Certain results associated with q-Agarwal fractional integral operators and a q-class of special functions

Abstract

This article investigates certain q-analogue of the fractional Agarwal integral operator and its application to a class of polynomials and a series of functions. By utilizing various types of q-Bessel functions, the fractional q-Agarwal integral has been discussed and formulated in a series expression form involving q-shifted factorials and gamma functions. Moreover, certain results and applications of the q-Bessel theory are reported by establishing suitable forms of the fractional integral. Furthermore, the fractional Agarwal integral has been evaluated for some multiple power series formulas. Meanwhile, some desirable results involving q-generating Heine’s series of the first type are provided. Over and above, certain conclusions associated with various exponential, hyperbolic sine and cosine functions are analysed.

Keywords:

• Agarwal fractional operator
• q-Jackson
• fractional q-integral
• q-Bessel functions
• q-generating Heines series

Mathematics Subject Classification 2010:

• Primary 54C40, 14E20
• Secondary 46E25, 20C20

1. Introduction

For more than three decades of years, the theory of quantum calculus (or q-calculus) has often been considered as one of the most interesting subjects to discuss in mathematics. Several research activities were given to the area of q-calculus due to its application in various fields of science including mathematics, mechanics and physical sciences as well. The most significance of the q-calculus concept may be explained by its wide diversity in its applications in gamma function theory, combinatorics, quantum mechanics, Mock theta functions, umbral calculus, hypergeometric functions, Sobolev spaces, theta functions, multiple hypergeometric functions, theory of finite differences, Bernoulli and Euler polynomials, analytic number theory, operator theory, and the theory of analytic univalent functions. In this theory, we are concerned in finding q-analogues that arise in some known results and their applications in various fields of mathematical and physical sciences as well.

Hereafter, we recall some well-known facts on fractional calculus and q-calculus, which can be found in literature (see, e.g. Al-Omari, 2016a, 2016b, 2017, 2020; Amini, Al-Omari, Fardi, & Nonlaopon, 2022; Amini, Fardi, Al-Omari, & Nonlaopon, 2022; Amini, Fardi, Al-Omari, & Nonlaopon, 2023; Araci, Bagdasaryan, Zel, & Srivastava, 2014; Atici & Eloe, 2007; Chandak, Suthar, Al-Omari, & Gulyaz-Ozyurt, 2021; Hahn, 1949; Jackson, 1905; Lavagno & Swamy, 2002; Nonlaopon, Jirakulchaiwong, Tariboon, Ntouyas, & Al-Omari, 2022; Purohit & Kalla, 2007; Rajkovic, Marinkovic, & Stankovic, 2007; Salem & Ucar, 2016; Srivastava, 2020; Ucar & Albayrak, 2011; Vyas, Al-Jarrah, Purohit, Araci, & Nisar, 2020). For an arbitrary function ψ and a real number q, $0<q<1,$ the q-difference operator of a function ψ is introduced by Jackson (1910) (1) $$\left(D_q\psi \right)\left(\zeta \right)=\frac{\psi \left(\zeta \right)-\psi \left(\zeta q\right)}{\zeta -\zeta q},\zeta \ne 0.$$(1)

The q-analogues of the positive integer k and its factorial are, respectively, given by Jackson (1910) $${\left[k\right]}_q=\frac{1-q^k}{1-q} \text{and} {\left[k\right]}_q!={\left[1\right]}_q{\left[2\right]}_q\dots {\left[k-1\right]}_q{\left[k\right]}_q,{\left[0\right]}_q!=1.$$

The q-shifted factorial of a complex number α is defined by Asper and Ahmen (1990) $${\left(\alpha ;q\right)}_n=\prod _{i=0}^{n-1}\left(1-\alpha q^i\right),{\left(\alpha ;q\right)}_0=1 \text{and} {\left(\alpha ;\mathrm{q}\right)}_{\infty }\text{=}\underset{\mathrm{n}\to \infty }{\mathrm{\text{lim}}}{\left(\alpha ;\mathrm{q}\right)}_{\mathrm{n}},\mathrm{n}\in \mathbb{N}.$$

