On q-Analogues of Laplace Type Integral Transforms of q2-Bessel Functions

Abstract

The present paper deals with the evaluation of the q-Analogues of Laplece transforms of a product of basic analogues of q2-special functions. We apply these transforms to three families of q-Bessel functions. Several special cases have been deducted.

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Alshibani, A. and Alkhairy, R. (2019) On q-Analogues of Laplace Type Integral Transforms of q2-Bessel Functions. Applied Mathematics, 10, 301-311. doi: 10.4236/am.2019.105021.

1. 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  and denoted by Ismail  as

${J}_{\mu }^{\left(1\right)}\left(z;q\right)={\left(\frac{z}{2}\right)}^{\mu }\underset{n=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{z}^{2}}{4}\right)}^{n}}{{\left(q,q\right)}_{\mu +n}{\left(q;q\right)}_{n}},|z|<2$ (1)

${J}_{\mu }^{\left(2\right)}\left(z;q\right)={\left(\frac{z}{2}\right)}^{\mu }\underset{n=0}{\overset{\infty }{\sum }}\frac{{q}^{n\left(n+\mu \right)}{\left(\frac{-{z}^{2}}{4}\right)}^{n}}{{\left(q,q\right)}_{\mu +n}{\left(q;q\right)}_{n}},z\in ℂ$ (2)

The third related q-Bessel function ${J}_{\mu }^{\left(3\right)}\left(z;q\right)$ was introduced in a full case as 

${J}_{\mu }^{\left(3\right)}\left(z;q\right)={z}^{\mu }\underset{n=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n}{q}^{\frac{n\left(n-1\right)}{2}}{\left(q{z}^{2}\right)}^{n}}{{\left(q,q\right)}_{\mu +n}{\left(q;q\right)}_{n}},z\in ℂ$ (3)

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

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  as

${}_{q}{L}_{2}\left(f\left(\xi \right);y\right)=\frac{1}{1-{q}^{2}}{\int }_{0}^{{y}^{-1}}\xi {E}_{{q}^{2}}\left({q}^{2}{y}^{2}{\xi }^{2}\right)f\left(\xi \right)\text{d}\xi$ (4)

and expressed in terms of series representation as

${}_{q}{L}_{2}\left(f\left(\xi \right);y\right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{{\left[2\right]}_{q}{y}^{2}}\underset{i=0}{\overset{\infty }{\sum }}\frac{{q}^{2i}}{{\left({q}^{2};{q}^{2}\right)}_{i}}f\left({q}^{i}{y}^{-1}\right).$ (5)

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

${}_{q}{\mathcal{l}}_{2}\left(f\left(\xi \right);y\right)=\frac{1}{1-{q}^{2}}{\int }_{0}^{\infty }\xi {e}_{{q}^{2}}\left(-{y}^{2}{\xi }^{2}\right)f\left(\xi \right){\text{d}}_{q}\xi$ (6)

whose q-series representation expressed as

${}_{q}{\mathcal{l}}_{2}\left(f\left(\xi \right);y\right)=\frac{1}{{\left[2\right]}_{2}{\left(-{y}^{2};{q}^{2}\right)}_{\infty }}\underset{i\in ℤ}{\sum }\text{ }{q}^{2i}f\left({q}^{i}\right){\left(-{y}^{2};{q}^{2}\right)}_{i}.$ (7)

In this paper we build upon analysis of  . Following  , we discuss the q-Laplace type integral transforms (4) and (7) on the q-Bessel functions ${J}_{\mu }^{\left(1\right)}\left(z;q\right)$ , ${J}_{\mu }^{\left(2\right)}\left(z;q\right)$ and ${J}_{\mu }^{\left(3\right)}\left(z;q\right)$ , 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 . For a complex number a, the q-analogue of a is introduced as ${\left[a\right]}_{q}=\frac{1-{q}^{a}}{1-q}$ . Also, by fixing $a\in ℂ$ , the q-shifted factorials are defined as

