A fractional q-integral operator associated with a certain class of q-Bessel functions and q-generating series

Abstract

This paper deals with Al-Salam fractional q-integral operator and its application to certain q-analogues of Bessel functions and power series. Al-Salam fractional q-integral operator has been applied to various types of q-Bessel functions and some power series of special type. It has been obtained for basic q-generating series, q-exponential and q-trigonometric functions as well. Various results and corollaries are provided as an application to this theory.

1 Introduction

The theory of q-calculus is an old subject centered on the idea of deriving q-analogous results without using limits. Jackson was the first to develop the q-calculus theory in systematic way [1]. He defined the concept of the q-integral and the concept of the q-difference operator in a generic manner. In excellence, the theory of q-calculus allows to deal with sets of non-differentiable functions, different classes of orthogonal polynomials, integral operators, and various classes of special functions including q-hypergeometric functions, q-Bessel functions, q-gamma and q-beta functions, and many others, to mention but a few. It connects mathematics and physics and plays a significant role in various fields of physical sciences such as cosmic strings [2], conformal quantum mechanics [3], and nuclear physics of high energy [4]. It, further, applies to topics in number theory, combinatorics, orthogonal polynomials, basic hypergeometric functions, quantum theory, mechanics, and the theory of relativity.

The q-integrals from 0 to ξ and from 0 to ∞ are, resp., defined by Jackson as [1]

$$ \int _{0}^{\xi }f ( t ) \,d_{q}t=\xi ( 1-q ) \sum_{j=0}^{\infty }q^{j}f \bigl( \xi q^{j} \bigr) $$

(1)

and

$$ \int _{0}^{\infty /A}f ( t ) \,d_{q}t= ( 1-q ) \sum_{j\in \mathbb{Z} }\frac{q^{j}}{A}f \biggl( \frac{q^{j}}{A} \biggr) . $$

(2)

The q-analogue of the Bessel function

$$ J_{\mu } ( \xi ) =\sum_{j=0}^{\infty } \frac{ ( -1 ) ^{j} ( \frac{\xi }{2} ) ^{\mu +2j}}{j!\Gamma ( \mu +j+1 ) } $$

(3)

of the first type, which was studied later by Hahn [5] and Ismail [6], is defined by [7] as

$$ J_{\mu }^{ ( 1 ) } ( \xi ;q ) = \biggl( \frac{\xi }{2} \biggr) ^{\mu }\sum_{j=0}^{\infty } \frac{ ( \frac{-\xi }{4}^{2} ) ^{j}}{ ( q,q ) _{\mu +j} ( q;q ) _{j}}, \quad \vert \xi \vert < 2. $$

(4)

Jackson defines the q-analogue of the Bessel function of the second type as [7]

$$ J_{\mu }^{ ( 2 ) } ( \xi ;q ) = \biggl( \frac{\xi }{2} \biggr) ^{\mu }\sum_{j=0}^{\infty } \frac{q^{j ( j+\mu ) } ( \frac{-\xi }{4}^{2} ) ^{j}}{ ( q;q ) _{\mu +j} ( q;q ) _{j}}, \quad \xi \in \mathbb{C} . $$

(5)

Hahn [8] and Exton [9] introduced the third type q-Bessel function (called Hahn–Exton q-Bessel function) as

$$ J_{\mu }^{ ( 3 ) } ( \xi ;q ) =\xi ^{\mu }\sum _{j=0}^{ \infty } \frac{ ( -1 ) ^{j}q^{\frac{j ( j-1 ) }{2}} ( q\xi ^{2} ) ^{j}}{ ( q;q ) _{\mu +j} ( q;q ) _{j}}, \quad \xi \in \mathbb{C} . $$

(6)

The q-shifted factorials are defined, in literature, by fixing $\xi \in \mathbb{C}$ as

$$ ( \xi ;q ) _{0}=1; \quad\quad ( \xi ;q ) _{n}= \prod _{j=0}^{n-1} \bigl( 1-\xi q^{k} \bigr) , \quad n=1,2,\ldots; \quad\quad ( \xi ;q ) _{\infty }= \underset{n \rightarrow \infty }{\lim } ( \xi ;q ) _{n}. $$

(7)

This indeed gives

$$ ( \xi ;q ) _{x}= \frac{ ( \xi ;q ) _{\infty }}{ ( \xi q^{x};q ) _{\infty }}, \quad x\in \mathbb{R} . $$

(8)

