EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 1, 2022, 158-168
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

A Note on the Infinite Sum of the Lerch function

Robert Reynolds1,∗, Allan Stauffer1

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

Abstract. We derive the infinite sum of the Lerch function in terms of the incomplete gamma
function and the Lerch function. Special cases are evaluated in terms of fundamentals constants.
All the results in this work are new.

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

Key Words and Phrases: Lerch function, incomplete gamma function, Catalan’s constant,
Apréy’s constant

1. Introduction

Infinite sums of special functions have been studied in the works of [1, 5, 6, 10, 13]. In
this present work we derive a new expression for the Lerch function in terms of the infinite
sum of the Incomplete gamma function given by

(1)

∞∑
n =1

2k−n+1
(
2−n

)k
em21−n

Φ
(
−e2

1−nm,−k, 2n−1 log(a) + 1
)

= −a−m(−m)−kΓ(k + 1,−m log(a))

m
− 2k+1e2mΦ

(
e2m,−k,

log(a)

2
+ 1

)
where the variables k, a,m are general complex numbers. This new expression is then
used to derive special cases in terms of fundamental constant and special functions. The
derivations follow the method used by us in [12]. 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.v15i1.4137

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

http://www.ejpam.com 158 © 2022 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 159

where y, w ∈ C and C is in general an open contour in the complex plane where the
bilinear concomitant [12] has the same value at the end points of the contour. This
method involves using a form of equation (2) then multiplies both sides by a function,
then takes the 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.

1.1. The Incomplete Gamma Function

The multivalued incomplete gamma functions [3], γ(s, z) and Γ(s, z), are defined by

γ(s, z) =

∫ z

0
ts−1e−tdt,

and

Γ(s, z) =

∫ ∞

z
ts−1e−tdt,

where Re(z) > 0. The incomplete gamma function has a recurrence relation given by

γ(s, z) + Γ(s, z) = Γ(s),

where z ̸= 0,−1,−2, ... The incomplete gamma function is continued analytically by

γ(a, ze2mπi) = e2πmiaγ(a, z),

and
Γ(s, ze2mπi) = e2πmisΓ(s, z) + (1− e2πmis)Γ(s),

where m ∈ Z. When z ̸= 0, Γ(s, z) is an entire function of s and γ(s, z) is meromorphic

with simple poles at s = −n for n = 0, 1, 2, ... with residue (−1)n

n! . These definitions are
listed in Section 8.2(i) and (ii) in [3].

2. The Lerch function

We use equation (1.11.3) in [2] where Φ(z, s, v) is the Lerch function which is a gen-
eralization 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 (3)

where |z|< 1, v ̸= 0,−1,−2,−3, .., and is continued analytically by its integral representa-
tion 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 (4)

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



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 160

3. Contour integral representation for the infinite sum of the Lerch
function

We use the method in [12]. The cut and contour are in the first quadrant of the
complex w-plane with 0 < Re(w + m) < 1. The cut approaches the origin from the
interior of the first quadrant and goes to infinity vertically 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 (2) we first replace y → log(a)+21−n(y+1) then multiply both
sides by (−1)yem21−n(y+1) and take the infinite sums over
y ∈ [0,∞) and n ∈ [1,∞) and simplify in terms of the Lerch function to get

(5)

∞∑
n =1

2k−n+1 (2−n)
k
em21−n

Φ
(
−e2

1−nm,−k, 2n−1 log(a) + 1
)

Γ(k + 1)

=
1

2πi

∞∑
n=1

∞∑
y=0

∫
C
2(−1)yaww−k−1e2

1−n(y+1)(m+w)dw

=
1

2πi

∫
C

∞∑
n=1

∞∑
y=0

2(−1)yaww−k−1e2
1−n(y+1)(m+w)dw

=
1

2πi

∫
C

∞∑
n=1

(
2−naww−k−1 tanh

(
2−n(m+ w)

)
+ 2−naww−k−1

)
dw

=
1

2πi

∫
C

(
aww−k−1 coth(m+ w)− aww−k−1

m+ w
+ aww−k−1

)
dw

from equations (5.3.8.4) in [11] and (1.232.1) in [4] where Re(w+m) > 0 and Im(w+m) >
0 in order for the sums to converge.