${\left(a;q\right)}_{0}=1;{\left(a,q\right)}_{n}=\underset{k=0}{\overset{n-1}{\prod }}\left(1-a{q}^{k}\right),n=1,2,\cdots ;{\left(a;q\right)}_{\infty }=\underset{n\to \infty }{\mathrm{lim}}{\left(a;q\right)}_{n}.$ (8)

This indeed lead to the conclusion

$\left({\left[n\right]}_{q}\right)!=\frac{{\left(q;q\right)}_{n}}{{\left(1-q\right)}^{n}},n\in ℕ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(a;q\right)}_{x}=\frac{{\left(a;q\right)}_{\infty }}{{\left(a{q}^{x};q\right)}_{\infty }}.$ (9)

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

${e}_{q}\left(x\right)=\underset{0}{\overset{\infty }{\sum }}\frac{{x}^{n}}{{\left(q;q\right)}_{n}}=\frac{1}{{\left(x;q\right)}_{\infty }},|x|<1.$ (10)

and

${E}_{q}\left(x\right)=\underset{0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n}{q}^{n\frac{n-1}{2}}{x}^{n}}{{\left(q;q\right)}_{n}},x\in ℂ.$ (11)

Indeed it has been shown that

${e}_{q}\left(x\right)=\frac{1}{{\left(x;q\right)}_{\infty }},|x|<1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{E}_{q}\left(x\right)={\left(x,q\right)}_{\infty },x\in ℂ$ (12)

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

${\int }_{0}^{x}f\left(t\right){\text{d}}_{q}t=x\left(1-q\right)\underset{k=0}{\overset{\infty }{\sum }}\text{ }{q}^{k}f\left(x{q}^{k}\right)$ (13)

and

${\int }_{0}^{\infty /A}f\left(t\right){\text{d}}_{q}t=\left(1-q\right)\underset{k\in ℤ}{\sum }\frac{{q}^{k}}{A}f\left(\frac{{q}^{k}}{A}\right).$ (14)

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

${\Gamma }_{q}\left(\alpha \right)={\int }_{0}^{1/\left(1-q\right)}{x}^{\alpha -1}{E}_{q}\left(q\left(1-q\right)x\right){\text{d}}_{q}x,\left(\alpha >0\right)$ (15)

and

${}_{q}\Gamma \left(\alpha \right)=K\left(A;\alpha \right){\int }_{0}^{\infty /A\left(1-q\right)}{x}^{\alpha -1}{e}_{q}\left(-\left(1-q\right)x\right){\text{d}}_{q}x$ (16)

where, ${\alpha }_{1}>0$ , where $K\left(A;\alpha \right)$ is the function given by

$K\left(A;\alpha \right)={A}^{\alpha -1}\frac{{\left(-q/\alpha ;q\right)}_{\infty }{\left(-\alpha ;q\right)}_{\infty }}{{\left(-{q}^{t}/\alpha ;q\right)}_{\infty }{\left(-\alpha {q}^{1-t};q\right)}_{\infty }}.$ (17)

Some useful results, for $x\ne 0,-1,-2,\cdots$ , we use here are given by

${\Gamma }_{q}\left(\alpha \right)=\frac{{\left(q;q\right)}_{\infty }}{{\left(1-q\right)}^{\alpha -1}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{k\alpha }}{{\left(q;q\right)}_{k}}=\frac{{\left(q;q\right)}_{\infty }}{{\left({q}^{\alpha };q\right)}_{\infty }}{\left(1-q\right)}^{1-x},$ (18)

and

${}_{q}\Gamma \left(\alpha \right)=\frac{K\left(A;\alpha \right)}{{\left(1-q\right)}^{\alpha -1}{\left(-\frac{1}{A};q\right)}_{\infty }}\underset{k\in ℤ}{\sum }\left(\frac{{q}^{k}}{A}\right){\left(-\frac{1}{A};q\right)}_{k}.$ (19)