Furthermore, the q-analogues of the exponential function are defined by Al-Omari (2016b), Al-Omari (2017), Al-Omari (2021b) and Ucar (2014a) (2) $$E_q\left(\zeta \right)=\sum _{i\ge 0}{\left(-1\right)}^n\frac{q^{\frac{n\left(n-1\right)}{2}}}{{\left(q;q\right)}_i}{\zeta }^n\left(\zeta \in \mathbb{N}\right)\text{and}{e}_q\left(\zeta \right)=\sum _{i\ge 0}\frac{{\zeta }^n}{{\left(q;q\right)}_{\infty }}\left(\left|\zeta \right|<1\right).$$(2)

Hence, it follows from definitions that $E_q(\zeta )={(\zeta ;q)}_{\infty },(\zeta \in \mathbb{N})$ and $e_q(\zeta )=\frac{1}{{(\zeta ;q)}_{\infty }},(\left|\zeta \right|<1).$ Due to Al-Omari (2016b), the definite and improper q-integrals are, respectively, given by (3) $${\int }_0^{\zeta }\psi \left(\xi \right)d_q\xi =\left(1-q\right)\sum _{i\ge 0}\psi \left(q^i\zeta \right)\zeta q^i$$(3) and $${\int }_0^{\frac{\infty }{\alpha }}\psi \left(\zeta \right)d_q\zeta =\left(1-q\right)\sum _{i\in \mathbb{Z}}\frac{q^i}{\alpha }\psi \left(\frac{q^i}{\alpha }\right).$$

This, indeed, leads to the conclusion that (4) $${\int }_{\zeta }^{\infty }\psi \left(\zeta \right)d_q\zeta ={\int }_0^{\infty }\psi \left(\zeta \right)d_q\zeta -{\int }_0^{\zeta }\psi \left(\zeta \right)d_q\zeta ,$$(4) where ${\int }_0^{\zeta }{D}_q\psi (\xi )d_q\xi =\psi (\zeta )-\psi (0).$ For $\alpha >0,$ the q-analogues of the gamma function are defined by Al-Omari (2016b), Al-Omari, Baleanu, and Purohit (2018), Miller and Ross (1993) and Youm (2000): (5) $${\Gamma }_q\left(\alpha \right)={\int }_0^{\frac{1}{1-q}}{\zeta }^{\alpha -1}{E}_q\left(q\left(1-q\right)\zeta \right)d_q\zeta ,$$(5) and (6) $${\Gamma }_q\left(\alpha \right)=k\left(T;\alpha \right){\int }_0^{\frac{\infty }{T\left(1-q\right)}}{\zeta }^{\alpha -1}{e}_q\left(-\left(1-q\right)\zeta \right)d_q\zeta ,$$(6) where (7) $$k\left(T;\alpha \right)=T^{\alpha -1}\frac{{\left(-q/T;q\right)}_{\infty }{\left(-T;q\right)}_{\infty }}{{\left(-q^{\alpha }/T;q\right)}_{\infty }{\left(-T{q}^{1-\alpha };q\right)}_{\infty }}.$$(7)

The useful derived identities for the gamma and beta functions are, respectively, defined by Asper and Ahmen (1990) (8) $${\Gamma }_q\left(\alpha \right)=\frac{{\left(q;q\right)}_{\infty }}{{\left(q^{\infty };q\right)}_{\infty }}{\left(1-q\right)}^{1-\alpha }\text{and}{\Gamma }_q\left(\alpha \right)=\frac{{\left(q;q\right)}_{\alpha -1}}{{\left(1-q\right)}^{\alpha -1}},$$(8) and $$B\left(t,\beta \right)={\int }_0^1{\zeta }^{t-1}{\left(1-q\right)}_q^{\beta -1}{d}_q\zeta \left(t,\beta >0\right).$$