For $\xi \in \mathbb{C}$, we mean

$$ [ \xi ] _{q}=\frac{1-q^{\xi }}{1-q}. $$

Hence, for $n\in \mathbb{N}$, we obtain

$$ \bigl( [ n ] _{q} \bigr) != \frac{ ( q;q ) _{n}}{ ( 1-q ) ^{n}}. $$

Due to [10, (1.5), (1.6)], we, resp., write

$$ \begin{bmatrix} n \\ k\end{bmatrix} _{q}= \frac{ [ n ] _{q}!}{ [ k ] _{q}! [ n-k ] _{q}!}= \frac{ ( q;q ) _{n}}{ ( q;q ) _{k} ( q;q ) _{n-k}} $$

(9)

and

$$ \begin{bmatrix} \alpha \\ k\end{bmatrix} _{q}= \frac{ ( q^{-\alpha };q ) _{k}}{ ( q;q ) _{k}} ( -1 ) ^{k}q^{\alpha k- \binom{ k}{ 2} }= \frac{\Gamma _{q} ( \alpha +1 ) }{\Gamma _{q} ( k+1 ) \Gamma _{q} ( \alpha -k ) }. $$

(10)

The q-analogue of the exponential function of the second type is given by

$$ e_{q} ( \xi ) =\sum_{j=0}^{\infty } \frac{\xi ^{j}}{ ( q;q ) _{j}}=\frac{1}{ ( \xi ;q ) _{\infty }}, \quad \vert \xi \vert < 1, $$

(11)

whereas the q-analogue of the exponential function of the first type is given by

$$ E_{q} ( \xi ) =\sum_{j=0}^{\infty } \frac{ ( -1 ) ^{j}q^{j\frac{ ( j-1 ) }{2}}\xi ^{j}}{ ( q;q ) _{j}}= ( \xi ;q ) _{\infty }, \quad \xi \in \mathbb{C} . $$

Consequently, the following formula holds:

$$ \bigl( q^{\xi +m};q \bigr) _{\infty }= \frac{ ( q^{\xi };q ) _{\infty }}{ ( q^{\xi };q ) _{m}},\quad m\in \mathbb{N} . $$

(12)

For real arguments t, the q-analogues of the gamma function are given by [11]

$$ \Gamma _{q} ( t ) = \int _{0}^{\frac{1}{1-q}}x^{t-1}E_{q} ( -qx ) \,d_{q}x\quad \text{and}\quad \hat{\Gamma }_{q} ( t ) = \int _{0}^{\infty }x^{t-1}e_{q} ( -x ) \,d_{q}x. $$

(13)

Henceforth, for $t\in \mathbb{R}$ and $n\in \mathbb{N}$, the following auxiliary results hold:

$$ \Gamma _{q} ( t+1 ) = [ t ] _{q}\Gamma _{q} ( t ) , \quad\quad \Gamma _{q} ( n+1 ) = [ n ] _{q}!\quad \text{and}\quad \Gamma _{q} ( t+1 ) =\frac{1-q^{t}}{1-q}\Gamma _{q} ( t ) . $$

(14)

The theory of fractional calculus was born in early 1695 due to a very deep question raised in a letter of L’Hospital to Leibniz [12–16]. During a long period of time (300 years), the fractional calculus has kept the attention of top level mathematicians. It has become a very useful tool for tackling dynamics of complex systems from various branches of science and engineering. The fractional q-calculus is the q-extension of the ordinary fractional calculus. Integral operators have attained their popularity due to their wide range of applications in various fields of science and engineering [17–22] and [23–34]. In [35, 36] Al-Salam and Agarwal studied certain q-fractional integrals and derivatives. Recently, perhaps due to explosion in research within the fractional calculus setting, new developments in the theory of fractional q-difference calculus, specifically, the q-analogues of the integral and the differential fractional operator properties were made, see, e.g., [37–39]. In [36, p. 966], Al-Salam defines a fractional q-integral operator in the form of the basic integral

$$ K_{q}^{\eta }f ( x ) = \frac{q^{-\eta }x^{\eta }}{\Gamma _{q} ( \alpha ) } \int _{x}^{\infty } ( y-x ) _{ \alpha -1}y^{-\eta -\alpha }f \bigl( yq^{1-\alpha } \bigr) \,d ( y;q ) , $$

(15)

provided $\alpha \neq 0,-1,-2,\ldots$ . With the aid of series definition (1), the above equation can be expressed as

$$ K_{q}^{\eta }f ( x ) = ( 1-q ) ^{\alpha }\sum _{k=0}^{\alpha } ( -1 ) ^{k}q^{k ( \eta + \alpha ) +\frac{1}{2}k ( k+1 ) } \begin{bmatrix} -\alpha \\ k\end{bmatrix} f \bigl( xq^{-\alpha -k} \bigr) . $$

(16)