3. Main Theorems

Theorem 1. Let ${J}_{2{\mu }_{1}}^{\left(1\right)}\left(2\sqrt{{a}_{1}t};{q}^{2}\right),\cdots ,{J}_{2{\mu }_{n}}^{\left(1\right)}\left(2\sqrt{{a}_{n}t};{q}^{2}\right)$ be a set of first kind of q2-Bessel functions, $f\left(t\right)={t}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{J}_{2{\mu }_{j}}^{\left(1\right)}\left(2\sqrt{{a}_{j}t};{q}^{2}\right)$ , where $\Delta$ , ${a}_{j}$ and ${\mu }_{j}$ for $j=1,2,\cdots ,n$ are constants; then the q-analogue of Lablace transformation ${}_{q}{L}_{2}$ of $f\left(t\right)$ is given as:

$\begin{array}{l}{}_{q}{L}_{2}\left(f\left(t\right);s\right)\\ ={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\end{array}$ (20)

and the q-analogue of Laplace transformation ${}_{q}{l}_{2}$ of $f\left(t\right)$ is given as:

$\begin{array}{l}{}_{q}{l}_{2}\left(f\left(t\right);s\right)\\ ={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{B}_{{m}_{j}}{\left({q}^{2}\right)}_{{q}^{2}}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}.\end{array}$ (21)

where $Re\left(s\right)>0$ , $Re\left(\Delta \right)>0$ and

${A}_{\Delta }=\frac{{\left(1-{q}^{2}\right)}^{\Delta /2}}{\left[2\right]{s}^{{}^{\Delta +1}}{\left({q}^{2};{q}^{2}\right)}_{\infty }},{B}_{{m}_{j}}\left({q}^{2}\right)=\frac{{\left({q}^{2{\mu }_{j}+{m}_{j}+2};{q}^{2}\right)}_{\infty }\left(1-{q}^{2}\right)\frac{{m}_{j}+{\mu }_{j}-1}{2}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}$

Proof. Now,

${}_{q}{L}_{2}\left(f\left(t\right);s\right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{2}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{2k}f\left({q}^{k}{s}^{-1}\right)}{{\left({q}^{2};{q}^{2}\right)}_{k}}$

since

${J}_{2{\mu }_{j}}^{\left(1\right)}\left(2\sqrt{{a}_{j}t};{q}^{2}\right)={\left(\frac{2\sqrt{{a}_{j}t}}{2}\right)}^{2{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{\left(2\sqrt{{a}_{j}t}\right)}^{2}}{4}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}$

so

$\begin{array}{l}{}_{q}{L}_{a}\left(f\left(t\right);s\right)\\ =\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{2}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{2k}}{{\left({q}^{2};{q}^{2}\right)}_{k}}{\left({q}^{k}{s}^{-1}\right)}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{\left(\sqrt{{a}_{j}{q}^{k}{s}^{-1}}\right)}^{2{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{\left({a}_{j}{q}^{k}{s}^{-1}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\\ =\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{\Delta +1}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{k\left(\Delta +1\right)}}{{\left({q}^{2};{q}^{2}\right)}_{k}}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}{q}^{k}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{\left(\frac{{a}_{j}{q}^{k}}{s}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\\ =\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{\Delta +1}}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{\left(\frac{{a}_{j}}{s}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{k\left(\Delta +1+{m}_{j}+{\mu }_{j}\right)}}{{\left({q}^{2};{q}^{2}\right)}_{k}}\end{array}$ (22)

Since

${\Gamma }_{{q}^{2}}\left(\alpha \right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{{\left(1-{q}^{2}\right)}^{\alpha -1}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{2k\alpha }}{{\left({q}^{2};{q}^{2}\right)}_{k}}$

putting $\alpha =\frac{1+\Delta +{m}_{j}+{\mu }_{j}}{2}$ , so (22) becomes:

${}_{q}{L}_{s}\left(f\left(t\right);s\right)=\frac{1}{\left[2\right]{s}^{\Delta +1}}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{\left(\frac{{a}_{j}}{s}\right)}^{{m}_{j}}{\left(1-{q}^{2}\right)}^{\frac{1+\Delta +{m}_{j}+{\mu }_{j}}{2}}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}.$

