EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 3, 2022, 916-923
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Triple Integral involving the Bessel-Integral Function
Jiv(z): Derivation and Evaluation

Robert Reynolds1,∗, Allan Stauffer1

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

Abstract. A triple integral involving the Bessel-integral function Jiv(z) is derived and evaluated
for certain real numbers of the parameters. The derived integral allows a representation in terms
of the product of the Hurwitz-Lerch zeta and Gamma functions with seven parameters. 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: Triple integral, Bessel-integral function, Hurwitz-Lerch zeta function,
Catalan’s constant

1. Significance Statement

The Bessel integral function Jiv(z) has been studied in many works namely; the oper-
ational solution of linear differential equations and properties of their solutions [11], and
the expansion of the works of by van der Pol were published in [7] and [5]. In the book
of Prudnikov et al. [9] section (3.3.2) some very interesting triple integrals containing the
Bessel function of order zero J0(z) are tabled without derivation. In this current work
we aim to expand on the Table of Prudnikov et al. by deriving and evaluating a triple
integral involving the Bessel integral function given in Table (3.37.4) in [1] and provide a
formal derivation by expressing the triple integral in terms of the Hurwitz-Lerch zeta and
Gamma functions.

2. Introduction

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

0

∫ ∞

0

∫ ∞

0

1

v2Γ(v)
xm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2 logk−1

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4259

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

https://www.ejpam.com 916 © 2022 EJPAM All rights reserved.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (3) (2022), 916-923 917(
ax

yz

)(
m log

(
ax

yz

)
+ k

)
1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dxdydz (1)

where the parameters k, a, b, c, v,m are general complex numbers and α ∈ R+, Re(b),
Re(c) > 0, Re(m) < 0 < Re(v) < 2. This definite integral will be used to derive special
cases in terms of special functions and fundamental constants. 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)

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 and z, then take a definite triple 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 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 [10]. The variable of integration in the contour integral is
s = w+m. The cut and contour are in the first quadrant of the complex s-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 triple integral by replacing y by log
(
ax
yz

)
and

multiplying by

(3)−
m2−vxm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2

1F2

(
v
2 ;

v
2 + 1, v + 1;−1

4x
2α2
)

v2Γ(v)

for the first equation and replacing y by log
(
ax
yz

)
and replacing k → k−1 and multiplying

by

(4)
2−vxm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2

1F2

(
v
2 ;

v
2 + 1, v + 1;−1

4x
2α2
)

v2Γ(v)

to form the second equation. Next we add both equations then take the definite triple
integral with respect to x ∈ [0,∞), y ∈ [0,∞) and z ∈ [0,∞) to obtain∫ ∞

0

∫ ∞

0

∫ ∞

0

1

v2Γ(v)Γ(k + 1)
xm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2

logk−1

(
ax

yz

)(
m log

(
ax

yz

)
+ k

)
1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dxdydz



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (3) (2022), 916-923 918

= − 1

2πi

∫ ∞

0

∫ ∞

0

∫ ∞

0

∫
C

1

v2Γ(v)
2−vaww−k−1(m+ w)xm+w−1(αx)ve−by2−cz2

y−m+v−w+1z−m−v−w+1
1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dwdxdydz

= − 1

2πi

∫
C

∫ ∞

0

∫ ∞

0

∫ ∞

0

1

v2Γ(v)
2−vaww−k−1(m+ w)xm+w−1(αx)ve−by2−cz2

y−m+v−w+1z−m−v−w+1
1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dxdydzdw

=
1

2πi

∫
C
πaww−k−12m+w−3α−m−wb

1
2
(m−v+w−2)c

1
2
(m+v+w−2)

csc

(
1

2
π(m+ v + w)

)
dw (5)

from equation (4) in [2], equation (3.37.4.1) in [1] and equation (3.326.2) in [4] where
0 < Re(w + m) < 2, Re(m) < Re(v) < 2, α ∈ R+ and using the reflection formula
(8.334.3) in [4] for the Gamma function. We are able to switch the order of integration
over x, y and z using Fubini’s theorem since the integrand is of bounded measure over the
space C× [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.

