EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 1, 2022, 30-35
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Extended Apelblat integrals for fractional calculus

Robert Reynolds1,∗, Allan Stauffer1

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

Abstract. A quadruple integral involving the logarithmic, exponential, polynomial and Gamma
functions is derived in terms of the Hurwitz-Lerch zeta function. Special cases of this integral are
evaluated in terms of special functions and fundamental constants. Almost all Hurwitz-Lerch zeta
functions have an asymmetrical zero-distribution. 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: Volterra function, Hurwitz-Lerch zeta function, Gamma function,
quadruple integral, contour integral, logarithmic function

1. Significance Statement

Apelblat [1] evaluated a triple integral involving the Volterra function in terms of the
Gamma function which is used in Fractional Calculus. In this work the authors expand
on Apelblat’s work by deriving a quadruple integral involving the Volterra function and
express it in terms of the Hurwitz-Lerch zeta function. Our hope is researchers will find
the results in this work useful where applicable.

2. Introduction

In this paper we derive the quadruple definite integral given by∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

ym−1z−me−bz−pxxα+u+y logk
(ay

z

)
Γ(u+ y + α+ 1)

dxdydudz (1)

where the parameters k, a, p, b, α and m are general complex numbers and the Volterra
function is given in equation (1.2.1) in [1]. This definite integral will be used to derive
special cases in terms of special functions and fundamental constants. The derivations

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i1.4181

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

http://www.ejpam.com 30 © 2022 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 30-35 31

follow the method used by us in [5]. 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)

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. We then multiply both sides by a
function of x, y, u and z, then take a definite quadruple 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 of x, y, u and z and take the infinite sums 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 [5]. The variable of integration in the contour integral is t =
w + m. The cut and contour are in the first quadrant of the complex t-plane. 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 form the quadruple integral by replacing y by log

(ay
z

)
and multiplying by ym−1z−me−bz−pxxα+u+y

Γ(u+y+α+1) then taking the definite integral with respect to

x ∈ [0,∞), y ∈ [0,∞), u ∈ [0,∞) and z ∈ [0,∞) to obtain

(3)

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

ym−1z−me−bz−pxxα+u+y logk
(ay

z

)
Γ(k + 1)Γ(u+ y + α+ 1)

dxdydudz

=
1

2πi

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫
C

aww−k−1ym+w−1z−m−we−bz−pxxα+u+y

Γ(u+ y + α+ 1)
dwdxdydudz

=
1

2πi

∫
C

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

aww−k−1ym+w−1z−m−we−bz−pxxα+u+y

Γ(u+ y + α+ 1)
dxdydudzdw

=
1

2πi

∫
C
πaww−k−1p−α−1bm+w−1 csc(π(m+ w)) log−m−w−1(p)dw

from equation (9.67.9) in [1] and equation (3.326.2) in [3] where Re(α) > −1, Re(p) >
1, Re(w+m) > 0 and using the reflection Formula (8.334.3) in [3] for the Gamma function.
We are able to switch the order of integration over t, x, y, u and z using Fubini’s theorem
since the integrand is of bounded measure over the space C×[0,∞)×[0,∞)×[0,∞)×[0,∞).

4. The Hurwitz-Lerch zeta Function and Infinite Sum of the Contour
Integral

In this section we use Equation (2) to derive the contour integral representations for
the Hurwitz-Lerch zeta function.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 30-35 32

4.1. The Hurwitz-Lerch zeta Function

The Hurwitz-Lerch zeta function (25.14) in [2] 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 Re(v) > 0, and either |z|≤ 1, z ̸= 1, Re(s) > 0, or z = 1, Re(s) > 1.

4.2. Infinite sum of the Contour Integral

Using Equation (2) and replacing y by log(a) + log(b)− log(log(p)) + iπ(2y + 1) then
multiplying both sides by −2iπbm−1p−α−1 log−m−1(p) taking the infinite sum over y ∈
[0,∞) and simplifying in terms of the Hurwitz-Lerch zeta function we obtain

(6)

− 1

Γ(k + 1)
(2iπ)k+1eiπmbm−1p−α−1 log−m−1(p)

Φ

(
e2imπ,−k,

−i log(a)− i log(b) + i log(log(p)) + π

2π

)
= − 1

2πi

∞∑
y=0

∫
C
2iπaww−k−1p−α−1bm+w−1eiπ(2y+1)(m+w) log−m−w−1(p)dw

= − 1

2πi

∫
C

∞∑
y=0

2iπaww−k−1p−α−1bm+w−1eiπ(2y+1)(m+w) log−m−w−1(p)dw

=
1

2πi

∫
C
πaww−k−1p−α−1bm+w−1 csc(π(m+ w)) log−m−w−1(p)dw

from equation (1.232.2) in [3] where Im(w +m) > 0 in order for the sum to converge.