${\Gamma }_{{q}^{2}}\left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)$ (23)

Since

${\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{{\left({q}^{2}{q}^{2{\mu }_{j}+{m}_{j}};{q}^{2}\right)}_{\infty }}$

so (23) becomes:

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)=\frac{{\left(1-{q}^{2}\right)}^{\Delta /2}}{\left[2\right]{s}^{\Delta +1}{\left({q}^{2};{q}^{2}\right)}_{\infty }}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{\left(\frac{{a}_{j}}{s}\right){\left(-1\right)}^{{m}_{j}}{\left({q}^{2{\mu }_{j}+{m}_{j}+2};{q}^{2}\right)}_{\infty }{\left(1-{q}^{2}\right)}^{\frac{{m}_{j}+{\mu }_{j}-1}{2}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}.\end{array}$

$\begin{array}{l}{\Gamma }_{{q}^{2}}\left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\\ ={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\left(\frac{{a}_{j}}{s}\right){\left(-1\right)}^{{m}_{j}}{B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\frac{\left({m}_{j}+{\mu }_{j}+\Delta +1\right)}{2}\end{array}$

Similarly we have

$\begin{array}{c}{}_{q}{l}_{2}\left(f\left(t\right);s\right)=\frac{1}{\left[2\right]}\frac{1}{{\left(-{s}^{2};{q}^{2}\right)}_{\infty }}\underset{k=0}{\overset{\infty }{\sum }}\text{ }{q}^{2k}{\left(-{s}^{2};{q}^{2}\right)}_{k}{\left({q}^{k}\right)}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{J}_{2{\mu }_{j}}^{\left(1\right)}\left(2\sqrt{{a}_{j}{q}^{k}};{q}^{2}\right)\\ =\frac{1}{\left[2\right]}\frac{1}{{\left(-{s}^{2};{q}^{2}\right)}_{\infty }}\underset{k=0}{\overset{\infty }{\sum }}\text{ }{q}^{2k}{\left(-{s}^{2};{q}^{2}\right)}_{k}{\left({q}^{k}\right)}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{\left({a}_{j}{q}^{k}\right)}^{{\mu }_{j}}.\end{array}$

$\begin{array}{l}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-{a}_{j}{q}^{k}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}+2{\mu }_{j}}\\ =\underset{j=1}{\overset{n}{\prod }}\frac{{\left({a}_{j}\right)}^{{\mu }_{j}}}{\left[2\right]}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-{a}_{j}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-{s}^{2};{q}^{2}\right)}_{k}{q}^{k\left({m}_{j}+{\mu }_{j}+\Delta +1\right)}}{{\left(-{s}^{2};{q}^{2}\right)}_{\infty }}\end{array}$

Now using

${}_{{q}^{2}}\Gamma \left(\alpha \right)=\frac{K\left(A;\alpha \right)}{{\left(1-{q}^{2}\right)}^{\alpha -1}{\left(-\frac{1}{A};{q}^{2}\right)}_{\infty }}\underset{k\in Z}{\sum }{\left(\frac{{q}^{K}}{A}\right)}^{\alpha }{\left(-\frac{1}{A};{q}^{2}\right)}_{K}$

with $A=\frac{1}{{s}^{2}}$ , $\alpha =\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}$ we get

$\begin{array}{l}{}_{q}{l}_{2}\left(f\left(t\right);s\right)\\ =\underset{j=1}{\overset{m}{\prod }}\frac{{\left({a}_{j}\right)}^{{\mu }_{j}}}{\left[2\right]{s}^{{\mu }_{j}+\Delta +1}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{\left(1-{q}^{2}\right)}^{\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}}{q}^{2}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right){\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\\ =\frac{{\left(1-{q}^{2}\right)}^{\frac{\Delta }{2}}}{\left[2\right]{s}^{\Delta +1}{\left({q}^{2};{q}^{2}\right)}_{\infty }}\underset{j=1}{\overset{m}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{\left(1-{q}^{2}\right)}^{\frac{{m}_{j}+{\mu }_{j}-1}{2}}{\left({q}^{{m}_{j}+2{\mu }_{j}+2};{q}^{2}\right)}_{\infty }}{K\left(\frac{1}{{s}^{2}},\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right){\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}.\end{array}$