3.1. Additional contour integral

Using a generalization of Cauchy’s integral formula (2) we first replace y → log(a) then
multiply both sides by 2−n and take the infinite sum over
n ∈ [1,∞) to get

(6)

−
∞∑

n =1

2−n logk(a)

Γ(k + 1)
= − 1

2πi

∞∑
n=1

∫
C
2−naww−k−1dw

= − 1

2πi

∫
C

∞∑
n=1

2−naww−k−1dw

= − 1

2πi

∫
C
aww−k−1dw



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 161

4. Contour integral representation for the Lerch and incomplete gamma
functions

Using a generalization of Cauchy’s integral formula (2) we first replace y → log(a) +
2(y + 1) then multiply both sides by e2m(y+1) and take the infinite sum over y ∈ [0,∞)
and simplify in terms of the Lerch function to get

(7)−
2k+1e2mΦ

(
e2m,−k, log(a)2 + 1

)
Γ(k + 1)

=
1

2πi

∫
C
aww−k−1 coth(m+ w) + aww−k−1dw

from equation (1.232.1) in [4] where Im(w +m) > 0 in order for the sum to converge.

4.1. Additional contour integral

Using a generalization of Cauchy’s integral formula (2) we first replace y → log(a) and
simplify to get

(8)
logk(a)

Γ(k + 1)
=

1

2πi

∫
C
aww−k−1dw

4.2. Contour integral representation for the incomplete gamma function

Using a generalization of Cauchy’s integral formula (2) we first replace y → y+ log(a)
then multiply both sides by emy and take the definite integral over y ∈ [0,∞) to get

(9)
a−m(−m)−k−1Γ(k + 1,−m log(a))

Γ(k + 1)
= − 1

2πi

∫
C

aww−k−1

m+ w
dw

from equation (3.383.4) in [4] where 0 < Re(w +m) < 1.

5. Infinite sum of the Lerch function in terms of the Lerch function

Theorem 1. For all k, a,m ∈ C,

∞∑
n=1

2k−n+1
(
2−n

)k
em21−n

Φ
(
−e2

1−nm,−k, 2n−1 log(a) + 1
)

= −a−m(−m)−kΓ(k + 1,−m log(a))

m
− 2k+1e2mΦ

(
e2m,−k,

log(a)

2
+ 1

)
(10)

or

∞∑
n=1

2k−n+1
(
2−n

)k
m2−n

Φ
(
−m2−n

,−k, 2n(a− 1) + 1
)

= −2k+1mΦ(m,−k, a)−
2k+1

(
e2(a−1)

)− log(m)
2 (− log(m))−kΓ(k + 1, (1− a) log(m))

log(m)
(11)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 162

Proof. Observe that the the addition of the right-hand sides of equations (5) and (6) is
equal to the addition of the right-hand sides of equation (7), (8) and (9) so we may equate
the left-hand sides and simplify the the Gamma function to yield the stated result.

6. Special cases

In this section we will evaluate equation (10) for various parameter ranges. We will
use the following functions for simplification; exponential integral function En(z) given in
section (8.19) in [3], and the exponential integral function Ei(z) given in equation (6.11.2)
[3], and the Riemann zeta function given in equation (25.2.1) in [3].

Example 1. The degenerate case.

∞∑
n=1

21−n

(
1− 1

em21−n + 1

)
=

∞∑
n=1

2−n
(
tanh

(
m2−n

)
+ 1
)
= − 1

m
+ coth(m) + 1 (12)

Proof. Use equation (10) and set k = 0 and simplify using entry (4) in table below
(64:12:7) in [9].

Example 2. Apéry’s constant ζ(3)

(13 )
∞∑

n =1

(−1)2
−n

4n−1Φ
(
(−1)1+2−n

, 3, 1 + 2n
)
=

1

16
(−4E3(−iπ)− 3ζ(3) + 4)

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next set k = −3, a = 2 and
simplify using equation (8.19.2) in [3] and entry (2) in table below (64:7) in [9].

