AMApplied Mathematics2152-7385Scientific Research Publishing10.4236/am.2019.105021AM-92259ArticlesPhysics&Mathematics On <i>q</i>-Analogues of Laplace Type Integral Transforms of <i>q</i><sup>2</sup>-Bessel Functions ArwaAlshibani1*ReemAlkhairy1Department of Mathematics, Faculty of Sciences, Imam Abdulrahman Bin Faisal University, Dammam, KSA06052019100530131111, March 20194, May 2019 7, May 2019© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

The present paper deals with the evaluation of the q-Analogues of Laplece transforms of a product of basic analogues of q 2-special functions. We apply these transforms to three families of q-Bessel functions. Several special cases have been deducted.

<i>q</i>-Extensions of Bessel Functions <i>q</i>-Analogues of Laplace Type Integrals Transforms <i>q</i>-Analogues of Gamma Function <i>q</i>-Shift Factorials1. Introduction

In the second half of twentieth century, there was a significant increase of activity in the area of the q-calculus mainly due to its application in mathematics, statistics and physics. In literature, several aspects of q-calculus were given to enlighten the strong inter disciplinary as well as mathematical character of this subject. Specifically, there have been many q-analogues and q-series representations of various kinds of special functions. In the case of q-Bessel function, there are two related q-Bessel functions introduced by Jackson [1] and denoted by Ismail [2] as

J μ ( 1 ) ( z ; q ) = ( z 2 ) μ ∑ n = 0 ∞ ( − z 2 4 ) n ( q , q ) μ + n ( q ; q ) n , | z | < 2 (1)

J μ ( 2 ) ( z ; q ) = ( z 2 ) μ ∑ n = 0 ∞ q n ( n + μ ) ( − z 2 4 ) n ( q , q ) μ + n ( q ; q ) n , z ∈ ℂ (2)

The third related q-Bessel function J μ ( 3 ) ( z ; q ) was introduced in a full case as [3]

J μ ( 3 ) ( z ; q ) = z μ ∑ n = 0 ∞ ( − 1 ) n q n ( n − 1 ) 2 ( q z 2 ) n ( q , q ) μ + n ( q ; q ) n , z ∈ ℂ (3)

A certain type of Laplace transforms, which is called L2-transform, was introduced by Yürekli and Sadek [4] . Then these transforms were studied in more details by Yürekli [5] , [6] . Purohit and Kalla applied the q-Laplace transforms to a product of basic analogues of the Bessel function [7] .

On the same manner, integral transforms have different q-analogues in the theory of q-calculus. The q-analogue of the Laplace type integral of the first kind is defined by [8] as

q L 2 ( f ( ξ ) ; y ) = 1 1 − q 2 ∫ 0 y − 1 ξ E q 2 ( q 2 y 2 ξ 2 ) f ( ξ ) d ξ (4)

and expressed in terms of series representation as

q L 2 ( f ( ξ ) ; y ) = ( q 2 ; q 2 ) ∞ [ 2 ] q y 2 ∑ i = 0 ∞ q 2 i ( q 2 ; q 2 ) i f ( q i y − 1 ) . (5)

On the other hand, the q-analogue of the Laplace type integral of the second kind is defined by [8] as

q l 2 ( f ( ξ ) ; y ) = 1 1 − q 2 ∫ 0 ∞ ξ e q 2 ( − y 2 ξ 2 ) f ( ξ ) d q ξ (6)

whose q-series representation expressed as

q l 2 ( f ( ξ ) ; y ) = 1 [ 2 ] 2 ( − y 2 ; q 2 ) ∞ ∑ i ∈ ℤ   q 2 i f ( q i ) ( − y 2 ; q 2 ) i . (7)

In this paper we build upon analysis of [8] . Following [9] , we discuss the q-Laplace type integral transforms (4) and (7) on the q-Bessel functions J μ ( 1 ) ( z ; q ) , J μ ( 2 ) ( z ; q ) and J μ ( 3 ) ( z ; q ) , respectively. In Section 2, we recall some notions and definitions from the q-calculus. In Section 3, we give the main results to evaluate the q-analogue of Laplace transformation of q2-Basel function. In Section 4, we discuss some special cases.