$\begin{array}{l}{}_{{q}^{2}}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\\ ={A}_{\Delta }\underset{j=1}{\overset{m}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{B}_{{m}_{j}}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left({m}_{j}+{\mu }_{j}+\Delta +1\right)\end{array}$

Theorem 2. Let ${J}_{2{\mu }_{1}}^{\left(2\right)}\left(2\sqrt{a,t};{q}^{2}\right),\cdots ,{J}_{2{\mu }_{n}}^{\left(2\right)}\left(2\sqrt{at};{q}^{2}\right)$ be a set of second order q2-Bessel function, $f\left(t\right)={t}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{J}_{2{\mu }_{j}}^{\left(2\right)}\left(2\sqrt{{a}_{j}t};{q}^{2}\right)$ where $\Delta ,{a}_{j}$ and ${\mu }_{j}$ for $j=1,2,\cdots ,n$ are constants then ${}_{q}{L}_{2}$ -transform of $f\left(t\right)$ is given as:

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right),s\right)={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}{\left(\frac{{a}_{j}}{s}\right)}^{{m}_{j}+{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left({m}_{j}+{\mu }_{j}+\Delta +1\right)\end{array}$ (24)

and the q-analogue of Laplace transformation ${}_{q}{l}_{2}$ of $f\left(t\right)$ is given as:

$\begin{array}{c}{}_{q}{l}_{2}\left(f\left(t\right);s\right)={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {B}_{{m}_{j}}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\end{array}$ (25)

Proof. Now,

${J}_{2{\mu }_{j}}^{\left(2\right)}\left(2\sqrt{{a}_{j}t};{q}^{2}\right)={\left(\frac{2\sqrt{{a}_{j}t}}{2}\right)}^{2{\mu }_{j}}\underset{{m}_{j}}{\overset{\infty }{\sum }}\frac{{\left(-\frac{{\left(2\sqrt{{a}_{j}t}\right)}^{2}}{4}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{a}_{j}\right)}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}$

so

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{2}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{2k}}{{\left({q}^{2};{q}^{2}\right)}_{k}}{\left({q}^{k}{s}^{-1}\right)}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{2\sqrt{{a}_{j}{q}^{k}{s}^{-1}}}{2}\right)}^{2{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-\frac{{\left(2\sqrt{{a}_{j}{q}^{k}{s}^{-1}}\right)}^{2}}{4}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}}{{\left({q}^{2};{q}^{2}\right)}_{2{\mu }_{j}+{m}_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\end{array}$ (26)

By the same argument we can write (26) as

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{\Delta +1}{\left({q}^{2};{q}^{2}\right)}_{\infty }}\underset{j=1}{\overset{n}{\prod }}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {\left(\frac{{a}_{j}}{s}\right)}^{{m}_{j}+{\mu }_{j}}{\left({q}^{2{\mu }_{j}+{m}_{j}+2};{q}^{2}\right)}_{\infty }\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{k\left({m}_{j}+{\mu }_{j}+1+\Delta \right)}}{{\left({q}^{2};{q}^{2}\right)}_{k}}\end{array}$

put $\alpha =\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}$ in ${\Gamma }_{{q}^{2}}\left(\alpha \right)$ , then

So (25) becomes:

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{s}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}{\left(\frac{{a}_{j}}{s}\right)}^{{m}_{j}+{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left({m}_{j}+{\mu }_{j}+\Delta +1\right)\end{array}$