In view of Ucar (2014b), the assigned q-analogues of the Bessel function of the first, second and third kinds are, respectively, defined by (9) $$J_{2b}^{\left(1\right)}\left(X;q\right)={\left(\frac{X}{2}\right)}^{2b}\sum _{n=0}^{\infty }\frac{{\left(\frac{-X^z}{4}\right)}^n}{{\left(q;q\right)}_{2b+n}{\left(q;q\right)}_n},$$(9) (10) $$J_{2b}^{\left(2\right)}\left(X;q\right)={\left(\frac{X}{2}\right)}^{2b}\sum _{n=0}^{\infty }\frac{q^{n\left(n+2b\right)}{\left(\frac{-X^2}{4}\right)}^n}{{\left(q;q\right)}_{2b+n}{\left(q;q\right)}_n},$$(10) and (11) $$J_{\mu }^{\left(3\right)}\left(X;q\right)=\frac{{\left(q^{\mu +1};q\right)}_{\infty }}{{\left(q;q\right)}_{\infty }}{X}^{\mu }{}_1{\varphi }_1\left[\begin{array}{c}0\hfill \\ q^{\mu +1}\hfill \end{array};q,q{X}^2\right],$$(11) where ${}_r{\varphi }_s$ is the hypergeometric function (Albayrak, Purohit, & Uçar, 2013) $${}_r{\varphi }_s\left[\begin{array}{c}{a}_1,a_2,\dots ,a_r\\ b_1,b_2,\dots ,b_s\end{array}|q,z\right]=\sum _0^{\infty }\frac{{\left(a_1,a_2,\dots ,a_r;q\right)}_n}{{\left(b_1,b_2,\dots ,b_s;q\right)}_n}\frac{z^n}{{\left(q;q\right)}_n}$$ where ${(a_1,a_2,\dots ,a_p;q)}_n={\prod }_0^p{(a_k,q)}_n.$ For $\alpha \in \mathbb{R}\backslash \left\{0,-1,-2,\dots \right\},$ the important identity is that (12) $${\Gamma }_q\left(\alpha \right)=\frac{G\left(q^{\alpha }\right)}{G\left(q\right)}{\left(1-q\right)}^{1-\alpha }-{\left(1-q\right)}_{\alpha -1}{\left(1-q\right)}^{1-\alpha },$$(12) where $G(q^{\alpha })=\frac{1}{{(q^{\alpha };q)}_{\infty }}.$

In recent decades, the area of fractional calculus has been employed in various topics including dynamical systems, modelling, mechanics, quantum physics, optimization and liquids, to mention but a few (Albayrak et al., 2013; Al-Omari, 2021a; Al-Salam & Verma, 1975; Exton, 1978; Jackson, 1910; Kac & Cheung, 2001; Kober, 1940; Serkan, Ugur, & Mehmet, 2019). N. Abel (Al-Salam, 1969) was the first to present an application to the fractional calculus in what is known tautochrone problem that caused attention and concerns of researchers from various fields of sciences. Despite no general definition has been made to the fractional operators, several authors have examined various mathematical definitions and properties of the fractional derivative. However, we intend to discuss in this article a q-analogue of the so-called Agarwal fractional integral operator. The q-analogue of the Agarwal fractional integral operator of order α of a function f is defined by Agarwal (1969) (13) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{-\eta -\alpha }}{{\Gamma }_q\left(\alpha \right)}{\int }_0^{\zeta }{\left(\zeta -\mathit{\text{tq}}\right)}_{\alpha -1}{t}^{\eta }f\left(t\right)d\left(t;q\right).$$(13)

Therefore, by using the concept of the q-integral, the q-series representation of the Agarwal fractional q-operator is given in Agarwal (1969) as $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{q}^k{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{k\eta }f\left(\zeta q^k\right).$$

This, indeed, can be simplified to yield (14) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right).$$(14)

The repeated fractional q-integral operator properties that we mention here are that (Agarwal, 1969) $$I_q^{\eta ,\lambda }{I}_q^{\eta ,\lambda ,\mu }f\left(\zeta \right)=I_q^{\eta ,\mu +\lambda }f\left(\zeta \right)=I_q^{\eta +\lambda ,\mu }f\left(\zeta \right)=I_q^{\eta ,\mu }{I}_q^{\eta +\mu ,\lambda }f\left(\zeta \right)=I_q^{\eta +\mu ,\lambda }{I}_q^{\eta ,\mu }f\left(\zeta \right).$$

However, we organize our results in this article as follows. Preliminaries and a brief review of fractional and quantum calculus are presented in Section 1. In Section 2, the Agarwal fractional q-integral operator has been applied to classes of Bessel functions. In Section 3, the Agarwal fractional integral has been studied on certain class of polynomials. In Section 4, some corollaries and applications of the fractional integral are provided.