4.1. The Hurwitz-Lerch zeta Function

The Hurwitz-Lerch zeta function (25.14) in [3] has a series representation given by

Φ(z, s, v) =
∞∑
n=0

(v + n)−szn (6)

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 (7)

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 (3) (2022), 916-923 919

4.2. Infinite sum of the Contour Integral

Using equation (2) and replacing y by log(a)−log(α)+ log(b)
2 + log(c)

2 + 1
2 iπ(2y+1)+log(2)

then multiplying both sides by −iπ2m−2α−mb
1
2
(m−v−2)c

1
2
(m+v−2)e

1
2
iπ(2y+1)(m+v) taking the

infinite sum over y ∈ [0,∞) and simplifying in terms of the Hurwitz-Lerch zeta function
we obtain

(8)

− 1

Γ(k + 1)
iπk+12m−2α−mb

1
2
(m−v−2)c

1
2
(m+v−2)e

1
2
iπ(k+m+v)

Φ

(
eiπ(m+v),−k,

−2i log(2a)− i log(b)− i log(c) + 2i log(α) + π

2π

)
= − 1

2πi

∞∑
y=0

∫
C
iπ2m−2aww−k−1α−mb

1
2
(m−v−2)c

1
2
(m+v−2)

exp

(
1

2
(w(−2 log(α) + log(b) + log(c) + log(4)) + iπ(2y + 1)(m+ v + w))

)
dw

= − 1

2πi

∫
C

∞∑
y=0

iπ2m−2aww−k−1α−mb
1
2
(m−v−2)c

1
2
(m+v−2)

exp

(
1

2
(w(−2 log(α) + log(b) + log(c) + log(4)) + iπ(2y + 1)(m+ v + w))

)
dw

=
1

2πi

∫
C
πaww−k−12m+w−3α−m−wb

1
2
(m−v+w−2)c

1
2
(m+v+w−2)

csc

(
1

2
π(m+ v + w)

)
dw

from equation (1.232.3) in [4] where Im
(
1
2π(m+ v + w)

)
> 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(b) > 0, Re(c) > 0, Re(m) < 0 < Re(v) < 2, α ∈ R+,∫ ∞

0

∫ ∞

0

∫ ∞

0

1

v2Γ(v)
xm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2

logk−1

(
ax

yz

)(
m log

(
ax

yz

)
+ k

)
1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dxdydz

= iπk+1α−m2m+v−2b
1
2
(m−v−2)c

1
2
(m+v−2)e

1
2
iπ(k+m+v)

Φ

(
eiπ(m+v),−k,

−2i log(2a)− i log(b)− i log(c) + 2i log(α) + π

2π

)
(9)



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (3) (2022), 916-923 920

Proof. The right-hand sides of relations (5) and (8) are identical; hence, the left-hand
sides of the same are identical too. Simplifying with the Gamma function yields the desired
conclusion.

Example 1. The degenerate case.∫ ∞

0

∫ ∞

0

∫ ∞

0

1

v2Γ(v)
mxm−1y−m+v+1z−m−v+1(αx)ve−by2−cz2

1F2

(
v

2
;
v

2
+ 1, v + 1;−1

4
x2α2

)
dxdydz

= πα−m
(
−2m+v−3

)
b
1
2
(m−v−2)c

1
2
(m+v−2) csc

(
1

2
π(m+ v)

)
(10)

Proof. Use equation (9) and set k = 0 and simplify using entry (2) in Table below
(64:12:7) in [8].

Example 2. The Hurwitz zeta function ζ(s, v)∫ ∞

0

∫ ∞

0

∫ ∞

0

1

25x5/3Γ
(
5
3

)9y10/3(αx)5/3e−by2−cz2
(
k − 2

3
log

(
ax

yz

))
logk−1

(
ax

yz

)
1F2

(
5

6
;
11

6
,
8

3
;−1

4
x2α2

)
dxdydz

=
iα2/32k−1e

1
2
iπ(k+1)πk+1ζ

(
−k, −2i log(2a)−i log(b)−i log(c)+2i log(α)+π

4π

)
b13/6

√
c

−
iα2/32k−1e

1
2
iπ(k+1)πk+1ζ

(
−k, 12

(
−2i log(2a)−i log(b)−i log(c)+2i log(α)+π

2π + 1
))

b13/6
√
c

(11)

Proof. Use equation (9) and set m = −2/3, v = 5/3 and simplify using entry (4) in
Table below (64:12:7) in [8].

Example 3.

(12 )

∫ ∞

0

∫ ∞

0

∫ ∞

0

y10/3e−y2−z2
(
2 log

(
− x

yz

)
+ 3
)

1F2

(
5
6 ;

11
6 ,

8
3 ;−x2

)
log2

(
− x

yz

) dxdydz

=
25

24
i(π − 4)Γ

(
5

3

)
Proof. Use equation (11) and set a = −1, b = c = α = 1 and simplify.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (3) (2022), 916-923 921

Example 4. The zeta function of Riemann ζ(s).∫ ∞

0

∫ ∞

0

∫ ∞

0
y10/3e−y2−z2

(
3k − 2 log

(
ix

2yz

))
logk−1

(
ix

2yz

)
1F2

(
5

6
;
11

6
,
8

3
;−x2

4

)
dxdydz

=
25

6

(
2k+1 − 1

)
e

iπk
2 πk+1Γ

(
5

3

)
ζ(−k) (13)

Proof. Use equation (11) and set a = i/2, b = c = α = 1 and simplify using entry (2)
in Table below (64:7) in [8].

