(x)−π2csch(x))
4x2+π2 −2(C − 1)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1132-1147 1146
Table 2: Table of definite integrals

f(x)
∫∞
0 f(x)dx

xcsch(x)
4x2+π2

1
8(π − 2)

xcsch(x)
x2+π2 log(2)− 1

2

x sinh(x)csch2
(2x)

x2+π2
8C−8+π(log(4)−1)

8π

(−4x2+π2 cosh(x)−π2)csch
2
(x)

2(4x2+π2)
1− C

(−4x2+π2 cosh(x)−π2)csch
2
(2x)

4x2+π2
1
8(−4C + 6− π log(2))

x tan−1(x) coth
(
πx
2

)
csch

(
πx
2

) −2+π+log

(
16Γ( 1

4)
4

Γ(− 1
4)

4

)
π

x tan−1(x) coth
(
πx
2

)
csch

(
πx
2

) −2+π+log

(
16Γ( 1

4)
4

Γ(− 1
4)

4

)
π

x tan−1
(

x√
π

)
coth

(√
πx
2

)
csch

(√
πx
2

)
−2 + π + log

(
16Γ( 1

4)
4

Γ(− 1
4)

4

)
x tan−1(x) coth(πx)csch(πx) log(2π)−1

2π

x tan−1
(

x√
2π

)
coth

(√
π
2x
)
csch

(√
π
2x
)

log(2π)− 1

27. Discussion

In this work the authors derived definite integrals used in physics along with some new
forms not previously published. Some of the integral forms were expressed in terms of
fundamental constants such as Catalan’s constant and π. The integral forms derived were
achieved by the use of our contour integral method [13] and the evaluation of integrals
using the Lerch function.

The Lerch function being multivariate with its analytic continuation properties allowed
the authors to widen the range of computation for the formulae derived. We used Wol-
fram’s Mathematica software for numerical evaluations for real and imaginary values of
the parameters involved. Derivations of definite integrals in [7, 8, 12] are useful as it will
assist in providing formal proofs and verifying whether the formulae in these books are
correct. We will be using our method to produce more tables of integrals in future work.



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