2. Agarwal fractional q-integral of q-Bessel functions

This section provides the results of the Agarwal fractional integral and its application to a specific type of Bessel functions.

Theorem 1.

Let $b_i,a_i,i=1,\dots ,r$ be a set of positive real numbers and $J_{2b_1}^{(1)}\left(2\sqrt{a_{1}t};q\right),\dots ,J_{2b_1}^{(1)}\left(2\sqrt{a_{r}t};q\right)$ be a class of first kind q-Bessel functions. Then, for $$f\left(t\right)=t^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(1\right)}\left(2\sqrt{a_{j}t};q\right),$$ we have $$\begin{array}{c}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\alpha -1}}{{\Gamma }_q\left(\zeta \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }{\left(a_j\zeta \right)}^n\frac{{\Gamma }_q\left(2b_j+n+1\right)}{{\left(q;q\right)}_n}\\ \sum _{k=0}^{\infty }{q}^{\left(1+\eta +b_j+n\right)k}\frac{{\Gamma }_q\left(k+1\right)}{{\Gamma }_q\left(\left(k+1\right)\alpha \right)}.\end{array}$$

Proof.

By following the q-series representation of the fractional q-operator (14)(14) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right).$$(14) we deduce $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right).$$

Hence, inserting the value given to the function f in the assumption yields (15) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\Delta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(1\right)}\left(2\sqrt{a_j\zeta q^k};q\right).$$(15)

Therefore, by employing the q-analogues of the Bessel function (9)(9) $$J_{2b}^{\left(1\right)}\left(X;q\right)={\left(\frac{X}{2}\right)}^{2b}\sum _{n=0}^{\infty }\frac{{\left(\frac{-X^z}{4}\right)}^n}{{\left(q;q\right)}_{2b+n}{\left(q;q\right)}_n},$$(9) (15)(15) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\Delta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(1\right)}\left(2\sqrt{a_j\zeta q^k};q\right).$$(15) can be written in the form: $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\Delta -1}\prod _{j=1}^r{\left(a_j\zeta q^k\right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta q^k\right)}^n}{{\left(q;q\right)}_{2b_j+n}}.$$

Thus, the above equation can be summarized to give (16) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}{\zeta }^{\Delta -1}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n}{{\left(q;q\right)}_{2b_j+n}}{\left(q;q\right)}_n.$$(16)

But, owing to the fact that (17) $${\left(1-q^{k+1}\right)}_{\alpha -1}=\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$(17) which follows from (12)(12) $${\Gamma }_q\left(\alpha \right)=\frac{G\left(q^{\alpha }\right)}{G\left(q\right)}{\left(1-q\right)}^{1-\alpha }-{\left(1-q\right)}_{\alpha -1}{\left(1-q\right)}^{1-\alpha },$$(12) , (16)(16) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}{\zeta }^{\Delta -1}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n}{{\left(q;q\right)}_{2b_j+n}}{\left(q;q\right)}_n.$$(16) can be expressed in the form (18) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}{\zeta }^{\alpha -1}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n}{{\left(q;q\right)}_{2b_j+n}{\left(q;q\right)}_n}\sum _{k=0}^{\infty }{q}^{\left(1+\eta +b_j+n\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$(18)

Hence, by virtue of the fact (Agarwal, 1969) (19) $${\left(a;q\right)}_{\alpha }=\frac{{\left(\alpha ;q\right)}_{\infty }}{{\left(a{q}^{\alpha };q\right)}_{\infty }},$$(19) the formula in $(18)$ can be expressed as (20) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\alpha -1}}{{\Gamma }_q\left(\zeta \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1};q\right)}_{\infty }}{{\left(q;q\right)}_n}\sum _{k=0}^{\infty }\frac{q^{\left(1+\eta +b_j+n\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$(20)