Consequently, by applying (9), (2) can be expressed as

$$ K_{q}^{\eta ,\alpha }f ( x ) = ( 1-q ) ^{ \alpha }\sum _{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta + \alpha ) +\frac{1}{2}k ( k+1 ) } \biggl( \frac{ ( q;q ) _{-\alpha }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \biggr) f \bigl( xq^{-\alpha -k} \bigr) . $$

Therefore, it follows that

$$ K_{q}^{\eta ,\alpha }f ( x ) = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}f \bigl( xq^{- \alpha -k} \bigr) . $$

(17)

In what follows, we discuss the Al-Salam fractional q-integral $( 15 )$ on some special functions. We apply it to various types of q-Bessel functions and some power series of special type. In Sect. 1, we already recalled some definitions and notations from the fractional q-calculus theory. In Sect. 2, we apply the Al-Salam fractional q-integral to a finite product of q-Bessel functions. In Sect. 3, we apply the Al-Salam fractional q-integral to a power series. We also include some new applications. In Sect. 4, we apply the Al-Salam q-integral operator to some q-generating series.

2 Main results

Theorem 1

Let $\{ J_{2\mu _{1}}^{ ( 1 ) } ( 2\sqrt{\delta _{1}t};q ) ,\ldots,J_{2\mu _{n}}^{ ( 1 ) } ( 2\sqrt{ \delta _{n}t};q ) \}$ be a set of first kind q-Bessel functions and

$$ f ( t ) =t^{\Delta -1}\prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 1 ) } ( 2\sqrt{\delta _{j}t};q ) . $$

(18)

Then, for some $B=q^{-\alpha ( \Delta -1 ) } \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}$, we have

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&B\prod_{j=1}^{n} \bigl( \delta _{j}xq^{-\alpha } \bigr) ^{\mu _{j}}\sum _{m=0}^{ \infty }\delta _{j}x^{m}q^{-\alpha m} \frac{ ( q^{2\mu _{j}+m+1};q ) _{\infty }}{\Gamma _{q} ( m+1 ) } \\ & {}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( -m+\eta +\alpha +\frac{1}{2}k+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

Proof

By employing (18), the fractional q-integral (17) reveals

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{k=0}^{\infty } ( -1 ) ^{k}\frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}f \bigl( xq^{- \alpha -k} \bigr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{k=0}^{\infty } \frac{ ( -1 ) ^{k}q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \bigl( xq^{{-\alpha -k}} \bigr) ^{ \Delta -1} \\ &{}\times \prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 1 ) } \bigl( 2 \sqrt{\delta _{j}xq^{-\alpha -k}};q \bigr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}\sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) - ( \alpha +k ) ( \Delta -1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \\ &{}\times \prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 1 ) } \bigl( 2 \sqrt{\delta _{j}xq^{-\alpha -k}};q \bigr) . \end{aligned}$$

By taking into account the definition of the Bessel function $J_{v}^{ ( 1 ) }$ given in (4), jointly with simple computations, the above equation reduces to yield

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&\frac{ ( q;q ) _{-\alpha }x^{\Delta -1}}{ ( 1-q ) ^{-\alpha }} \sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) - ( \alpha +k ) ( \Delta -1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \prod_{j=1}^{n} \bigl( a_{j}xq^{-\alpha -k} \bigr) ^{\mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } \frac{ ( \delta _{j}xq^{-\alpha -k} ) ^{m}}{ ( q,q ) _{2\mu _{j}+m} ( q,q ) _{m}} \\ =&q^{-\alpha ( \Delta -1 ) } \frac{ ( q;q ) _{-\alpha }x^{\Delta -1}}{ ( 1-q ) ^{-\alpha }} \prod_{j=1}^{n} \bigl( \delta _{j}xq^{-\alpha } \bigr) ^{ \mu _{j}}\sum _{m=0}^{\infty } \frac{ ( \delta _{j}xq^{-\alpha } ) ^{m}q^{-km}}{ ( q;q ) _{2\mu _{j}+m} ( q;q ) _{m}} \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{-km+k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) -k ( \Delta -1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}. \end{aligned}$$

Hence, by the fact [40, Equ. (8)]

$$ ( \zeta ;q ) _{x}= \frac{ ( \zeta ;q ) _{\infty }}{ ( \zeta q^{x};q ) _{\infty }}, $$

(19)

we obtain

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&q^{-\alpha ( \Delta -1 ) } \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1} \prod_{j=1}^{n} \delta _{j}^{\mu _{j}} \mu _{j}xq^{\alpha \mu _{j}-\alpha }q^{\mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } \frac{\delta _{j}^{m}x^{m}q^{-\alpha m} ( q^{2\mu _{j}+m+1};q ) _{\infty }}{ ( q;q ) _{m}} \sum _{k=0}^{\infty } \frac{ ( -1 ) ^{k}q^{k ( -m+\eta +\alpha +\frac{1}{2}k+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

(20)

This completes the proof of the theorem. □

Now the identity

$$ ( q;q ) _{\alpha }=\Gamma _{q} ( \alpha +1 ) $$

(21)

leads to the following useful remark.

