Arwa Alshibani, Reem Alkhairy
Department of Mathematics, Faculty of Sciences, Imam Abdulrahman Bin Faisal University, Dammam, KSA.
DOI: 10.4236/am.2019.105021   PDF    HTML   XML   928 Downloads   2,175 Views   Citations

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.

Keywords

q-Extensions of Bessel Functions, q-Analogues of Laplace Type Integrals Transforms, q-Analogues of Gamma Function, q-Shift Factorials

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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 [1] and denoted by Ismail [2] as

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

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

The third related q-Bessel function $J_{\mu }^{\left(3\right)}\left(z\mathrm{<mo>\; ;\; </mo>}q\right)$ was introduced in a full case as [3]

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

A certain type of Laplace transforms, which is called L₂-transform, was introduced by Yürekli and Sadek [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

${}_q{L}_2\left(f\left(\xi \right)\mathrm{<mo>\; ;\; </mo>}y\right)=\frac{1}{1-q^2}{\displaystyle {\int }_0^{y^{-1}}}\xi E_{q^2}\left(q^2{y}^2{\xi }^2\right)f\left(\xi \right)\text{d}\xi$ (4)

and expressed in terms of series representation as

${}_q{L}_2\left(f\left(\xi \right)\mathrm{<mo>\; ;\; </mo>}y\right)=\frac{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{{\left[2\right]}_q{y}^2}{\displaystyle \underset{i=0}{\overset{\infty }{\sum }}}\frac{q^{2i}}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_i}f\left(q^i{y}^{-1}\right)\mathrm{.}$ (5)

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

${}_q{\mathcal{l}}_2\left(f\left(\xi \right)\mathrm{<mo>\; ;\; </mo>}y\right)=\frac{1}{1-q^2}{\displaystyle {\int }_0^{\infty }}\xi e_{q^2}\left(-y^2{\xi }^2\right)f\left(\xi \right){\text{d}}_q\xi$ (6)

whose q-series representation expressed as

${}_q{\mathcal{l}}_2\left(f\left(\xi \right)\mathrm{<mo>\; ;\; </mo>}y\right)=\frac{1}{{\left[2\right]}_2{\left(-y^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{\displaystyle \underset{i\in \mathbb{Z}}{\sum }}{q}^{2i}f\left(q^i\right){\left(-y^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_i\mathrm{.}$ (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 $J_{\mu }^{\left(1\right)}\left(z\mathrm{<mo>\; ;\; </mo>}q\right)$ , $J_{\mu }^{\left(2\right)}\left(z\mathrm{<mo>\; ;\; </mo>}q\right)$ and $J_{\mu }^{\left(3\right)}\left(z\mathrm{<mo>\; ;\; </mo>}q\right)$ , 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 $0<q<1$ . 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

${\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_0=\mathrm{1\; <mo>\; ;\; </mo>}{\left(a\mathrm{,}q\right)}_n={\displaystyle \underset{k=0}{\overset{n-1}{\prod }}}\left(1-a{q}^k\right)\mathrm{,}n=\mathrm{1,2,}\cdots \mathrm{<mo>\; ;\; </mo>}{\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }=\underset{n\to \infty }{\lim }{\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_n\mathrm{.}$ (8)

This indeed lead to the conclusion

$\left({\left[n\right]}_q\right)\mathrm{<mo>\; !\; </mo>}=\frac{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_n}{{\left(1-q\right)}^n}\mathrm{,}n\in \mathbb{N}\text{and}{\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_x=\frac{{\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}{{\left(a{q}^x\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}\mathrm{.}$ (9)

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

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

and

$E_q\left(x\right)={\displaystyle \underset{0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^n{q}^{n\frac{n-1}{2}}{x}^n}{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_n}\mathrm{,}x\in ℂ\mathrm{.}$ (11)

Indeed it has been shown that

$e_q\left(x\right)=\frac{1}{{\left(x\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}\mathrm{,}\left|x\right|<1\text{and}{E}_q\left(x\right)={\left(x\mathrm{,}q\right)}_{\infty }\mathrm{,}x\in ℂ$ (12)