Once again, by taking into account (Agarwal, 1969), we have $${\left(\zeta ;q\right)}_{\infty }=E_q\left(\zeta \right)\text{and}\frac{1}{{\left(\zeta ;q\right)}_{\infty }}{e}_q\left(\zeta \right).$$

Therefore, we deduce that $$\begin{array}{c}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\alpha -1}}{{\Gamma }_q\left(\zeta \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }{\left(a_j\zeta \right)}^n\frac{{\Gamma }_q\left(2b_j+n+1\right)}{{\left(q;q\right)}_n}\\ \sum _{k=0}^{\infty }{q}^{\left(1+\eta +b_j+n\right)k}\frac{{\Gamma }_q\left(k+1\right)}{{\Gamma }_q\left(\left(k+1\right)\alpha \right)}.\end{array}$$

The proof is finished.

Theorem 2.

Let $b_i,a_i,i=1,\dots ,r$ be a set of positive real numbers and $J_{2b_1}^{(2)}\left(2\sqrt{a_{j}t};q\right),\dots ,J_{2b_r}^{(2)}\left(2\sqrt{a_{r}t};q\right)$ be a set of second kind q-Bessel functions. If (21) $$f\left(t\right)=t^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(2\right)}\left(2\sqrt{a_{j}t};q\right),$$(21) then $$\begin{array}{c}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)x^{\beta -1}}{{\Gamma }_q\left(\alpha \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^j\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1};q\right)}_{\infty }{q}^{n\left(n+2b_j\right)}}{{\left(q;q\right)}_n}\\ \sum _{k=0}^{\infty }{q}^{\left(1+\eta +\beta -1+j+n\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.\end{array}$$

Proof.

By making use of the q-series representation of the fractional q-operator (14)(14) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right).$$(14) we obtain (22) $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right)\\ =\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(2\right)}\left(2\sqrt{a_j\zeta q^k};q\right).\end{array}$$(22)

Hence, by invoking the definition of $J_b^{(2)},$ presented in (10)(10) $$J_{2b}^{\left(2\right)}\left(X;q\right)={\left(\frac{X}{2}\right)}^{2b}\sum _{n=0}^{\infty }\frac{q^{n\left(n+2b\right)}{\left(\frac{-X^2}{4}\right)}^n}{{\left(q;q\right)}_{2b+n}{\left(q;q\right)}_n},$$(10) we write (22)(22) $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right)\\ =\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(2\right)}\left(2\sqrt{a_j\zeta q^k};q\right).\end{array}$$(22) into the form (23) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}\prod _{j=1}^r{\left(a_j\zeta q^k\right)}^j\sum _{n=0}^{\infty }\frac{q^{n\left(n+2b_j\right)}{\left(a_j\zeta q^k\right)}^n}{{\left(q;q\right)}_{2b_j+n}{\left(q;q\right)}_n}.$$(23)

The fact that ${(a,q)}_{\alpha }=\frac{{(a;q)}_{\infty }}{{(a{q}^{\alpha };q)}_{\infty }}(\text{see}\left[1,(1.2)\right])$ then reveals $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q{\zeta }^{\beta -1}}{{\Gamma }_q\left(\alpha \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^j\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1}\right)}_{\infty }{q}^{n\left(n+2b_j\right)}}{{\left(q;q\right)}_n}\\ \sum _{k=0}^{\infty }{q}^{\left(1+\eta +\beta -1+j+n\right)k}{\left(1-q^{k+1}\right)}_{\alpha -1}.\end{array}$$

But, by invoking the identity $${\left(1-q^{k+1}\right)}_{\alpha -1}=\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$ we obtain (24) $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\beta -1}}{{\Gamma }_q\left(\alpha \right){\left(q;q\right)}_{\infty }}\prod _{j=1}^r{\left(a_j\zeta \right)}^j\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1};q\right)}_{\infty }{q}^{n\left(n+2b_j\right)}}{{\left(q;q\right)}_n}\\ \sum _{k=0}^{\infty }{q}^{\left(1+\eta +\beta -1+j+n\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.\end{array}$$(24)

The proof is finished.

Following is a theorem which investigates the case when the third type q-Bessel functions are invoked.

Theorem 3.