Similarly

$\begin{array}{c}{}_{q}{l}_{2}\left(f\left(t\right);s\right)=\frac{1}{\left[2\right]}\frac{1}{{\left(-{s}^{2};{q}^{2}\right)}_{\infty }}\underset{k=0}{\overset{\infty }{\sum }}{q}^{2k}{\left(-{s}^{2};{q}^{2}\right)}_{k}{\left({q}^{k}\right)}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{\left({a}_{j}{q}^{k}\right)}^{{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-{a}_{j}{q}^{k}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\end{array}$

Put $A=\frac{1}{{s}^{2}},\alpha =\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}$ we get

$\begin{array}{l}{}_{q}{l}_{2}\left(f\left(t\right);s\right)\\ =\frac{1}{\left[2\right]}\underset{j=1}{\overset{n}{\prod }}{\left({a}_{j}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-{a}_{j}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}{\left(1-{q}^{2}\right)}^{\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}}{q}^{2}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right){s}^{{m}_{j}+{\mu }_{j}+\Delta +1}}\\ ={A}_{\Delta }\underset{{m}_{j}=0}{\overset{\infty }{\prod }}\frac{{\left(\frac{-{a}_{j}}{s}\right)}^{{m}_{j}}{q}^{2{m}_{j}\left({m}_{j}+2{\mu }_{j}\right)}}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{B}_{{m}_{j}}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\end{array}$

Theorem 3. Let ${J}_{2{\mu }_{j}}^{\left(3\right)}\left(\sqrt{{q}^{-1}{a}_{1}t};{q}^{2}\right),\cdots ,{J}_{2{\mu }_{n}}^{\left(3\right)}\left(\sqrt{{q}^{-1}{a}_{n}t};{q}^{2}\right)$ be s set of q2-Bessel functions, $f\left(t\right)={t}^{\Delta -1}\underset{j=1}{\overset{n}{\prod }}{J}_{2{\mu }_{j}}^{\left(3\right)}\left(\sqrt{{q}^{-1}{a}_{j}t};{q}^{2}\right)$ where $\Delta ,{a}_{j}$ and ${\mu }_{j}$ for $j=1,2,\cdots ,n$ are constants. Then we have

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{qs}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}{\left(\frac{{a}_{j}q}{s}\right)}^{{m}_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\end{array}$ (27)

and the q-analogue of Laplace transformation ${}_{q}{l}_{2}$ of $f\left(t\right)$ is given by:

$\begin{array}{c}{}_{q}{l}_{2}\left(f\left(t\right);s\right)={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{qs}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}q}{s}\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot {B}_{{m}_{j}}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \frac{\left({m}_{j}+{\mu }_{j}+\Delta +1\right)}{2}\end{array}$ (28)

Proof. Now

${J}_{2{\mu }_{j}}^{\left(3\right)}\left(\sqrt{{a}_{j}{q}^{k-1}{s}^{-1}};{q}^{2}\right)={\left(\sqrt{{a}_{j}{q}^{k-1}{s}^{-1}}\right)}^{2{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{{m}_{j}}\frac{{q}^{\frac{2{m}_{j}\left({m}_{j}-1\right)}{2}}{\left({q}^{2}{a}_{j}{q}^{k-1}{s}^{-1}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}$

$\begin{array}{c}{}_{q}{L}_{2}\left(f\left(t\right);s\right)=\frac{{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left[2\right]{s}^{2}}\underset{k=0}{\overset{\infty }{\sum }}\frac{{q}^{2k}{\left({q}^{k}{s}^{-1}\right)}^{\Delta -1}}{{\left({q}^{2};{q}^{2}\right)}_{k}}\underset{j=1}{\overset{n}{\prod }}{\left({a}_{j}{q}^{k-1}{s}^{-1}\right)}^{{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}{\left({q}^{2}{a}_{j}{q}^{k-1}{s}^{-1}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+2{\mu }_{j}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\end{array}$

put $\alpha =\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}$ , we get

$\begin{array}{l}{}_{q}{L}_{2}\left(f\left(t\right);s\right)\\ ={A}_{\Delta }\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{qs}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}{\left(\frac{{a}_{j}q}{s}\right)}^{{m}_{j}}{B}_{{m}_{j}}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)\end{array}$