2. Definitions and Preliminaries

In this section, we recall some usual notions and notations used in the q-theory. It is assumed in this paper wherever it appears that 0 < q < 1 . For a complex number a, the q-analogue of a is introduced as [ a ] q = 1 − q a 1 − q . Also, by fixing a ∈ ℂ , the q-shifted factorials are defined as

( a ; q ) 0 = 1 ; ( a , q ) n = ∏ k = 0 n − 1 ( 1 − a q k ) , n = 1,2, ⋯ ; ( a ; q ) ∞ = lim n → ∞ ( a ; q ) n . (8)

This indeed lead to the conclusion

( [ n ] q ) ! = ( q ; q ) n ( 1 − q ) n , n ∈ ℕ     and     ( a ; q ) x = ( a ; q ) ∞ ( a q x ; q ) ∞ . (9)

The q-analogue of the exponential function of first and second type are respectively given in [10] by

e q ( x ) = ∑ 0 ∞ x n ( q ; q ) n = 1 ( x ; q ) ∞ , | x | < 1. (10)

and

E q ( x ) = ∑ 0 ∞ ( − 1 ) n q n n − 1 2 x n ( q ; q ) n , x ∈ ℂ . (11)

Indeed it has been shown that

e q ( x ) = 1 ( x ; q ) ∞ , | x | < 1     and     E q ( x ) = ( x , q ) ∞ , x ∈ ℂ (12)

The finite q-Jackson and improper integrals are respectively defined by [11]

∫ 0 x f ( t ) d q t = x ( 1 − q ) ∑ k = 0 ∞   q k f ( x q k ) (13)

and

∫ 0 ∞ / A f ( t ) d q t = ( 1 − q ) ∑ k ∈ ℤ q k A f ( q k A ) . (14)

The q-analogues of the gamma function of first and second type are respectively defined in [9] as

Γ q ( α ) = ∫ 0 1 / ( 1 − q ) x α − 1 E q ( q ( 1 − q ) x ) d q x , ( α > 0 ) (15)

and

q Γ ( α ) = K ( A ; α ) ∫ 0 ∞ / A ( 1 − q ) x α − 1 e q ( − ( 1 − q ) x ) d q x (16)

where, α 1 > 0 , where K ( A ; α ) is the function given by

K ( A ; α ) = A α − 1 ( − q / α ; q ) ∞ ( − α ; q ) ∞ ( − q t / α ; q ) ∞ ( − α q 1 − t ; q ) ∞ . (17)

Some useful results, for x ≠ 0, − 1, − 2, ⋯ , we use here are given by

Γ q ( α ) = ( q ; q ) ∞ ( 1 − q ) α − 1 ∑ k = 0 ∞ q k α ( q ; q ) k = ( q ; q ) ∞ ( q α ; q ) ∞ ( 1 − q ) 1 − x , (18)

and

q Γ ( α ) = K ( A ; α ) ( 1 − q ) α − 1 ( − 1 A ; q ) ∞ ∑ k ∈ ℤ ( q k A ) ( − 1 A ; q ) k . (19)

3. Main Theorems

Theorem 1. Let J 2 μ 1 ( 1 ) ( 2 a 1 t ; q 2 ) , ⋯ , J 2 μ n ( 1 ) ( 2 a n t ; q 2 ) be a set of first kind of q2-Bessel functions, f ( t ) = t Δ − 1 ∏ j = 1 n J 2 μ j ( 1 ) ( 2 a j t ; q 2 ) , where Δ , a j and μ j for j = 1 , 2 , ⋯ , n are constants; then the q-analogue of Lablace transformation q L 2 of f ( t ) is given as:

q L 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − a j s ) m j B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 ) (20)

and the q-analogue of Laplace transformation q l 2 of f ( t ) is given as:

q l 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − a j s ) m j B m j ( q 2 ) q 2 Γ ( m j + μ j + Δ + 1 2 ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) . (21)

where R e ( s ) > 0 , R e ( Δ ) > 0 and

A Δ = ( 1 − q 2 ) Δ / 2 [ 2 ] s Δ + 1 ( q 2 ; q 2 ) ∞ , B m j ( q 2 ) = ( q 2 μ j + m j + 2 ; q 2 ) ∞ ( 1 − q 2 ) m j + μ j − 1 2 ( q 2 ; q 2 ) m j