Let $b_i,a_i,i=1,\dots ,r$ be a set of positive real numbers and $J_{2b_1}^{(3)}\left(\sqrt{q^{-1}{a}_{1}t};q\right),\dots ,J_{2b_r}^{(3)}\left(\sqrt{q^{-1}{a}_{n}t};q\right)$ be a set of third kind q-Bessel functions. If $$f\left(t\right)=t^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(3\right)}\left(\sqrt{q^{-1}{a}_{j}t};q\right),$$ then $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\beta -1}}{{\Gamma }_q\left(\alpha \right)}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{q}^{\frac{n\left(n-1\right)}{2}}{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1};q\right)}_{\infty }}{{\left(q;q\right)}_n{\left(q;q\right)}_{\infty }}\\ \sum _{k=0}^{\infty }{q}^{k\left(\eta +\beta +b_j+n\right)}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.\end{array}$$

Proof.

By the q-series representation of the fractional q-operator and the definition of the q-analogue (11)(11) $$J_{\mu }^{\left(3\right)}\left(X;q\right)=\frac{{\left(q^{\mu +1};q\right)}_{\infty }}{{\left(q;q\right)}_{\infty }}{X}^{\mu }{}_1{\varphi }_1\left[\begin{array}{c}0\hfill \\ q^{\mu +1}\hfill \end{array};q,q{X}^2\right],$$(11) , we write (25) $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right)\\ =\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}\prod _{j=1}^r{J}_{2b_j}^{\left(3\right)}\left(\sqrt{a_j\zeta q^{k-1}};q\right).\end{array}$$(25)

That is, (26) $$\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\zeta }^{\beta -1}{q}^{k\beta -k}\prod _{j=1}^r{\left(\sqrt{a_j\zeta q^{k-1}}\right)}^{2b_j}\sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{q^{\frac{n\left(n+1\right)}{2}}{\left(q{a}_j\zeta q^{k-1}\right)}^n}{{\left(q,q\right)}_{2b_j+n}{\left(q;q\right)}_n}.$$(26)

Hence, we have (27) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q{\zeta }^{\beta -1}}{{\Gamma }_q\left(\alpha \right)}\prod _{j=1}^r{\left(a_j\zeta q^{-1}\right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{q}^{\frac{n\left(n+1\right)}{2}}{\left(a_j\zeta \right)}^n}{{\left(q;q\right)}_{2b_j+n}{\left(q;q\right)}_n}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k+k\beta -k+b_{j}k+\mathit{\text{nk}}}.$$(27)

Therefore, by $(19),$ $(27)$ can be expressed as $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right){\zeta }^{\beta -1}}{{\Gamma }_q\left(\alpha \right)}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{q}^{\frac{n\left(n-1\right)}{2}}{\left(a_j\zeta \right)}^n{\left(q^{2b_j+n+1};q\right)}_{\infty }}{{\left(q;q\right)}_n{\left(q;q\right)}_{\infty }}\\ \sum _{k=0}^{\infty }{q}^{k\left(\eta +\beta +b_j+n\right)}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.\end{array}$$

The proof is finished.

3. Application to power series

This section aims to investigate some results involving Agarwal fractional q-integral and certain class of power series.

Theorem 4.

Let β be a positive real number and $f(\zeta )={\sum }_{i=0}^{\infty }{r}_i{\zeta }^i$ be a convergent power series. Then, we have $$I_q^{\eta ,\alpha }f\left(\zeta \right)= I_q^{\eta ,\alpha }\left({\zeta }^{\beta -1}g\right)\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right){\left(q^{\alpha };q\right)}_{\infty }}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{k\alpha };q\right)}_{\infty }}.$$

Proof.