Similarly

$\begin{array}{l}{}_{q}{l}_{2}\left(f\left(t\right);s\right)\\ =\frac{1}{\left[2\right]}\frac{1}{{\left(-{s}^{2};{q}^{2}\right)}_{\infty }}\underset{j=1}{\overset{n}{\prod }}{\left({q}^{k-1}\right)}^{{\mu }_{j}}{\left({a}_{j}\right)}^{{\mu }_{j}}\underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}\left({q}^{k{m}_{j}}\right){\left(q{a}_{j}\right)}^{{m}_{j}}}{{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}+{\mu }_{2}}{\left({q}^{2};{q}^{2}\right)}_{{m}_{j}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{k=0}{\overset{\infty }{\sum }}\text{ }\text{ }{q}^{k\left(\Delta +1\right)}{\left(-{s}^{2};{q}^{2}\right)}_{k}.\end{array}$

Put $\alpha =\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}$ , $A=\frac{1}{{s}^{2}}$ we get

$\begin{array}{c}{}_{q}{l}_{2}\left(f\left(t\right);s\right)=\frac{{\left(1-{q}^{2}\right)}^{\frac{\Delta }{2}}}{\left[2\right]{s}^{\Delta +1}{\left({q}^{2};{q}^{2}\right)}_{\infty }}\underset{j=1}{\overset{n}{\prod }}{\left(\frac{{a}_{j}}{qs}\right)}^{{\mu }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \underset{{m}_{j}=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-{a}_{j}q}{s}\right)}^{{m}_{j}}{q}^{{m}_{j}\left({m}_{j}-1\right)}{B}_{{m}_{j}}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}{K\left(\frac{1}{{s}^{2}};\frac{{m}_{j}+{\mu }_{j}+\Delta +1}{2}\right)}.\end{array}$

4. Special Cases

1) Let $n=1$ , ${\mu }_{1}=\mu$ , ${a}_{1}=a$ in above theorems, respectively we have:

$\begin{array}{l}{}_{q}{L}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}{\left(\frac{-a}{s}\right)}^{m}{B}_{m}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (29)

$\begin{array}{l}{}_{q}{l}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-a}{s}\right)}^{m}}{K\left(\frac{1}{{s}^{2}};\frac{m+\mu +\Delta +1}{2}\right)}{B}_{m}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (30)

$\begin{array}{l}{}_{q}{L}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(2\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{m}{q}^{2m\left(m+2\mu \right)}{B}_{m}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (31)

$\begin{array}{l}{}_{q}{l}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(2\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-a}{s}\right)}^{m}{q}^{2m\left(m+2\mu \right)}}{K\left(\frac{1}{{s}^{2}};\frac{m+\mu +\Delta +1}{2}\right)}{B}_{m}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (32)

$\begin{array}{l}{}_{q}{L}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{a{q}^{-1}t};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{qs}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{m}{q}^{m\left(m-1\right)}{\left(\frac{aq}{s}\right)}^{m}{B}_{m}\left({q}^{2}\right){\Gamma }_{{q}^{2}}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (33)

$\begin{array}{l}{}_{q}{l}_{2}\left({t}^{\Delta -1}{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{a{q}^{-1}t};{q}^{2}\right);s\right)\\ ={A}_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }\underset{m=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{aq}{s}\right)}^{m}{q}^{m\left(m-1\right)}}{K\left(\frac{1}{{s}^{2}};\frac{m+\mu +\Delta +1}{2}\right)}{B}_{m}\left({q}^{2}\right){}_{{q}^{2}}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (34)

2) Put $\Delta -1=\mu$ in part (29) above, then

${}_{q}{L}_{2}\left({t}^{\mu }{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)=\frac{{\left(1-{q}^{2}\right)}^{\frac{\mu +1}{2}}}{\left[2\right]{s}^{\mu +2}{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left(\frac{a}{s}\right)}^{\mu }$