Remark 2

Let $\{ J_{2\mu _{1}}^{ ( 1 ) } ( 2\sqrt{\delta _{1}t};q ) ,\ldots,J_{2\mu _{n}}^{ ( 1 ) } ( 2\sqrt{ \delta _{n}t};q ) \}$ be a set of first kind q-Bessel functions and $f ( t ) =t^{\Delta -1}\prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 1 ) } ( 2\sqrt{\delta _{j}t};q )$. Then, for some $B=q^{-\alpha ( \Delta -1 ) } \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}$, we have

$$\begin{aligned} K_{q}^{\eta ,\alpha }f(x) =&B\prod_{j=1}^{n} \bigl( \delta _{j}xq^{- \alpha } \bigr) ^{\mu _{j}}\sum _{m=0}^{\infty }\delta _{j}x^{m}q^{- \alpha m} \frac{ ( q^{2\mu _{j}+m+1};q ) _{\infty }}{\Gamma _{q} ( m+1 ) } \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( -m+\eta +\alpha +\frac{1}{2}k+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

Proof

Indeed, from Theorem 1 and (21), we have

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&\frac{q^{-\alpha ( \Delta -1 ) }x^{\Delta -1}}{ ( q;q ) _{\infty } ( 1-q ) ^{-\alpha }}\Gamma _{q} ( 1-\alpha ) \prod_{j=1}^{n} \bigl( \delta _{j}xq^{-\alpha \mu _{j}-\alpha } \bigr) ^{\mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } \frac{ ( \delta _{j}xq^{-\alpha } ) ^{m} ( q^{\mu _{j}+m+1};q ) _{\infty }}{\Gamma _{q} ( m+1 ) } \sum _{k=0}^{\infty } \frac{ ( -1 ) ^{k}q^{k ( -m+\eta +\alpha +\frac{1}{2}k+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( 1-\alpha -k ) } \\ =&B\prod_{j=1}^{n} \bigl( \delta _{j}xq^{-\alpha } \bigr) ^{ \mu _{j}}\sum _{m=0}^{\infty }\delta _{j}x^{m}q^{-\alpha m} \frac{ ( q^{2\mu _{j}+m+1};q ) _{\infty }}{\Gamma _{q} ( m+1 ) } \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( -m+\eta +\alpha +\frac{1}{2}k+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

This completes the proof of the remark. □

Theorem 3

Let $J_{2\mu _{1}}^{ ( 2 ) } ( 2\sqrt{\delta _{1}t};q ) ,\ldots,J_{2\mu _{n}}^{ ( 2 ) } ( 2\sqrt{\delta _{n}t};q )$ and $f ( t ) =t^{\Delta -1}\prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 2 ) } ( 2\sqrt{\delta _{j}t};q )$. Then, for some $A= \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}q^{-\alpha ( \Delta -1 ) }$, we have

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&A\prod_{j=1}^{n} \delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{-\alpha \mu _{j}}\sum _{m=0}^{ \infty }q^{m ( m+\mu _{j} ) } \frac{ ( -\delta _{j}xq^{-\alpha -k} ) ^{m} ( q^{\mu _{i}+m+1};q_{\infty } ) }{ ( q;q ) _{\infty }\Gamma _{q} ( m+k ) } \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +k+\mu _{j}+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( 1+k ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

Proof

Let the hypothesis of the theorem be satisfied. Then we have

$$ K_{q}^{\eta ,\alpha }f ( x ) = ( 1-q ) ^{ \alpha }\sum _{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta + \alpha ) +\frac{1}{2}k ( k+1 ) } \begin{bmatrix} -\alpha \\ k\end{bmatrix} f \bigl( xq^{-\alpha -k} \bigr). $$