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

${\displaystyle {\int }_0^x}f\left(t\right){\text{d}}_{q}t=x\left(1-q\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{q}^{k}f\left(x{q}^k\right)$ (13)

and

${\displaystyle {\int }_0^{\infty \mathrm{<mo>\; /\; </mo>}A}}f\left(t\right){\text{d}}_{q}t=\left(1-q\right){\displaystyle \underset{k\in \mathbb{Z}}{\sum }}\frac{q^k}{A}f\left(\frac{q^k}{A}\right)\mathrm{.}$ (14)

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

${\Gamma }_q\left(\alpha \right)={\displaystyle {\int }_0^{1/\left(1-q\right)}}{x}^{\alpha -1}{E}_q\left(q\left(1-q\right)x\right){\text{d}}_{q}x\mathrm{,}\left(\alpha >0\right)$ (15)

and

${}_q\Gamma \left(\alpha \right)=K\left(A\mathrm{<mo>\; ;\; </mo>}\alpha \right){\displaystyle {\int }_0^{\infty \mathrm{<mo>\; /\; </mo>}A\left(1-q\right)}}{x}^{\alpha -1}{e}_q\left(-\left(1-q\right)x\right){\text{d}}_{q}x$ (16)

where, ${\alpha }_1>0$ , where $K\left(A\mathrm{<mo>\; ;\; </mo>}\alpha \right)$ is the function given by

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

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

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

and

${}_q\Gamma \left(\alpha \right)=\frac{K\left(A\mathrm{<mo>\; ;\; </mo>}\alpha \right)}{{\left(1-q\right)}^{\alpha -1}{\left(-\frac{1}{A}\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}{\displaystyle \underset{k\in \mathbb{Z}}{\sum }}\left(\frac{q^k}{A}\right){\left(-\frac{1}{A}\mathrm{<mo>\; ;\; </mo>}q\right)}_k\mathrm{.}$ (19)

3. Main Theorems

Theorem 1. Let $J_{2{\mu }_1}^{\left(1\right)}\left(2\sqrt{a_{1}t}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)\mathrm{,}\cdots \mathrm{,}{J}_{2{\mu }_n}^{\left(1\right)}\left(2\sqrt{a_{n}t}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)$ be a set of first kind of q²-Bessel functions, $f\left(t\right)=t^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{J}_{2{\mu }_j}^{\left(1\right)}\left(2\sqrt{a_{j}t};q^2\right)$ , 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:

$\begin{array}{l}{}_q{L}_2\left(f\left(t\right);s\right)\\ =A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(\frac{-a_j}{s}\right)}^{m_j}{B}_{m_j}\left(q^2\right){\Gamma }_{q^2}\left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)\end{array}$ (20)

and the q-analogue of Laplace transformation ${}_q{l}_2$ of $f\left(t\right)$ is given as:

$\begin{array}{l}{}_q{l}_2\left(f\left(t\right);s\right)\\ =A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_j}{s}\right)}^{m_j}{B}_{m_j}{\left(q^2\right)}_{q^2}\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}.\end{array}$ (21)

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

$A_{\Delta }=\frac{{\left(1-q^2\right)}^{\Delta /2}}{\left[2\right]s^{{}^{\Delta +1}}{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}\mathrm{,}{B}_{m_j}\left(q^2\right)\mathrm{<mo>\; =\; </mo>}\frac{{\left(q^{2{\mu }_j+m_j+2}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }\left(1-q^2\right)\frac{m_j+{\mu }_j-1}{2}}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{m_j}}$

Proof. Now,

${}_q{L}_2\left(f\left(t\right)\mathrm{<mo>\; ;\; </mo>}s\right)=\frac{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{\left[2\right]s^2}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{2k}f\left(q^k{s}^{-1}\right)}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_k}$

since

$J_{2{\mu }_j}^{\left(1\right)}\left(2\sqrt{a_{j}t}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)={\left(\frac{2\sqrt{a_{j}t}}{2}\right)}^{2{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-{\left(2\sqrt{a_{j}t}\right)}^2}{4}\right)}^{m_j}}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{2{\mu }_j+m_j}{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{m_j}}$