Proof. Now,

q L 2 ( f ( t ) ; s ) = ( q 2 ; q 2 ) ∞ [ 2 ] s 2 ∑ k = 0 ∞ q 2 k f ( q k s − 1 ) ( q 2 ; q 2 ) k

since

J 2 μ j ( 1 ) ( 2 a j t ; q 2 ) = ( 2 a j t 2 ) 2 μ j ∑ m j = 0 ∞ ( − ( 2 a j t ) 2 4 ) m j ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j

so

q L a ( f ( t ) ; s ) = ( q 2 ; q 2 ) ∞ [ 2 ] s 2 ∑ k = 0 ∞ q 2 k ( q 2 ; q 2 ) k ( q k s − 1 ) Δ − 1 ∏ j = 1 n ( a j q k s − 1 ) 2 μ j     ⋅ ∑ m j = 0 ∞ ( − 1 ) m j ( a j q k s − 1 ) m j ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j = ( q 2 ; q 2 ) ∞ [ 2 ] s Δ + 1 ∑ k = 0 ∞ q k ( Δ + 1 ) ( q 2 ; q 2 ) k ∏ j = 1 n ( a j q k s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j ( a j q k s ) m j ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j = ( q 2 ; q 2 ) ∞ [ 2 ] s Δ + 1 ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j ( a j s ) m j ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j ∑ k = 0 ∞ q k ( Δ + 1 + m j + μ j ) ( q 2 ; q 2 ) k (22)

Since

Γ q 2 ( α ) = ( q 2 ; q 2 ) ∞ ( 1 − q 2 ) α − 1 ∑ k = 0 ∞ q 2 k α ( q 2 ; q 2 ) k

putting α = 1 + Δ + m j + μ j 2 , so (22) becomes:

q L s ( f ( t ) ; s ) = 1 [ 2 ] s Δ + 1 ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j ( a j s ) m j ( 1 − q 2 ) 1 + Δ + m j + μ j 2 ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j .

Γ q 2 ( m j + μ j + Δ + 1 2 ) (23)

Since

( q 2 ; q 2 ) 2 μ j + m j = ( q 2 ; q 2 ) ∞ ( q 2 q 2 μ j + m j ; q 2 ) ∞

so (23) becomes:

q L 2 ( f ( t ) ; s ) = ( 1 − q 2 ) Δ / 2 [ 2 ] s Δ + 1 ( q 2 ; q 2 ) ∞ ∏ j = 1 n ( a j s ) μ j     ⋅ ∑ m j = 0 ∞ ( a j s ) ( − 1 ) m j ( q 2 μ j + m j + 2 ; q 2 ) ∞ ( 1 − q 2 ) m j + μ j − 1 2 ( q 2 ; q 2 ) m j .

Γ q 2 ( m j + μ j + Δ + 1 2 ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( a j s ) ( − 1 ) m j B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 ) 2

Similarly we have

q l 2 ( f ( t ) ; s ) = 1 [ 2 ] 1 ( − s 2 ; q 2 ) ∞ ∑ k = 0 ∞   q 2 k ( − s 2 ; q 2 ) k ( q k ) Δ − 1 ∏ j = 1 n J 2 μ j ( 1 ) ( 2 a j q k ; q 2 ) = 1 [ 2 ] 1 ( − s 2 ; q 2 ) ∞ ∑ k = 0 ∞   q 2 k ( − s 2 ; q 2 ) k ( q k ) Δ − 1 ∏ j = 1 n ( a j q k ) μ j .

∑ m j = 0 ∞ ( − a j q k ) m j ( q 2 ; q 2 ) m j + 2 μ j = ∏ j = 1 n ( a j ) μ j [ 2 ] ∑ m j = 0 ∞ ( − a j ) m j ( q 2 ; q 2 ) m j + 2 μ j ( q 2 ; q 2 ) m j ∑ k = 0 ∞ ( − s 2 ; q 2 ) k q k ( m j + μ j + Δ + 1 ) ( − s 2 ; q 2 ) ∞