Therefore, in view of (18) and (3), we write

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =& ( 1-q ) ^{ \alpha }\sum _{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta + \alpha ) +\frac{1}{2}k ( k+1 ) } \frac{ ( q;q ) _{-\alpha }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}f \bigl( xq^{- \alpha -k} \bigr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \bigl( xq^{-\alpha -k} \bigr) ^{\Delta -1} \\ &{}\times \prod_{j=1}^{n}J_{2\mu _{j}}^{ ( 2 ) } \bigl( \sqrt[2]{\delta _{j}xq^{-\alpha -k}};q \bigr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \bigl( xq^{-\alpha -k} \bigr) ^{\Delta -1} \\ &{}\times \prod_{j=1}^{n} \bigl( \delta _{j}xq^{-\alpha -k} \bigr) ^{ \mu _{j}} \sum _{m=0}^{\infty } \frac{q^{m ( m+\mu _{j} ) } ( -\delta _{j}xq^{-\alpha -k} ) ^{m}}{ ( q;q ) _{\mu _{j}+m} ( q;q ) _{m}} \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}q^{-\alpha ( \Delta -1 ) }\sum _{k=0}^{ \infty } ( -1 ) ^{q} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) -k ( \Delta -1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \\ &{}\times \prod_{j=1}^{n}\delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{- ( \alpha +k ) \mu _{j}}\sum _{m=0}^{\infty } \frac{q^{m ( m+\mu _{j} ) } ( -\delta _{j}xq^{-\alpha -k} ) ^{m}}{ ( q;q ) _{\mu _{j}+m} ( q,q ) _{m}}. \end{aligned}$$

Hence, it yields

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) = &\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}x^{\Delta -1}q^{-\alpha ( \Delta -1 ) } \prod_{j=1}^{n}a_{j}^{\mu _{j}}x^{\mu _{j}}q^{- ( \alpha +k ) \mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } \frac{q^{m ( m+\mu _{j} ) } ( -a_{j}xq^{-\alpha -k} ) ^{m}}{ ( q;q ) _{\mu _{j}+m} ( q;q ) _{m}} \sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) -k ( \Delta -1 ) +\mu _{j}k}}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}. \end{aligned}$$

(22)

By the fact $( q;q ) _{k}=\Gamma _{q} ( 1+k )$ and the identity

$$ ( \zeta ;q ) _{x}= \frac{ ( q;q ) _{\infty }}{ ( \zeta q^{x};q ) _{\infty }}, $$

(23)

we write

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&A\prod_{j=1}^{n} \delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{-\alpha \mu _{j}}\sum _{m=0}^{ \infty } \frac{q^{m ( m+\mu _{j} ) } ( \delta _{j}xq^{-\alpha -k} ) ^{m}}{ ( q;q ) _{\infty } ( q;q ) _{m}} \bigl( q^{ \mu _{i}+m+1};q \bigr) _{\infty } \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) -k ( \Delta -1 ) +\mu _{j}k}}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \\ =&A\prod_{j=1}^{n}\delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{- \alpha \mu _{j}}\sum _{m=0}^{\infty }q^{m ( m+\mu _{j} ) } \frac{ ( -\delta _{j}xq^{-\alpha -k} ) ^{m} ( q^{\mu _{i}+m+1};q ) _{\infty }}{ ( q;q ) _{\infty }\Gamma _{q} ( m+k ) } \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +k+\mu _{j}+\frac{3}{2}-\Delta ) }}{\Gamma _{q} ( 1+k ) \Gamma _{q} ( 1-\alpha -k ) }. \end{aligned}$$

This completes the proof of the theorem. □

Theorem 4

Let $J_{2\mu _{1}}^{ ( 3 ) } ( 2\sqrt{q^{-1}\delta _{1}t};q ) ,\ldots,J_{2\mu _{n}}^{ ( 3 ) } ( 2\sqrt{q^{-1}\delta _{n}t};q )$ be n q-Bessel functions and

$$ f ( t ) =t^{\Delta -1}\prod_{j=1}^{n} \delta ^{ \mu _{j}}J_{2\mu _{j}}^{ ( 3 ) } \bigl( 2 \sqrt{q^{-1} \delta _{n}t};q \bigr). $$

Then we have

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&\frac{x^{\Delta -1}\Gamma _{q} ( 1-\alpha ) ( 1-q ) ^{\alpha }}{ ( q;q ) _{\infty }}\prod _{j=1}^{n}\delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{ ( -\alpha -k ) \mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } ( -1 ) ^{m} \frac{q^{m\frac{ ( m-1 ) }{2}+m ( -\alpha ) }x^{m}\delta _{j}^{m} ( q^{2\mu _{j}+m+1};q ) _{\infty }}{ ( q;q ) _{m}} \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) + ( \Delta -1 ) ( -\alpha -k ) -mk}}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( -\alpha -k ) }. \end{aligned}$$