so

$\begin{array}{l}{}_q{L}_a\left(f\left(t\right);s\right)\\ =\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^2}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{2k}}{{\left(q^2;q^2\right)}_k}{\left(q^k{s}^{-1}\right)}^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\sqrt{a_j{q}^k{s}^{-1}}\right)}^{2{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{\left(a_j{q}^k{s}^{-1}\right)}^{m_j}}{{\left(q^2;q^2\right)}_{2{\mu }_j+m_j}{\left(q^2;q^2\right)}_{m_j}}\\ =\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^{\Delta +1}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{k\left(\Delta +1\right)}}{{\left(q^2;q^2\right)}_k}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j{q}^k}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{\left(\frac{a_j{q}^k}{s}\right)}^{m_j}}{{\left(q^2;q^2\right)}_{2{\mu }_j+m_j}{\left(q^2;q^2\right)}_{m_j}}\\ =\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^{\Delta +1}}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{\left(\frac{a_j}{s}\right)}^{m_j}}{{\left(q^2;q^2\right)}_{2{\mu }_j+m_j}{\left(q^2;q^2\right)}_{m_j}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{k\left(\Delta +1+m_j+{\mu }_j\right)}}{{\left(q^2;q^2\right)}_k}\end{array}$ (22)

Since

${\Gamma }_{q^2}\left(\alpha \right)=\frac{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{{\left(1-q^2\right)}^{\alpha -1}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{2k\alpha }}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_k}$

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

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

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

Since

${\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{2{\mu }_j+m_j}=\frac{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{{\left(q^2{q}^{2{\mu }_j+m_j}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}$

so (23) becomes:

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=\frac{{\left(1-q^2\right)}^{\Delta /2}}{\left[2\right]s^{\Delta +1}{\left(q^2;q^2\right)}_{\infty }}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{\left(\frac{a_j}{s}\right){\left(-1\right)}^{m_j}{\left(q^{2{\mu }_j+m_j+2};q^2\right)}_{\infty }{\left(1-q^2\right)}^{\frac{m_j+{\mu }_j-1}{2}}}{{\left(q^2;q^2\right)}_{m_j}}.\end{array}$

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

Similarly we have

$\begin{array}{c}{}_q{l}_2\left(f\left(t\right);s\right)=\frac{1}{\left[2\right]}\frac{1}{{\left(-s^2;q^2\right)}_{\infty }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{q}^{2k}{\left(-s^2;q^2\right)}_k{\left(q^k\right)}^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{J}_{2{\mu }_j}^{\left(1\right)}\left(2\sqrt{a_j{q}^k};q^2\right)\\ =\frac{1}{\left[2\right]}\frac{1}{{\left(-s^2;q^2\right)}_{\infty }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{q}^{2k}{\left(-s^2;q^2\right)}_k{\left(q^k\right)}^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(a_j{q}^k\right)}^{{\mu }_j}.\end{array}$

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

Now using

${}_{q^2}\Gamma \left(\alpha \right)=\frac{K\left(A\mathrm{<mo>\; ;\; </mo>}\alpha \right)}{{\left(1-q^2\right)}^{\alpha -1}{\left(-\frac{1}{A}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{\displaystyle \underset{k\in Z}{\sum }}{\left(\frac{q^K}{A}\right)}^{\alpha }{\left(-\frac{1}{A}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_K$

with $A=\frac{1}{s^2}$ , $\alpha =\frac{m_j+{\mu }_j+\Delta +1}{2}$ we get

$\begin{array}{l}{}_q{l}_2\left(f\left(t\right);s\right)\\ ={\displaystyle \underset{j=1}{\overset{m}{\prod }}}\frac{{\left(a_j\right)}^{{\mu }_j}}{\left[2\right]s^{{\mu }_j+\Delta +1}}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_j}{s}\right)}^{m_j}{\left(1-q^2\right)}^{\frac{m_j+{\mu }_j+\Delta +1}{2}}{q}^2\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right){\left(q^2;q^2\right)}_{m_j+2{\mu }_j}{\left(q^2;q^2\right)}_{m_j}}\\ =\frac{{\left(1-q^2\right)}^{\frac{\Delta }{2}}}{\left[2\right]s^{\Delta +1}{\left(q^2;q^2\right)}_{\infty }}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_j}{s}\right)}^{m_j}{\left(1-q^2\right)}^{\frac{m_j+{\mu }_j-1}{2}}{\left(q^{m_j+2{\mu }_j+2};q^2\right)}_{\infty }}{K\left(\frac{1}{s^2},\frac{m_j+{\mu }_j+\Delta +1}{2}\right){\left(q^2;q^2\right)}_{m_j}}.\end{array}$