Example 3. Fundamental constant ζ(5)

(14 )
∞∑

n =1

(−1)2
−n

16n−1Φ
(
(−1)1+2−n

, 5, 1 + 2n
)
=

1

256
(−16E5(−iπ)− 15ζ(5) + 16)

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next set k = −5, a = 2 and
simplify using equation (8.19.2) in [3] and entry (2) in table below (64:7) in [9].

Example 4. Catalan’s constant C

∞∑
n=1

(−1)2
−n

2n−1Φ
(
(−1)1+2−n

, 2, 1 + 2n−1
)
= −2C + 3 +

1

2
πΓ

(
0,− iπ

2

)
(15)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 163

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next set k = −2, a = 3/2
and simplify using equation (8.19.2) in [3], equation (64:4:1) in [9] and equation (2.2.1.2.7)
in [7].

Example 5. The exponential integral function En(z)

∞∑
n=1

(−1)2
−n

Φ
(
(−1)1+2−n

, 1, 1 + 2n−1
)
=

1

2

(
2iEi

(
iπ

2

)
+ 4 + π

)
(16)

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next we apply l’Hopital’s
rule as k → −1 and simplify using equation (8.19.2) in [3] and equation (64:4:1) in [9].

Example 6.

∞∑
n=1

(−1)2
−n (

2−n
)3/2

Φ

(
(−1)1+2−n

,−1

2
, 1 + 2n

)
= −E− 1

2
(−iπ) +

(
2
√
2− 1

)
ζ

(
−1

2

)
+ 1 (17)

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next set a = 2 and simplify
in terms of the Riemann zeta function ζ(s) using entry (2) in table below (64:7) in [9].
Next set k = −1/2 and simplify.

Example 7.

∞∑
n=1

(−1)2
−n√

2−nΦ

(
(−1)1+2−n

,
1

2
, 1 + 2n

)
= −E 1

2
(−iπ) +

(√
2− 1

)
ζ

(
1

2

)
+ 1 (18)

Proof. Use equation (11) and set m = −1 and simplify in terms of the Hurwitz zeta
function ζ(s, a) using entry (4) in table below (64:12:7) in [9]. Next set a = 2 and simplify
in terms of the Riemann zeta function ζ(s) using entry (2) in table below (64:7) in [9].
Next set k = 1/2 and simplify.

7. Infinite sum of the Lerch function in terms of the Lerch
transformation

Theorem 2. For Re(s) < 0, 0 < v ≤ 1, 0 ≤ Im(α) < 2π,

∞∑
n=1

eα(−2−n)2−n−s+1
(
2−n

)−s
Φ
(
−e−2−nα, s, 2n(v − 1) + 1

)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 164

= 21−sαs−1
(
e2(v−1)

)α/2
Γ(1− s, (v − 1)α)− e−α21−s

(
αs−1Γ(1− s)eαv

−i(2π)s−1Γ(1− s)eαv
(
ei(

πs
2
+2πv)Φ

(
e2iπv, 1− s, 1− iα

2π

)
− e−i(πs

2
+2πv)Φ

(
e−2iπv, 1− s,

iα

2π
+ 1

)))
(19)

Proof. Use equation (12) in [8] and substitute into equation (11) and simplify.

8. Extended Vălean Infinite Sums and Infinite Products

In the book of Vălean [14] section (3.59) has some very interesting infinite sums of
trigonometric functions. In this section of the present paper we will expand on the sums
published by Vălean using a special case of equation (11) for various values of the param-
eters involved.

Lemma 1.

∞∑
n=1

(−1)2
−n

2k−n+1
(
2−n

)k
Φ
(
(−1)1+2−n

,−k, 2n(a− 1) + 1
)

= 2k+1

(
2kζ

(
−k,

a

2

)
− 2kζ

(
−k,

a+ 1

2

))
+i
(
e2(a−1)

)− iπ
2
(−i)−k2k+1π−k−1Γ(k+1, i(1−a)π)

(20)

Proof. Use equation (11) and set m = 1 and simplify using entry (5) in Table below
(64:12:7) in [9].