Proof

By (2) and (6), we obtain

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =& ( 1-q ) ^{ \alpha }\sum _{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta + \alpha ) +\frac{1}{2}k ( k+1 ) } \begin{bmatrix} -\alpha \\ k\end{bmatrix} f \bigl( xq^{-\alpha -k} \bigr) \\ =& ( 1-q ) ^{\alpha }\sum_{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) } \begin{bmatrix} -\alpha \\ k\end{bmatrix} \bigl( xq^{-\alpha -k} \bigr) ^{\Delta -1} \\ &{}\times \prod_{j=1}^{n}q^{\mu j}J_{2\mu _{j}}^{ ( 3 ) } \bigl( \sqrt{q^{-1}\delta _{j}xq^{-\alpha -k}};q \bigr) \\ =& ( 1-q ) ^{\alpha }\sum_{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) } \begin{bmatrix} -\alpha \\ k\end{bmatrix} \bigl( xq^{-\alpha -k} \bigr) ^{\Delta -1} \\ &{}\times \prod_{j=1}^{n}q^{\mu j} \bigl( q^{-1}\delta _{j}xq^{- \alpha -k} \bigr) ^{\mu j} \sum_{m=0}^{\infty } ( -1 ) ^{m} \frac{q^{m\frac{ ( m-1 ) }{2}} ( qq^{-1}\delta _{j}q^{-\alpha -k} ) ^{m}}{ ( q;q ) _{2\mu _{j}+m} ( q;q ) _{m}}. \end{aligned}$$

Equations (10), (21), and simple simplifications reveal

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x ) =&x^{\Delta -1} ( 1-q ) ^{\alpha }\sum_{k=0}^{\infty } ( -1 ) ^{k}q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) + ( \Delta -1 ) ( -\alpha -k ) } \frac{\Gamma _{q} ( 1-\alpha ) }{\Gamma _{q} ( k+1 ) \Gamma _{q} ( -\alpha -k ) } \\ &{}\times \prod_{j=1}^{n}\delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{ ( -\alpha -k ) \mu _{j}}\sum _{m=0}^{\infty } ( -1 ) ^{m} \frac{q^{m\frac{ ( m-1 ) }{2}+m ( -\alpha -k ) }x^{m}\delta _{j}^{m} ( q^{2\mu _{j}+m+1};q ) _{\infty }}{ ( q;q ) _{\infty } ( q;q ) _{m}} \\ =&\frac{x^{\Delta -1}\Gamma _{q} ( 1-\alpha ) ( 1-q ) ^{\alpha }}{ ( q;q ) _{\infty }}\prod_{j=1}^{n} \delta _{j}^{\mu _{j}}x^{\mu _{j}}q^{ ( -\alpha -k ) \mu _{j}} \\ &{}\times \sum_{m=0}^{\infty } ( -1 ) ^{m} \frac{q^{m\frac{ ( m-1 ) }{2}+m ( -\alpha ) }x^{m}\delta _{j}^{m} ( q^{2\mu _{j}+m+1};q ) _{\infty }}{ ( q;q ) _{m}} \\ &{}\times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) + ( \Delta -1 ) ( -\alpha -k ) -mk}}{\Gamma _{q} ( k+1 ) \Gamma _{q} ( -\alpha -k ) }. \end{aligned}$$

This completes the proof of the theorem. □

3 The fractional q-integral of the power series

This section is briefly devoted to the application of the fractional q-integral to functions of a power series form. Some corollaries associated with polynomials and unit functions are also deduced.

Theorem 5

Let $g ( x ) =\sum_{i=0}^{\infty }r_{i}x^{i}$ be a power series and β be a positive real number. If $f ( x ) = ( x^{\beta -1}g ) ( x )$, then we have

$$ K_{q}^{\eta ,\alpha }f ( x ) = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k+\frac{1}{2}-i}}{\Gamma _{q} ( k ) \Gamma _{q} ( -\alpha -k ) }. $$

Proof

Let $g ( x ) =\sum_{i=0}^{\infty }r_{i}x^{i}$ be a power series and β be a positive real number. From (26) it follows

$$\begin{aligned} K_{q}^{\eta ,\alpha }f ( x )& = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{k=0}^{\infty } ( -1 ) ^{k}\frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}f \bigl( xq^{- \alpha -k} \bigr) \\ &= \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \bigl( xq^{-\alpha -k} \bigr) ^{\beta -1} \sum_{i=0}^{\infty }r_{i} \bigl( xq^{-\alpha -k} \bigr) ^{i}. \end{aligned}$$

(24)

Interchanging the order of summation in (24) leads to

$$ K_{q}^{\eta ,\alpha }f ( x ) = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) -k_{i}}}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}}. $$

Employing (21) indeed gives

$$ K_{q}^{\eta ,\alpha }f ( x ) = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k+\frac{1}{2}-i}}{\Gamma _{q} ( k ) \Gamma _{q} ( -\alpha -k ) }. $$

Hence, the proof of the theorem is completed. □

Corollary 6

Let $\beta >0$ be a real number. Then we have

$$ K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1} \bigr) = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1}}{ ( 1-q ) ^{-\alpha }} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( ( \eta +\alpha ) +\frac{1}{2}k+\frac{1}{2} ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( -\alpha -k ) }. $$

This result follows from setting $r_{0}=1$ and $r_{i}=0$ for $i=1,2,3,\ldots$ .