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

Theorem 2. Let $J_{2{\mu }_1}^{\left(2\right)}\left(2\sqrt{a\mathrm{,}t}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)\mathrm{,}\cdots \mathrm{,}{J}_{2{\mu }_n}^{\left(2\right)}\left(2\sqrt{at}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)$ be a set of second order q²-Bessel function, $f\left(t\right)=t^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{J}_{2{\mu }_j}^{\left(2\right)}\left(2\sqrt{a_{j}t};q^2\right)$ where $\Delta \mathrm{,}{a}_j$ and ${\mu }_j$ for $j=1,2,\cdots ,n$ are constants then ${}_q{L}_2$ -transform of $f\left(t\right)$ is given as:

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right),s\right)=A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}{\left(\frac{a_j}{s}\right)}^{m_j+{\mu }_j}\\ \cdot B_{m_j}\left(q^2\right){\Gamma }_{q^2}\left(m_j+{\mu }_j+\Delta +1\right)\end{array}$ (24)

and the q-analogue of Laplace transformation ${}_q{l}_2$ of $f\left(t\right)$ is given as:

$\begin{array}{c}{}_q{l}_2\left(f\left(t\right);s\right)=A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_j}{s}\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}\\ \cdot B_{m_j}\left(q^2\right){}_{q^2}\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)\end{array}$ (25)

Proof. Now,

$J_{2{\mu }_j}^{\left(2\right)}\left(2\sqrt{a_{j}t};q^2\right)={\left(\frac{2\sqrt{a_{j}t}}{2}\right)}^{2{\mu }_j}{\displaystyle \underset{m_j}{\overset{\infty }{\sum }}}\frac{{\left(-\frac{{\left(2\sqrt{a_{j}t}\right)}^2}{4}\right)}^{m_j}{q}^{2m_j\left(m_j+2a_j\right)}}{{\left(q^2;q^2\right)}_{2{\mu }_j+m_j}{\left(q^2;q^2\right)}_{m_j}}$

so

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^2}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{2k}}{{\left(q^2;q^2\right)}_k}{\left(q^k{s}^{-1}\right)}^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{2\sqrt{a_j{q}^k{s}^{-1}}}{2}\right)}^{2{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-\frac{{\left(2\sqrt{a_j{q}^k{s}^{-1}}\right)}^2}{4}\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}}{{\left(q^2;q^2\right)}_{2{\mu }_j+m_j}{\left(q^2;q^2\right)}_{m_j}}\end{array}$ (26)

By the same argument we can write (26) as

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^{\Delta +1}{\left(q^2;q^2\right)}_{\infty }}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}}{{\left(q^2;q^2\right)}_{m_j}}\\ \cdot {\left(\frac{a_j}{s}\right)}^{m_j+{\mu }_j}{\left(q^{2{\mu }_j+m_j+2}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{k\left(m_j+{\mu }_j+1+\Delta \right)}}{{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_k}\end{array}$

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

So (25) becomes:

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{s}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}{\left(\frac{a_j}{s}\right)}^{m_j+{\mu }_j}\\ \cdot B_{m_j}\left(q^2\right){\Gamma }_{q^2}\left(m_j+{\mu }_j+\Delta +1\right)\end{array}$

Similarly

$\begin{array}{c}{}_q{l}_2\left(f\left(t\right);s\right)=\frac{1}{\left[2\right]}\frac{1}{{\left(-s^2;q^2\right)}_{\infty }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{q}^{2k}{\left(-s^2;q^2\right)}_k{\left(q^k\right)}^{\Delta -1}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(a_j{q}^k\right)}^{{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-a_j{q}^k\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}}{{\left(q^2;q^2\right)}_{m_j+2{\mu }_j}{\left(q^2;q^2\right)}_{m_j}}\end{array}$