Example 5.∫ ∞

0

∫ ∞

0

∫ ∞

0

1

log2
(

x
yz

)xv−1e−y2−z2y−m−p+v+1z−m−p−v+1

(
ymzmxp

(
p log

(
x

yz

)
− 1

)
+ xmypzp

(
1−m log

(
x

yz

)))
1F2

(v
2
;
v

2
+ 1, v + 1;−x2

)
dxdydz

=
1

2
vΓ(v + 1)

(
tanh−1

(
e

1
2
iπ(p+v)

)
− tanh−1

(
e

1
2
iπ(m+v)

))
(14)

Proof. Use equation (9) and form a second equation by replacing m → p and take
their difference. Next set k = −1, a = b = c = 1, α = 2 and simplify using entry (3) in
Table below (64:12:7) in [8].

Example 6.∫ ∞

0

∫ ∞

0

∫ ∞

0

1

3
√
x log2

(
x
yz

)y44/15z4/15e−y2−z2

((
10 15

√
y 15
√
z − 9 15

√
x
)
log

(
x

yz

)
− 15

(
15
√
x− 15

√
y 15
√
z
))

1F2

(
2

3
;
5

3
,
7

3
;−x2

)
dxdydz

= −5Γ

(
7

3

)
tanh−1

(
2 +

3

sin
(
2π
15

)
− 2

)
(15)

Proof. Use equation (14) and set m = −2/3, p = −3/5, v = 4/3 and simplify.



R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 15 (3) (2022), 916-923 922

Example 7. The Polylogarithm function Lin(z).∫ ∞

0

∫ ∞

0

∫ ∞

0
e−y2−z2xm+v−1y−m+v+1z−m−v+1 logk−1

(
ix

2yz

)
(
k +m log

(
ix

2yz

))
1F2

(
v

2
;
v

2
+ 1, v + 1;−x2

4

)
dxdydz

= iπk+1v2m+v−2Γ(v + 1)e−
1
2
iπ(−k+m+v)Li−k

(
eiπ(m+v)

)
(16)

Proof. Use equation (9) and set a = i/2, b = c = α = 1 and simplify using equation
(64:12:2) in [8].

Example 8. Catalan’s constant C.∫ ∞

0

∫ ∞

0

∫ ∞

0

1
√
x log3

(
ix
2yz

)√zy2v+
1
2 e−y2−z2

(
−2 +

(
1

2
− v

)
log

(
ix

2yz

))

1F2

(
v

2
;
v

2
+ 1, v + 1;−x2

4

)
dxdydz

=
ie

3iπ
4

(
−π2

48 + iC
)
vΓ(v + 1)

2
√
2π

(17)

Proof. Use equation (16) and set k = −2,m = 1/2 − v and simplify using equation
(2.2.1.2.7) in [6].

Example 9. The Hypergeometric function s−1
2F1 (1, s, 1 + s; z).∫ ∞

0

∫ ∞

0

∫ ∞

0

1
√
x log2

(
− x

2yz

)√zy2v+
1
2 e−y2−z2

((
1

2
− v

)
log

(
− x

2yz

)
− 1

)

1F2

(
v

2
;
v

2
+ 1, v + 1;−x2

4

)
dxdydz

=

(
1

2
− i

2

)(
−1 + 2F1

(
1

2
, 1;

3

2
; i

))
vΓ(v + 1) (18)

Proof. Use equation (9) and set k = −1, a = −1/2, b = c = α = 1,m = 1/2 − v and
simplify using equation (9.559) in [4].



REFERENCES 923

6. Discussion

In this paper, we have presented a novel method for deriving a new integral transform
involving the Bessel integral function Jiv(z) along with some interesting definite integrals
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.

References

[1] Yu. A. Brychkov, O. I. Marichev, and N. V. Savischenko. Handbook of Mellin tran-
forms. CRC Press., 2019.

[2] Yong-Kum Cho, Seok-Young Chung, and Hera Yun. An extension of positivity for
integrals of bessel functions and buhmann’s radial basis functions. Proceedings of the
American Mathematical Society, Series B, 5:25–39, 2018.

[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] P. Humbert. Bessel-integral functions. Proceedings of the Edinburgh Mathematical
Society, 4:276–285, 1933.

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

[7] F. Oberhettinger. On some expansions for bessel integral functions. Journal of Re-
search of the National Bureau of Standards, 59, 1957.

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

[9] Anatolĭı Platonovich Prudnikov, ÎUrĭı Aleksandrovich Brychkov, and Oleg Igorevich
Marichev. Integrals and Series: Special functions Volume 2. CRC Press, 1986.

[10] 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.

[11] B. van der Pol. On the operational solution of linear differential equations and an
investigation of the properties of their solutions. Phil. Mag., 8:861–898, 1929.