Corollary 7

We have

$$ K_{q}^{\eta ,\alpha } ( 1 ) = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2} ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( -\alpha -k ) }. $$

(25)

4 $K_{q}^{\eta ,\alpha }$ of q-generating Heines series

The basic q-generating series of the first type is defined by [41] as

$$ _{r}\phi _{s} ( \delta _{1},\ldots,\delta _{r};b_{1},\ldots,b_{s},q, \zeta ) =\sum _{i\geq 0} \frac{ ( \delta _{1};q ) _{1},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( b_{1};q ) _{i},\ldots, ( b_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{1+s-r}\zeta ^{i}, $$

where

$$ \bigl( 2^{i} \bigr) =\frac{i ( i-1 ) }{2},\quad\quad r>s+1,\quad\quad q>0. $$

(26)

The basic q-generating series of the second type is given as

$$ _{r}\psi _{s} ( \delta _{1},\ldots,\delta _{r};\grave{\delta }_{1},\ldots,\grave{\delta }_{s},q,\zeta ) =\sum_{i\geq 0}^{\infty } \frac{ ( \delta _{1};q ) _{1},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( \grave{\delta }_{1};q ) _{i},\ldots, ( \grave{\delta }_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{s-r}\zeta ^{i}. $$

(27)

The parameters $b_{1},\ldots,b_{s}$ are given so that the denominator factors in terms of the series are never zero, and the basic series terminates when one of its numerator parameters is of type $q^{-n}$, $n=0,1,2,\ldots$ .

Theorem 8

Let β and γ be real numbers. Then, provided $\beta >0$, we have

$$\begin{aligned}& K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1} \bigr) _{r} \phi _{s} ( \delta _{1},\ldots,\delta _{r}; \grave{\delta }_{1},\ldots,\grave{\delta }_{s};q, \gamma x ) \\& \quad = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }} \\& \quad\quad {} \times \sum_{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{ \infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. \end{aligned}$$

Proof

Let β and γ be real numbers. Then, by (17), we write

$$\begin{aligned}& K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1} \bigr) _{r} \phi _{s} ( \delta _{1},\ldots,\delta _{r}; \grave{\delta }_{1},\ldots,\grave{\delta }_{s};q, \gamma x ) \\& \quad = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) +\frac{1}{2}k ( k+1 ) }}{ ( q;q ) _{k} ( q;q ) _{-\alpha -k}} \\& \quad\quad {}\times f \bigl( xq^{-\alpha -k} \bigr) \bigl( xq^{-\alpha -k} \bigr) _{{}}^{ \beta -1}{ }_{r}\phi _{s} \bigl( \delta _{1},\ldots,\delta _{r}; \grave{\delta }_{1},\ldots,\grave{\delta }_{s};q,\gamma xq^{-\alpha -k} \bigr) . \end{aligned}$$

On the other hand, we have

$$\begin{aligned}& _{r}\phi _{s} \bigl( \delta _{1},\ldots, \delta _{r};\grave{\delta }_{1},\ldots,\grave{ \delta }_{s};q,\gamma xq^{-\alpha -k} \bigr) \sum _{i \geq 0}^{\infty } \frac{ ( \delta _{1};q ) _{i},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( \grave{\delta }_{1};q ) _{i},\ldots, ( \grave{\delta }_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{s-r} \\& \quad {} \times \bigl( \gamma xq^{-\alpha -k} \bigr) ^{i}=\sum _{i \geq 0}^{\infty }r_{i}x^{i}, \end{aligned}$$

where

$$ r_{i}= \frac{ ( \delta _{1};q ) _{i},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( \grave{\delta }_{1};q ) _{i},\ldots, ( \grave{\delta }_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{s-r}\gamma ^{i}q^{ ( -\alpha -k ) i}. $$

(28)

Therefore, by Theorem 5 we get

$$\begin{aligned} K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1} \bigr) _{r} \phi _{s} ( \delta _{1},\ldots,\delta _{r}; \grave{\delta }_{1},\ldots,\grave{\delta }_{s};q, \gamma x ) =& \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{ \infty }r_{i}q^{-\alpha i}x^{i} \\ &{} \times \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. \end{aligned}$$

(29)

This completes the proof of the theorem. □

Theorem 9

Let $\beta >0$ and r be real numbers. Then we have

$$\begin{aligned} K_{q}^{\eta ,\alpha } \bigl( x_{r}^{\beta -1}\psi _{s} ( \delta _{1},\ldots, \delta _{r}; \grave{\delta }_{1},\ldots,\grave{\delta }_{s};q,\gamma x ) \bigr) =&\frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \\ &{} \times \sum_{k=0}^{\infty } ( -1 ) ^{k}\frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. \end{aligned}$$