Now using

q 2 Γ ( α ) = K ( A ; α ) ( 1 − q 2 ) α − 1 ( − 1 A ; q 2 ) ∞ ∑ k ∈ Z ( q K A ) α ( − 1 A ; q 2 ) K

with A = 1 s 2 , α = m j + μ j + Δ + 1 2 we get

q l 2 ( f ( t ) ; s ) = ∏ j = 1 m ( a j ) μ j [ 2 ] s μ j + Δ + 1 ∑ m j = 0 ∞ ( − a j s ) m j ( 1 − q 2 ) m j + μ j + Δ + 1 2 q 2 Γ ( m j + μ j + Δ + 1 2 ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) ( q 2 ; q 2 ) m j + 2 μ j ( q 2 ; q 2 ) m j = ( 1 − q 2 ) Δ 2 [ 2 ] s Δ + 1 ( q 2 ; q 2 ) ∞ ∏ j = 1 m ( a j s ) μ j ∑ m j = 0 ∞ ( − a j s ) m j ( 1 − q 2 ) m j + μ j − 1 2 ( q m j + 2 μ j + 2 ; q 2 ) ∞ K ( 1 s 2 , m j + μ j + Δ + 1 2 ) ( q 2 ; q 2 ) m j .

q 2 Γ ( m j + μ j + Δ + 1 2 ) = A Δ ∏ j = 1 m ( a j s ) μ j ∑ m j = 0 ∞ ( − a j s ) m j K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 )

Theorem 2. Let J 2 μ 1 ( 2 ) ( 2 a , t ; q 2 ) , ⋯ , J 2 μ n ( 2 ) ( 2 a t ; q 2 ) be a set of second order q2-Bessel function, f ( t ) = t Δ − 1 ∏ j = 1 n J 2 μ j ( 2 ) ( 2 a j t ; q 2 ) where Δ , a j and μ j for j = 1 , 2 , ⋯ , n are constants then q L 2 -transform of f ( t ) is given as:

q L 2 ( f ( t ) , s ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j q 2 m j ( m j + 2 μ j ) ( a j s ) m j + μ j     ⋅ B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 ) (24)

and the q-analogue of Laplace transformation q l 2 of f ( t ) is given as:

q l 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − a j s ) m j q 2 m j ( m j + 2 μ j ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 )     ⋅ B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 ) (25)

Proof. Now,

J 2 μ j ( 2 ) ( 2 a j t ; q 2 ) = ( 2 a j t 2 ) 2 μ j ∑ m j ∞ ( − ( 2 a j t ) 2 4 ) m j q 2 m j ( m j + 2 a j ) ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j

so

q L 2 ( f ( t ) ; s ) = ( q 2 ; q 2 ) ∞ [ 2 ] s 2 ∑ k = 0 ∞ q 2 k ( q 2 ; q 2 ) k ( q k s − 1 ) Δ − 1 ∏ j = 1 n ( 2 a j q k s − 1 2 ) 2 μ j     ⋅ ∑ m j = 0 ∞ ( − ( 2 a j q k s − 1 ) 2 4 ) m j q 2 m j ( m j + 2 μ j ) ( q 2 ; q 2 ) 2 μ j + m j ( q 2 ; q 2 ) m j (26)

By the same argument we can write (26) as

q L 2 ( f ( t ) ; s ) = ( q 2 ; q 2 ) ∞ [ 2 ] s Δ + 1 ( q 2 ; q 2 ) ∞ ∏ j = 1 n ∑ m j = 0 ∞ ( − 1 ) m j q 2 m j ( m j + 2 μ j ) ( q 2 ; q 2 ) m j     ⋅ ( a j s ) m j + μ j ( q 2 μ j + m j + 2 ; q 2 ) ∞ ∑ k = 0 ∞ q k ( m j + μ j + 1 + Δ ) ( q 2 ; q 2 ) k

put α = m j + μ j + Δ + 1 2 in Γ q 2 ( α ) , then

So (25) becomes:

q L 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j q 2 m j ( m j + 2 μ j ) ( a j s ) m j + μ j     ⋅ B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 )

Similarly

q l 2 ( f ( t ) ; s ) = 1 [ 2 ] 1 ( − s 2 ; q 2 ) ∞ ∑ k = 0 ∞ q 2 k ( − s 2 ; q 2 ) k ( q k ) Δ − 1 ∏ j = 1 n ( a j q k ) μ j     ⋅ ∑ m j = 0 ∞ ( − a j q k ) m j q 2 m j ( m j + 2 μ j ) ( q 2 ; q 2 ) m j + 2 μ j ( q 2 ; q 2 ) m j