$\begin{array}{l}\underset{m=0}{\overset{\infty }{\sum }}{\left(\frac{a}{s}\right)}^{m}\frac{{\left({q}^{2\mu +m+2};{q}^{2}\right)}_{\infty }{\left(1-{q}^{2}\right)}^{\frac{m+\mu -1}{2}}}{{\left({q}^{2};{q}^{2}\right)}_{m}}{\Gamma }_{{q}^{2}}\left(\frac{m+2\mu +2}{2}\right)\\ =\frac{{\left(\frac{a}{s}\right)}^{\mu }}{\left[2\right]{s}^{\mu +2}}\underset{m=0}{\overset{\infty }{\sum }}\frac{{\left(\frac{-a}{s}\right)}^{m}}{{\left({q}^{2};{q}^{2}\right)}_{m}}=\frac{{\left(a\right)}^{\mu }}{\left[2\right]{s}^{2\mu +2}}{e}_{{q}^{2}}\left(\frac{-a}{s}\right).\end{array}$

3) Put $\mu =0$ we get

${}_{q}{L}_{2}\left({J}_{0}^{\left(1\right)}\left(2\sqrt{at};{q}^{2}\right);s\right)=\frac{1}{\left[2\right]{s}^{2}}{e}_{{q}^{2}}\left(\frac{-a}{s}\right).$

which is the same result cited by  .

4) Put $\Delta -1$ in (33), then

${}_{q}{L}_{2}\left({t}^{\mu }{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{{q}^{-1}at}\right);s\right)=\frac{{\left(1-{q}^{2}\right)}^{\frac{\mu +1}{2}}}{\left[2\right]{s}^{\mu +2}{\left({q}^{2};{q}^{2}\right)}_{\infty }}{\left(\frac{a}{qs}\right)}^{\mu }.$

$\begin{array}{l}\underset{m=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{m}\frac{{q}^{m\left(m-1\right)}{\left(\frac{aq}{s}\right)}^{m}\left({q}^{2\mu +m+2};{q}^{2}\right){\left(1-{q}^{2}\right)}^{\frac{m+\mu -1}{2}}{\Gamma }_{{q}^{2}}\left(\frac{m+2\mu +2}{2}\right)}{{\left({q}^{2};{q}^{2}\right)}_{m}}\\ =\frac{{\left(\frac{a}{q}\right)}^{\mu }}{\left[2\right]{s}^{2\mu +2}}\underset{m=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{m}\frac{{\left(\frac{aq}{s}\right)}^{m}{q}^{2m\frac{m-1}{2}}}{{\left({q}^{2};{q}^{2}\right)}_{m}}=\frac{{\left(\frac{a}{q}\right)}^{\mu }}{\left[2\right]{s}^{2\mu +2}}{E}_{{q}^{2}}\left(\frac{aq}{s}\right).\end{array}$

5) Let $\mu =0$ and $a=0$ in (34), then

${}_{q}{L}_{2}\left({t}^{\Delta -1};s\right)=\frac{{\left(1-{q}^{2}\right)}^{\frac{\Delta }{2}}}{\left[2\right]{s}^{\Delta +1}}\frac{1}{K\left(\frac{1}{{s}^{2}};\frac{\Delta +1}{2}\right)}{\left(1-{q}^{2}\right)}^{-\frac{1}{2}}{\Gamma }_{{q}^{2}}\left(\frac{\Delta +1}{2}\right)$

replacing $\Delta -1$ by $\alpha$ , we get

${}_{q}{L}_{2}\left({t}^{\alpha };s\right)=\frac{{\left(1-{q}^{2}\right)}^{\frac{\alpha }{2}}}{\left[2\right]{s}^{\alpha +2}}\frac{1}{K\left(\frac{1}{{s}^{2}};1+\frac{\alpha }{2}\right)}{\Gamma }_{{q}^{2}}\left(1+\frac{\alpha }{2}\right)$

which is the same result in  .

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.

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