5. Definite Integral in terms of the Hurwitz-Lerch zeta function

Theorem 1. For all k, a ∈ C, Re(p) > 1, Re(b) > 0, Re(α) > −1, 1/2 < Re(m) > 1,

(7 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

ym−1z−me−bz−pxxα+u+y logk
(ay

z

)
Γ(u+ y + α+ 1)

dxdydudz

= (2iπ)k+1eiπm
(
−bm−1

)
p−α−1 log−m−1(p)

Φ

(
e2imπ,−k,

−i log(a)− i log(b) + i log(log(p)) + π

2π

)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 30-35 33

Proof. Observe the right-hand side of equation (3) is equal to the right-hand side of
equation (6) so we may equate the left-hand sides to yield the stated result.

6. Special Cases

In this section we will evaluate equation (7) for various parameter values in terms of
the Riemann zeta function ζ(s), equation (25.2.1) in [2], Aprey’s constant ζ(3), equation
(25.6.9) in [2], log(2) and π.

Example 1. The degenerate case.

(8 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

ym−1z−me−bz−pxxα+u+y

Γ(u+ y + α+ 1)
dxdydudz

= πbm−1 csc(πm)p−α−1 log−m−1(p)

Proof. Use equation (7) and set k = 0 and simplify using entry (2) in Table below
(64:12:7) in [4].

Example 2.

(9 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−ex−zxu+y+1 logk
(
−y

z

)
√
y
√
zΓ(u+ y + 2)

dxdydudz

= −
ik+2

(
1− 2k+1

)
(2π)k+1ζ(−k)

e2

Proof. Use equation (7) and set a = −1, p = e, b = 1,m = 1/2, α = 1 and simplify
using entry (2) in Table below (64:7) and entry (4) in table below (64:12:7) in [4].

Example 3.

(10 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−ex−zxu+y+1

√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)dxdydudz =
log(2)

e2π

and

(11 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−ex−zxu+y+1 log
(y
z

)
√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)dxdydudz = 0

Proof. Use equation (9) and apply l’Hopital’s rule to the right-hand side as k → −1
simplify using equation (25.6.11) in [2] and rationalize the denominator and equate real
and imaginary parts.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (1) (2022), 30-35 34

Example 4.

(12 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−ex−zxu+y+1
(
π2 − 3 log2

(y
z

))
√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)3dxdydudz =
3ζ(3)

16e2π3

and

(13 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−ex−zxu+y+1 log
(y
z

) (
log2

(y
z

)
− 3π2

)
√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)3 dxdydudz = 0

Proof. Use equation (9) and set k = −3 using entry (2) in Table below (64:7) and
rationalize the denominator and simplify into real and imaginary parts.

Example 5.

(14 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−pxp−zxu+y+1 logk
(
−y

z

)
√
y
√
zΓ(u+ y + 2)

dxdydudz

= −
ik
(
2k+1 − 1

)
(2π)k+1ζ(−k)

p2 log2(p)

Proof. Use equation (7) and set a = −1, b = log(p),m = 1/2, α = 1 and simplify using
entry (2) in Table below (64:7) and entry (4) in Table below (64:12:7) in [4].

Example 6.

(15 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−2x2−zxu+y+1

√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)dxdydudz =
1

4π log(2)

and

(16 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−2x2−zxu+y+1 log
(y
z

)
√
y
√
zΓ(u+ y + 2)

(
log2

(y
z

)
+ π2

)dxdydudz = 0

Proof. Use equation (14) and apply l’Hopital’s rule to the right-hand side as k → −1
and set p = 2 and rationalize the denominator and simplify into real and imaginary parts.

Example 7.∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

z−m−ne−bx−z log(b) (zmyn − ymzn)xα+u+y

y log
(y
z

)
Γ(u+ y + α+ 1)

dxdydudz

= b−α−1 log−m−n−2(b)
(
2 tanh−1

(
eiπm

)
logm+n(b)− 2 tanh−1

(
eiπn

)
logm+n(b)

)
(17)

Proof. Use equation(7) and form a second equation by replacing m → n and taking
their difference. Next set k = −1, a = 1, b = log(b), p = b and simplify using entry (3) in
Table below (64:12:7) in [4].



REFERENCES 35

Example 8.

(18 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−2x2−z
(

12
√
z − 12

√
y
)
xu+y+2

3
√
yz3/4Γ(u+ y + 3) log

(y
z

) dxdydudz =
log

(
9− 6

√
2
)

16 log2(2)

Proof. Use equation (17) and set b = 2, α = 2, n = 2/3,m = 3/4 and simplify.

Example 9.

(19 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫ ∞

0

e−2x2−z
(

4
√
y − 4

√
z
)
xu+y+1

√
yz3/4Γ(u+ y + 2) log

(y
z

) dxdydudz =
coth−1

(√
2
)

4 log2(2)

Proof. Use equation (17) and set b = 2, α = 1, n = 1/2,m = 3/4 and simplify.

7. Discussion

In this paper, we have presented a novel method for deriving a quadruple integral
involving the Volterra function along with some interesting definite integrals using our
contour integration method. The results presented were numerically verified for both real
and imaginary complex values of the parameters in the integrals using Mathematica by
Wolfram.

Acknowledgements

This research is supported by NSERC Canada under grant 504070.

References

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lishers, Inc., 12 2010.

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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.

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

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

[5] 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.