By the assistance of $(14)$ we derive (28) $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right)\\ =\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}g\left(\zeta q^k\right)\\ =\frac{1-q}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}^{\beta -1}\sum _{i=0}^{\infty }{r}_i{\left(\zeta q^k\right)}^i.\end{array}$$(28)

Hence, interchanging the order of summations gives (29) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}{\left(1-q^{k+1}\right)}_{\alpha -1}.$$(29)

By formula (17)(17) $${\left(1-q^{k+1}\right)}_{\alpha -1}=\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$(17) , (29)(29) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}{\left(1-q^{k+1}\right)}_{\alpha -1}.$$(29) suggests to have (30) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$(30)

Now, by applying (18)(18) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\zeta \right)}{\zeta }^{\alpha -1}\prod _{j=1}^r{\left(a_j\zeta \right)}^{b_j}\sum _{n=0}^{\infty }\frac{{\left(a_j\zeta \right)}^n}{{\left(q;q\right)}_{2b_j+n}{\left(q;q\right)}_n}\sum _{k=0}^{\infty }{q}^{\left(1+\eta +b_j+n\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$(18) we set (30)(30) $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$(30) in the form $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right){\left(q^{\alpha };q\right)}_{\infty }}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{k\alpha };q\right)}_{\infty }}.$$

This finishes the proof.

Corollary 5.

For an arbitrary real number $\beta >0,$ we have $$I_q^{\eta ,\alpha }\left({\zeta }^{\beta -1}\right)=\frac{{\zeta }^{\beta -1}\left(1-q\right){\left(q^{\alpha };q\right)}_{\infty }}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{q}^{\left(\eta +\beta +i\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{k\alpha };q\right)}_{\infty }}.$$

Proof.

The proof of this result follows from setting $r_0=1$ and that r_i = 0 for $i=1,2,3,\dots .$

Corollary 6.

Let $f(\zeta )=1,$ then we have $$I_q^{\eta ,\alpha }\left(1\right)=\frac{\left(1-q\right){\left(q^{\alpha };q\right)}_{\infty }}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{q}^{\left(\eta +2\right)k}\frac{{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{k\alpha };q\right)}_{\infty }}.$$

The proof of this corollary is straightforward. Hence details are, therefore, deleted.

4. The fractional q-integral $I_q^{\eta ,\alpha }$ of q-generating series

In the course of the following analysis, we investigate the fractional q-integral $I_q^{\eta ,\alpha }$ on a class of q-hypergeometric basic series. The q-analogue of the generating Heine’s series of the first type (the q-hypergeometric series) is defined by Al-Omari, Suthar, and Araci (2021) (31) $${}_r{\varphi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,X\right)=\sum _{i\ge 0}\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{X}^i,$$(31) where $(2^i)=\frac{i(i-1)}{2},$ q > 0 and $r>s+1.$ The q-analogue of the generating Heine’s series of the second type is defined by Asper and Ahmen (1990) (32) $${}_r{\psi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,X\right)=\sum _{i\ge 0}\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{s-r}{X}^i,$$(32) provided the denominator factors $d_1,\dots ,d_s$ are never zero and if one of the parameters of its numerator is of the type $q^{-n},n=0,1,2,\dots ,$ the basic series terminates.

Theorem 7.

Let the real numbers β and γ be given arbitrary such that $\beta >0.$ Then, we have $$I_q^{\eta ,\alpha }\left(x^{\beta -1}{}_{\gamma }{\varphi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,\gamma \zeta \right)\right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +\beta +i\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$

Proof.

By $(32)$ and $(14)$ we write $$I_q^{\eta ,\alpha }\left({\zeta }^{\beta -1}{}_{\gamma }{\varphi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,\gamma \zeta \right)\right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}f\left(\zeta q^k\right)$$ (33) $$=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{k=0}^{\infty }{\left(1-q^{k+1}\right)}_{\alpha -1}{q}^{\left(1+\eta \right)k}{\left(\zeta q^k\right)}_r^{\beta -1}{\varphi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,\gamma \zeta q^k\right).$$(33)

But, indeed, we have (34) $$\begin{array}{l}{}_{\gamma }{\varphi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,\gamma \zeta q^k\right)=\sum _{i=0}^{\infty }\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(b_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}{\zeta }^i\\ =\sum _{i=0}^{\infty }{r}_i{\zeta }^i,\end{array}$$(34) where (35) $$r_i=\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}.$$(35)

Hence, from Theorem 4, we obtain $$I_q^{\eta ,\alpha }\left(f\left(\zeta \right)\right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +\beta +i\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$ where r_i has the significance of $(35).$

This finishes the proof.