Put $A=\frac{1}{s^2},\alpha =\frac{m_j+{\mu }_j+\Delta +1}{2}$ we get

$\begin{array}{l}{}_q{l}_2\left(f\left(t\right);s\right)\\ =\frac{1}{\left[2\right]}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(a_j\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-a_j\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}{\left(1-q^2\right)}^{\frac{m_j+{\mu }_j+\Delta +1}{2}}{q}^2\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}{{\left(q^2;q^2\right)}_{m_j+2{\mu }_j}K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)s^{m_j+{\mu }_j+\Delta +1}}\\ =A_{\Delta }{\displaystyle \underset{m_j=0}{\overset{\infty }{\prod }}}\frac{{\left(\frac{-a_j}{s}\right)}^{m_j}{q}^{2m_j\left(m_j+2{\mu }_j\right)}}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}{B}_{m_j}\left(q^2\right){}_{q^2}\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)\end{array}$

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

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{qs}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}{\left(\frac{a_{j}q}{s}\right)}^{m_j}\\ \cdot B_{m_j}\left(q^2\right){\Gamma }_{q^2}\left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)\end{array}$ (27)

and the q-analogue of Laplace transformation ${}_q{l}_2$ of $f\left(t\right)$ is given by:

$\begin{array}{c}{}_q{l}_2\left(f\left(t\right);s\right)=A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{qs}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_{j}q}{s}\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}\\ \cdot B_{m_j}\left(q^2\right){}_{q^2}\Gamma \frac{\left(m_j+{\mu }_j+\Delta +1\right)}{2}\end{array}$ (28)

Proof. Now

$J_{2{\mu }_j}^{\left(3\right)}\left(\sqrt{a_j{q}^{k-1}{s}^{-1}};q^2\right)={\left(\sqrt{a_j{q}^{k-1}{s}^{-1}}\right)}^{2{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^{m_j}\frac{q^{\frac{2m_j\left(m_j-1\right)}{2}}{\left(q^2{a}_j{q}^{k-1}{s}^{-1}\right)}^{m_j}}{{\left(q^2;q^2\right)}_{m_j+2{\mu }_j}{\left(q^2;q^2\right)}_{m_j}}$

$\begin{array}{c}{}_q{L}_2\left(f\left(t\right);s\right)=\frac{{\left(q^2;q^2\right)}_{\infty }}{\left[2\right]s^2}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}\frac{q^{2k}{\left(q^k{s}^{-1}\right)}^{\Delta -1}}{{\left(q^2;q^2\right)}_k}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(a_j{q}^{k-1}{s}^{-1}\right)}^{{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}{\left(q^2{a}_j{q}^{k-1}{s}^{-1}\right)}^{m_j}}{{\left(q^2;q^2\right)}_{m_j+2{\mu }_j}{\left(q^2;q^2\right)}_{m_j}}\end{array}$

put $\alpha =\frac{m_j+{\mu }_j+\Delta +1}{2}$ , we get

$\begin{array}{l}{}_q{L}_2\left(f\left(t\right);s\right)\\ =A_{\Delta }{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{qs}\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}{\left(\frac{a_{j}q}{s}\right)}^{m_j}{B}_{m_j}\left(q^2\right){\Gamma }_{q^2}\left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)\end{array}$

Similarly

$\begin{array}{l}{}_q{l}_2\left(f\left(t\right);s\right)\\ =\frac{1}{\left[2\right]}\frac{1}{{\left(-s^2;q^2\right)}_{\infty }}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(q^{k-1}\right)}^{{\mu }_j}{\left(a_j\right)}^{{\mu }_j}{\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}\left(q^{k{m}_j}\right){\left(q{a}_j\right)}^{m_j}}{{\left(q^2;q^2\right)}_{m_j+{\mu }_2}{\left(q^2;q^2\right)}_{m_j}}\\ \cdot {\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{q}^{k\left(\Delta +1\right)}{\left(-s^2;q^2\right)}_k.\end{array}$