Proof

By taking into account (20), we write

$$\begin{aligned} K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1}{ }_{r}\psi _{s} \bigr) =& \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum _{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha ) + \frac{1}{2}k+\frac{1}{2}}}{ ( q;q ) _{k} ( q;q ) _{-\alpha ,k}}f \bigl( xq^{-\alpha -k} \bigr) _{{}}^{\beta -1} \\ &{} \times f \bigl( xq^{-\alpha -k} \bigr) _{{}}^{\beta -1}{ } _{r} \psi _{s} \bigl( \delta _{1},\ldots, \delta _{r};\grave{\delta }_{1},\ldots,\grave{ \delta }_{s};q^{-\alpha -k} \bigr) . \end{aligned}$$

(30)

However,

$$\begin{aligned} _{r}\psi _{s} \bigl( \delta _{1},\ldots, \delta _{r};\grave{\delta }_{1},\ldots,\grave{ \delta }_{s};q^{-\alpha -k} \bigr) =&\sum _{i\geq 0}^{ \infty }\frac{ ( \delta _{1};q ) _{i},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( \grave{\delta }_{1};q ) _{i},\ldots, ( \grave{\delta }_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{s-r} \\ & {} \times \bigl( \gamma xq^{-\alpha -k} \bigr) ^{i} \\ =&\sum_{i\geq 0}^{\infty }r_{i}x^{i}, \end{aligned}$$

where

$$ r_{i}= \frac{ ( \delta _{1};q ) _{i},\ldots, ( \delta _{r};q ) _{i}}{ ( q;q ) _{i}, ( \grave{\delta }_{1};q ) _{i},\ldots, ( \grave{\delta }_{s};q ) _{i}} \bigl( ( -1 ) ^{i}q^{ ( 2^{i} ) } \bigr) ^{s-r}\gamma ^{i}q^{ ( -\alpha -k ) i}. $$

(31)

Hence, by Theorem 5 it follows

$$ K_{q}^{\eta ,\alpha } \bigl( x^{\beta -1}{ }_{r}\psi _{s} \bigr) = \frac{q^{-\alpha \beta +\alpha }x^{\beta -1} ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. $$

This completes the proof of the theorem. □

Corollary 10

Let γ be a real number. Then we have

$$ K_{q}^{\eta ,\alpha } \bigl( E_{q} ( \gamma x ) \bigr) = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{ \infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. $$

Proof

By setting $\beta =0$, $r=0$, and $s=0$, the result easily follows from Theorem 8. The proof is completed. □

Corollary 11

Let γ be a real number. Then we have

$$ \bigl( K_{q}^{\eta ,\alpha }e_{q} ( \gamma x ) \bigr) = \frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{\infty }r_{i}q^{-\alpha i}x^{i} \sum_{k=0}^{ \infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k+\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }. $$

Proof

By setting $\beta =1$, $r=0$, and $s=0$, Theorem 8 completes the proof of the corollary. □

The proof of the following corollary is straightforward. Details are therefore deleted.

Corollary 12

Let γ be a real number. Then we have

$$\begin{aligned} ( i ) K_{q}^{\eta ,\alpha } \bigl( \sinh _{q} ( \gamma x ) \bigr) =&K_{q}^{\eta ,\alpha } \biggl( \frac{E_{q} ( \gamma x ) -E_{q} ( -\gamma x ) }{2} \biggr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{\infty }r_{i}q^{-\alpha i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}k-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }x^{i} \frac{ ( 1+ ( -1 ) ^{i+1} ) }{2}. \end{aligned}$$

$$\begin{aligned} ( \mathit{ii} ) K_{q}^{\eta ,\alpha } \bigl( \cosh _{q} ( \gamma x ) \bigr) =&K_{q}^{\eta ,\alpha } \biggl( \frac{E_{q} ( \gamma x ) -E_{q} ( -\gamma x ) }{2} \biggr) \\ =&\frac{ ( q;q ) _{-\alpha }}{ ( 1-q ) ^{-\alpha }}\sum_{i=0}^{ \infty }r_{i}q^{-\alpha i} \sum_{k=0}^{\infty } ( -1 ) ^{k} \frac{q^{k ( \eta +\alpha +\frac{1}{2}-i ) }}{\Gamma _{q} ( k ) \Gamma _{q} ( _{-\alpha -k} ) }x^{i} \frac{ ( 1+ ( -1 ) ^{i} ) }{2}. \end{aligned}$$