Example 8.

∞∑
n=1

21−2n
(
1 + α2n

(
1 + eiπ2

−n
))

cos (π2−n) + 1
= 2α+ 4α2

(
e2α
)− iπ

2 E−1(−iπα) + 1 (21)

Proof. Use equation (20) and set k = 1 and replace a → α+ 1 and simplify.

Example 9.
∞∑
n=1

8−n sin4
(
π2−n−1

)
csc3

(
π2−n

)
=

1

π3
(22)

Proof. Use equation (20) and set k = 2, a = 1 and simplify.

Example 10.
∞∑
n=1

2−4n
(
cos
(
π2−n

)
− 2
)
sec4

(
π2−n−1

)
=

48

π4
− 1 (23)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 165

Proof. Use equation (20) and set k = 3, a = 1 and simplify.

Example 11. The polylogarithm function Lin(z) and Riemann zeta function ζ(s)

∞∑
n=1

−2k−n+1
(
2−n

)k
Li−k

(
(−1)1+2−n

)
= 2k+1

(
−
(
2k+1 − 1

)
ζ(−k) + i(−i)−kπ−k−1Γ(k + 1)

)
(24)

Proof. Use equation (20) and set a = 1 and simplify using entry (1) in Table below
(64:7) in [9].

Example 12. The polylogarithm function Lin(z)

∞∑
n=1

√
21−nLi 1

2

(
(−1)1+2−n

)
= −

(√
2− 2

)
ζ

(
1

2

)
+ (−1− i) (25)

Proof. Use equation (24) and set k = −1/2 and simplify.

Example 13.
∞∑
n=1

4−n sec2
(
π2−n−1

)
= 1− 4

π2
(26)

Proof. Use equation (10) and set k = 1, a = 1,m = πi/2 and simplify using entry (3)
in table below (64:12:7) in [9].

8.1. Definite Integral of Infinite Sum

In this section we extend upon the results in [14]. We take the definite integral of
specific infinite sums in terms of hyperbolic trigonometric functions. These definite in-
tegrals result in the sum of the logarithm of hyperbolic trigonometric functions which is
equivalent to the infinite product of the logarithm of hyperbolic trigonometric functions.

Example 14.

Using equation (10) and setting k = 1, a = 1,m = y and

simplifying using entry (3) in table below (64:12:7) in [9] we get
∞∑
n=1

4−n sech2
(
2−ny

)
=

1

y2
− csch2(y)

Next we take the definite integral over y ∈ [x, x/2] to get∫ x

x/2

( ∞∑
n=1

4−n sech2
(
2−ny

)
=

1

y2
− csch2(y)

)
dy



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 158-168 166

We reverse the order of integration and summation and again take

the definite integral over x ∈ [z, z/2] to get∫ z

z/2

( ∞∑
n=1

−2−n
(
tanh

(
2−n−1x

)
− tanh

(
2−nx

))
=

1

x
− coth

(x
2

)
+ coth(x)

)
dx

Reversing the order of the sum and integral and simplifying we get
∞∑
n=1

log
(
cosh2

(
2−n−2z

)
cosh

(
2−nz

)
sech3

(
2−n−1z

))
= log

(
tanh2

(z
4

)
+ 1
)

whence upon replacing z → x we obtain that
∞∏
n=1

cosh2
(
2−2−nx

)
cosh (2−nx)

cosh3 (2−1−nx)
= 1 + tanh2

(x
4

)
Example 15.

∞∏
n=1

cosh
(
2−nx

) (
tanh

(
2−n−1x

)
+ 1
)
e2

−n tanh(2−n−1x) sech(2−nx) =
1

2
(ex + 1) e

1
x
−coth(x

2 )+coth(x)

(27)

Proof. Use the process in Example (14) with k = 1, a = e,m = x and simplify.