Put $\alpha =\frac{m_j+{\mu }_j+\Delta +1}{2}$ , $A=\frac{1}{s^2}$ we get

$\begin{array}{c}{}_q{l}_2\left(f\left(t\right);s\right)=\frac{{\left(1-q^2\right)}^{\frac{\Delta }{2}}}{\left[2\right]s^{\Delta +1}{\left(q^2;q^2\right)}_{\infty }}{\displaystyle \underset{j=1}{\overset{n}{\prod }}}{\left(\frac{a_j}{qs}\right)}^{{\mu }_j}\\ \cdot {\displaystyle \underset{m_j=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a_{j}q}{s}\right)}^{m_j}{q}^{m_j\left(m_j-1\right)}{B}_{m_j}\left(q^2\right){}_{q^2}\Gamma \left(\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}{K\left(\frac{1}{s^2};\frac{m_j+{\mu }_j+\Delta +1}{2}\right)}.\end{array}$

4. Special Cases

1) Let $n=1$ , ${\mu }_1=\mu$ , $a_1=a$ in above theorems, respectively we have:

$\begin{array}{l}{}_q{L}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{\left(\frac{-a}{s}\right)}^m{B}_m\left(q^2\right){\Gamma }_{q^2}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (29)

$\begin{array}{l}{}_q{l}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a}{s}\right)}^m}{K\left(\frac{1}{s^2};\frac{m+\mu +\Delta +1}{2}\right)}{B}_m\left(q^2\right){}_{q^2}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (30)

$\begin{array}{l}{}_q{L}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(2\right)}\left(2\sqrt{at};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^m{q}^{2m\left(m+2\mu \right)}{B}_m\left(q^2\right){\Gamma }_{q^2}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (31)

$\begin{array}{l}{}_q{l}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(2\right)}\left(2\sqrt{at};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{-a}{s}\right)}^m{q}^{2m\left(m+2\mu \right)}}{K\left(\frac{1}{s^2};\frac{m+\mu +\Delta +1}{2}\right)}{B}_m\left(q^2\right){}_{q^2}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (32)

$\begin{array}{l}{}_q{L}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{a{q}^{-1}t};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{qs}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{\left(-1\right)}^m{q}^{m\left(m-1\right)}{\left(\frac{aq}{s}\right)}^m{B}_m\left(q^2\right){\Gamma }_{q^2}\left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (33)

$\begin{array}{l}{}_q{l}_2\left(t^{\Delta -1}{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{a{q}^{-1}t};q^2\right);s\right)\\ =A_{\Delta }{\left(\frac{a}{s}\right)}^{\mu }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}\frac{{\left(\frac{aq}{s}\right)}^m{q}^{m\left(m-1\right)}}{K\left(\frac{1}{s^2};\frac{m+\mu +\Delta +1}{2}\right)}{B}_m\left(q^2\right){}_{q^2}\Gamma \left(\frac{m+\mu +\Delta +1}{2}\right)\end{array}$ (34)

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

${}_q{L}_2\left(t^{\mu }{J}_{2\mu }^{\left(1\right)}\left(2\sqrt{at}\mathrm{<mo>\; ;\; </mo>}{q}^2\right)\mathrm{<mo>\; ;\; </mo>}s\right)=\frac{{\left(1-q^2\right)}^{\frac{\mu +1}{2}}}{\left[2\right]s^{\mu +2}{\left(q^2\mathrm{<mo>\; ;\; </mo>}{q}^2\right)}_{\infty }}{\left(\frac{a}{s}\right)}^{\mu }$

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

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 [7] .

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

${}_q{L}_2\left(t^{\mu }{J}_{2\mu }^{\left(3\right)}\left(2\sqrt{q^{-1}at}\right);s\right)=\frac{{\left(1-q^2\right)}^{\frac{\mu +1}{2}}}{\left[2\right]s^{\mu +2}{\left(q^2;q^2\right)}_{\infty }}{\left(\frac{a}{qs}\right)}^{\mu }.$

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

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

${}_q{L}_2\left(t^{\alpha }\mathrm{<mo>\; ;\; </mo>}s\right)=\frac{{\left(1-q^2\right)}^{\frac{\alpha }{2}}}{\left[2\right]s^{\alpha +2}}\frac{1}{K\left(\frac{1}{s^2}\mathrm{<mo>\; ;\; </mo>1}+\frac{\alpha }{2}\right)}{\Gamma }_{q^2}\left(1+\frac{\alpha }{2}\right)$

which is the same result in [8] .

Acknowledgements

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