Theorem 8.

Let the real numbers β and γ be given arbitrarily such that $\beta >0.$ Then, we have $$\begin{array}{l}{I}_q^{\eta ,\alpha }f\left(\zeta \right)=I_q^{\eta ,\alpha }\left({\zeta }^{\beta -1}{}_{\gamma }{\psi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,\gamma \zeta \right)\right)\\ =\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +\beta +i\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},\end{array}$$ where r_i has the meaning of (35)(35) $$r_i=\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}.$$(35) .

Proof.

On the aid of (32)(32) $${}_r{\psi }_s\left(c_1,\dots ,c_r;d_1,\dots ,d_s;q,X\right)=\sum _{i\ge 0}\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{s-r}{X}^i,$$(32) we establish that $${}_{\gamma }{\psi }_s\left(c_1,\dots .,c_r,\dots ,d_s;q,\gamma \zeta q^k\right)=\sum _{i=0}^{\infty }{r}_i{\zeta }^i,$$ where r_i is given by (35)(35) $$r_i=\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}.$$(35) Hence, applying Theorem 4 gives rise to the conclusion that $$I_q^{\eta ,\alpha }f\left(\zeta \right)=\frac{{\zeta }^{\beta -1}\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +\beta +i\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$ where r_i has the meaning of (35)(35) $$r_i=\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}.$$(35) .

This finishes the proof.

Corollary 9.

Let γ($\gamma \in \mathbb{R}$) be an arbitrary real number, then we have $$I_q^{\eta ,\alpha }{E}_q\left(\gamma \zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +i+1\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}.$$

Proof.

The proof of this result follows from Theorem 7, by setting $\beta =1,\gamma =0,s=0.$

Corollary 10.

Let γ be an arbitrary real number, then we have $$I_q^{\eta ,\alpha }{e}_q\left(\gamma \zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i{\zeta }^i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +i+1\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }},$$ where r_i has the meaning of (35)(35) $$r_i=\frac{{\left(c_1;q\right)}_i\dots {\left(c_r;q\right)}_i}{{\left(q;q\right)}_i{\left(d_1;q\right)}_i\dots {\left(d_s;q\right)}_i}{\left({\left(-1\right)}^i{q}^{\left(2^i\right)}\right)}^{1+s-r}{\gamma }^i{q}^{\mathit{\text{ki}}}.$$(35) .

Now, we state without proof the following result. The proof is straightforward. Hence, we delete the details.

Corollary 11.

Let γ be an arbitrary real number, then we have $$\begin{array}{l}\left(i\right)I_q^{\eta ,\alpha }{\mathrm{\text{sinh}}}_q\left(\gamma \zeta \right)=\frac{1}{2}{I}_q^{\eta ,\alpha }\left(E_q\left(\gamma \zeta \right)-E_q\left(-\gamma \zeta \right)\right)\\ =\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +i+1\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}\frac{\left(1+{\left(-1\right)}^{i+1}\right)}{2}{\zeta }^i.\\ \left(\mathit{\text{ii}}\right)I_q^{\eta ,\alpha }{\mathrm{\text{cosh}}}_q\left(\gamma \zeta \right)=\frac{\left(1-q\right)}{{\Gamma }_q\left(\alpha \right)}\sum _{i=0}^{\infty }{r}_i\sum _{k=0}^{\infty }\frac{q^{\left(\eta +i+1\right)k}{\left(q^{k+1};q\right)}_{\infty }}{{\left(q^{\left(k+1\right)\alpha };q\right)}_{\infty }}\frac{\left(1+{\left(-1\right)}^i\right)}{2}{\zeta }^i.\end{array}$$

5. Concluding remark

In this article, the fractional Agarwal q-integral has been applied to various classes of special functions. Various results including Bessel functions are obtained by making use of the fractional integral series form. In addition, several remarks and corollaries associated with q-generating series are discussed by using the idea of the q-shifted factorials and the q-gamma functions. Over and above, some computations involving q-hyperbolic and exponential functions are also established.

Acknowledgements

Not applicable.

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No potential conflict of interest was reported by the authors.

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Funding

This study has not received any funding in any form.