Example 16.
∞∏
n=1

cosh7
(
2−1−nx

)
cosh8

(
2−3−nx

)
cosh (2−nx) cosh14 (2−2−nx)

=
cosh6

(
x
4

)
cosh8

(
x
8

)
cosh

(
x
2

) (28)

Proof. Use the process in Example (14) with k = 2, a = 1,m = x and simplify.

Example 17.
∞∏
n=1

cosh
(
2−nx

) (
tanh

(
2−n−1x

)
+ 1
)
=

1

2
ex/2 sinh(x) csch

(x
2

)
(29)

Proof. Use the process in Example (14) with k = 1, a = e2a,m = x and simplify.

Example 18.
∞∏
n=1

cosh (2−nx)

cosh (2−1−nx)
=

sinh(x)

2 sinh
(
x
2

) (30)

Proof. Use the process in Example (14) with k = 1, a = ea,m = x and simplify.

Example 19.
∞∏
n=1

cosh2
(
2−2−nx

)
cosh (2−nx)

cosh3 (2−1−nx)
=

cosh
(
x
2

)
cosh2

(
x
4

) (31)

Proof. Use the process in Example (14) with k = 1, a = ea,m = x by forming a second
equation with replacing a → −a and simplify.



REFERENCES 167

9. Discussion

The authors construct an infinite sum of the Lerch function Φ(k, a,m) in terms of
the incomplete gamma function Γ(k, a) and the Lerch function, where the parameter con-
straints are wide. The infinite sum of the Lerch function in terms of the Lerch transforma-
tion was also derived. The derivations used fundamental constants and special functions,
and the infinite sum allowed for a wide range of the parameters. We checked the outcome
numerically using Wolfram Mathematica.

References

[1] D.H. Bailey and J.M. Borwein. Crandall’s computation of the incomplete gamma
function and the hurwitz zeta function, with applications to dirichlet l-series. Applied
Mathematics and Computation, 268:462–477, 10 2015.

[2] Harry Bateman. Higher Transcendental Functions V.1. McGraw-Hill, 1953.

[3] Nist digital library of mathematical functions. F. W. J. Olver, A. B. Olde Daalhuis,
D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, B. R. Miller, B. V.
Saunders, H. S. Cohl, and M. A. McClain, eds.

[4] I. S. Gradshteyn and I. M. Ryzhik. Table of integrals, series, and products. Else-
vier/Academic Press, Amsterdam, seventh edition, 2007.

[5] S. Kanemitsu, H. Kumagai, H.M. Srivastava, and M. Yoshimoto. Some integral and
asymptotic formulas associated with the hurwitz zeta function. Applied Mathematics
and Computation, 154:641–664, 07 2004.

[6] S Kanemitsu, Y Tanigawa, H Tsukada, and M Yoshimoto. Contributions to the
theory of the hurwitz zeta-function. Hardy-Ramanujan Journal, Volume 30, 01 2007.

[7] Leonard Lewin. Polylogarithms and Associated Functions. North Holland, 1981.

[8] F. Oberhettinger. Note on the lerch zeta function. Pacific Journal of Mathematics,
6(1):117 – 120, 1956.

[9] Keith B. Oldham, Jan Myland, and Jerome Spanier. An Atlas of Functions: with
Equator, the Atlas Function Calculator. Springer Science & Business Media, 07 2010.

[10] R. B. Paris. The stokes phenomenon associated with the hurwitz zeta function ζ(s, a).
Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences,
461:297–304, 01 2005.

[11] Anatolĭı Platonovich Prudnikov, Yu A. Brychkov, and Oleg Igorevich Marichev. In-
tegrals and Series: More special functions. Gordon and Breach Science Publishers,
1986.



REFERENCES 168

[12] Robert Reynolds and Allan Stauffer. A method for evaluating definite integrals in
terms of special functions with examples. International Mathematical Forum, 15:235–
244, 2020.

[13] Robert Reynolds and Allan Stauffer. Infinite sum of the incomplete gamma function
expressed in terms of the hurwitz zeta function. Mathematics, 9:1952, 08 2021.

[14] Cornel Ioan Vălean. (Almost) Impossible Integrals, Sums, and Series. Springer, 05
2019.