Put A = 1 s 2 , α = m j + μ j + Δ + 1 2 we get

q l 2 ( f ( t ) ; s ) = 1 [ 2 ] ∏ j = 1 n ( a j ) μ j ∑ m j = 0 ∞ ( − a j ) m j q 2 m j ( m j + 2 μ j ) ( 1 − q 2 ) m j + μ j + Δ + 1 2 q 2 Γ ( m j + μ j + Δ + 1 2 ) ( q 2 ; q 2 ) m j + 2 μ j K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) s m j + μ j + Δ + 1 = A Δ ∏ m j = 0 ∞ ( − a j s ) m j q 2 m j ( m j + 2 μ j ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 )

Theorem 3. Let J 2 μ j ( 3 ) ( q − 1 a 1 t ; q 2 ) , ⋯ , J 2 μ n ( 3 ) ( q − 1 a n t ; q 2 ) be s set of q2-Bessel functions, f ( t ) = t Δ − 1 ∏ j = 1 n J 2 μ j ( 3 ) ( q − 1 a j t ; q 2 ) where Δ , a j and μ j for j = 1 , 2 , ⋯ , n are constants. Then we have

q L 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j q s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j q m j ( m j − 1 ) ( a j q s ) m j     ⋅ B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 ) (27)

and the q-analogue of Laplace transformation q l 2 of f ( t ) is given by:

q l 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j q s ) μ j ∑ m j = 0 ∞ ( − a j q s ) m j q m j ( m j − 1 ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 )     ⋅ B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 ) 2 (28)

Proof. Now

J 2 μ j ( 3 ) ( a j q k − 1 s − 1 ; q 2 ) = ( a j q k − 1 s − 1 ) 2 μ j ∑ m j = 0 ∞ ( − 1 ) m j q 2 m j ( m j − 1 ) 2 ( q 2 a j q k − 1 s − 1 ) m j ( q 2 ; q 2 ) m j + 2 μ j ( q 2 ; q 2 ) m j

q L 2 ( f ( t ) ; s ) = ( q 2 ; q 2 ) ∞ [ 2 ] s 2 ∑ k = 0 ∞ q 2 k ( q k s − 1 ) Δ − 1 ( q 2 ; q 2 ) k ∏ j = 1 n ( a j q k − 1 s − 1 ) μ j     ⋅ ∑ m j = 0 ∞ ( − 1 ) m j q m j ( m j − 1 ) ( q 2 a j q k − 1 s − 1 ) m j ( q 2 ; q 2 ) m j + 2 μ j ( q 2 ; q 2 ) m j

put α = m j + μ j + Δ + 1 2 , we get

q L 2 ( f ( t ) ; s ) = A Δ ∏ j = 1 n ( a j q s ) μ j ∑ m j = 0 ∞ ( − 1 ) m j q m j ( m j − 1 ) ( a j q s ) m j B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 )

Similarly

q l 2 ( f ( t ) ; s ) = 1 [ 2 ] 1 ( − s 2 ; q 2 ) ∞ ∏ j = 1 n ( q k − 1 ) μ j ( a j ) μ j ∑ m j = 0 ∞ ( − 1 ) m j q m j ( m j − 1 ) ( q k m j ) ( q a j ) m j ( q 2 ; q 2 ) m j + μ 2 ( q 2 ; q 2 ) m j     ⋅ ∑ k = 0 ∞     q k ( Δ + 1 ) ( − s 2 ; q 2 ) k .

Put α = m j + μ j + Δ + 1 2 , A = 1 s 2 we get

q l 2 ( f ( t ) ; s ) = ( 1 − q 2 ) Δ 2 [ 2 ] s Δ + 1 ( q 2 ; q 2 ) ∞ ∏ j = 1 n ( a j q s ) μ j     ⋅ ∑ m j = 0 ∞ ( − a j q s ) m j q m j ( m j − 1 ) B m j ( q 2 ) Γ q 2 ( m j + μ j + Δ + 1 2 ) K ( 1 s 2 ; m j + μ j + Δ + 1 2 ) .

4. Special Cases

1) Let n = 1 , μ 1 = μ , a 1 = a in above theorems, respectively we have:

q L 2 ( t Δ − 1 J 2 μ ( 1 ) ( 2 a t ; q 2 ) ; s ) = A Δ ( a s ) μ ∑ m = 0 ∞ ( − a s ) m B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (29)

q l 2 ( t Δ − 1 J 2 μ ( 1 ) ( 2 a t ; q 2 ) ; s ) = A Δ ( a s ) μ ∑ m = 0 ∞ ( − a s ) m K ( 1 s 2 ; m + μ + Δ + 1 2 ) B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (30)

q L 2 ( t Δ − 1 J 2 μ ( 2 ) ( 2 a t ; q 2 ) ; s ) = A Δ ( a s ) μ ∑ m = 0 ∞ ( − 1 ) m q 2 m ( m + 2 μ ) B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (31)

q l 2 ( t Δ − 1 J 2 μ ( 2 ) ( 2 a t ; q 2 ) ; s ) = A Δ ( a s ) μ ∑ m = 0 ∞ ( − a s ) m q 2 m ( m + 2 μ ) K ( 1 s 2 ; m + μ + Δ + 1 2 ) B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (32)

q L 2 ( t Δ − 1 J 2 μ ( 3 ) ( 2 a q − 1 t ; q 2 ) ; s ) = A Δ ( a q s ) μ ∑ m = 0 ∞ ( − 1 ) m q m ( m − 1 ) ( a q s ) m B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (33)

q l 2 ( t Δ − 1 J 2 μ ( 3 ) ( 2 a q − 1 t ; q 2 ) ; s ) = A Δ ( a s ) μ ∑ m = 0 ∞ ( a q s ) m q m ( m − 1 ) K ( 1 s 2 ; m + μ + Δ + 1 2 ) B m ( q 2 ) Γ q 2 ( m + μ + Δ + 1 2 ) (34)

2) Put Δ − 1 = μ in part (29) above, then

q L 2 ( t μ J 2 μ ( 1 ) ( 2 a t ; q 2 ) ; s ) = ( 1 − q 2 ) μ + 1 2 [ 2 ] s μ + 2 ( q 2 ; q 2 ) ∞ ( a s ) μ

∑ m = 0 ∞ ( a s ) m ( q 2 μ + m + 2 ; q 2 ) ∞ ( 1 − q 2 ) m + μ − 1 2 ( q 2 ; q 2 ) m Γ q 2 ( m + 2 μ + 2 2 ) = ( a s ) μ [ 2 ] s μ + 2 ∑ m = 0 ∞ ( − a s ) m ( q 2 ; q 2 ) m = ( a ) μ [ 2 ] s 2 μ + 2 e q 2 ( − a s ) .

3) Put μ = 0 we get

q L 2 ( J 0 ( 1 ) ( 2 a t ; q 2 ) ; s ) = 1 [ 2 ] s 2 e q 2 ( − a s ) .

which is the same result cited by [7] .

4) Put Δ − 1 in (33), then

q L 2 ( t μ J 2 μ ( 3 ) ( 2 q − 1 a t ) ; s ) = ( 1 − q 2 ) μ + 1 2 [ 2 ] s μ + 2 ( q 2 ; q 2 ) ∞ ( a q s ) μ .

∑ m = 0 ∞ ( − 1 ) m q m ( m − 1 ) ( a q s ) m ( q 2 μ + m + 2 ; q 2 ) ( 1 − q 2 ) m + μ − 1 2 Γ q 2 ( m + 2 μ + 2 2 ) ( q 2 ; q 2 ) m = ( a q ) μ [ 2 ] s 2 μ + 2 ∑ m = 0 ∞ ( − 1 ) m ( a q s ) m q 2 m m − 1 2 ( q 2 ; q 2 ) m = ( a q ) μ [ 2 ] s 2 μ + 2 E q 2 ( a q s ) .

5) Let μ = 0 and a = 0 in (34), then

q L 2 ( t Δ − 1 ; s ) = ( 1 − q 2 ) Δ 2 [ 2 ] s Δ + 1 1 K ( 1 s 2 ; Δ + 1 2 ) ( 1 − q 2 ) − 1 2 Γ q 2 ( Δ + 1 2 )

replacing Δ − 1 by α , we get

q L 2 ( t α ; s ) = ( 1 − q 2 ) α 2 [ 2 ] s α + 2 1 K ( 1 s 2 ; 1 + α 2 ) Γ q 2 ( 1 + α 2 )

which is the same result in [8] .

Acknowledgements

The authors are thankful to Professor S. K. Al-Omari for his suggestions in this paper.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Al-Shibani, A. and Al-Khairy, R.T. (2019) On q-Analogues of Laplace Type Integral Transforms of q2-Bessel Functions. Applied Mathematics, 10, 301-311. https://doi.org/10.4236/am.2019.105021

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