We begin with the following theorem, which shows the integral representation of the generalized q-Mittag-Leffler function.
Theorem 6.1
(Integral representation)
For the generalized q-Mittag-Leffler function, we have
$$ E_{\eta ,\kappa }^{(\sigma ;c)}(u;q ) = \frac{1}{B_{q}(\sigma , c-\sigma )} \int _{0}^{1}t^{\sigma - 1} \frac{(tq;q)_{\infty }}{(tq^{c-\sigma };q)_{\infty }} E_{\eta ,\kappa }^{(c)}(tu;q) \,d_{q}t, $$
(6.1)
provided that \(\eta , \kappa , \sigma \in \mathbb{C}\), \(\Re (\eta )>0\), \(\Re (\kappa )>0\), \(\Re (\sigma )>0 \), and \(\sigma \neq c \).
Proof
By the definition of q-analogue of beta function, we can rewrite equation (3.2) as follows:
$$\begin{aligned} E_{\eta ,\kappa }^{(\sigma ;c)}(u;q ) ={}& \sum_{m=0}^{\infty } \biggl\{ \int _{0}^{1}t^{\sigma +m-1} \frac{(tq;q)_{\infty }}{(tq^{c-\sigma };q)_{\infty }} \,d_{q}t \biggr\} \frac{1}{B_{q}(\sigma , c-\sigma )} \\ &{}\times \frac{(q^{c};q)_{m}}{\Gamma _{q}(\eta m + \kappa )} \frac{u^{m}}{(q;q)_{m}} \\ = {}&\frac{1}{B_{q}(\sigma , c-\sigma )}\sum_{m=0}^{\infty } \biggl\{ \int _{0}^{1}t^{\sigma -1} \frac{(tq;q)_{\infty }}{(tq^{c-\sigma };q)_{\infty }} \,d_{q}t \biggl( \frac{(q^{c};q)_{m}}{(q;q)_{m}} \frac{{tu}^{m}}{\Gamma _{q}(\eta m + \kappa )} \biggr) \biggr\} , \end{aligned}$$
which leads to the required result (6.1). □
Theorem 6.2
For \(\eta , \kappa , \sigma \in \mathbb{C}\), \(\Re (\eta )>0\), \(\Re (\kappa )>0\), \(\Re (\sigma )>0\), \(c\neq \sigma \), then for any \(m \in \mathbb{N}\), we have
$$ D_{q}^{m} \bigl[u^{\kappa - 1} E_{\eta , \kappa }^{(\sigma ;c)} \bigl(\lambda u^{ \eta };q \bigr) \bigr] = u^{\kappa - m - 1} E_{\eta , \kappa - m}^{(\sigma ;c)} \bigl( \lambda u^{\eta };q \bigr). $$
(6.2)
Proof
By considering the function
$$ f(u) = u^{\kappa -1} E_{\eta , \kappa }^{(\sigma ;c)} \bigl(\lambda u^{\eta };q \bigr). $$
In view of (2.11) and using definition (3.2), we obtain
$$\begin{aligned}& \begin{aligned} D_{q} \bigl[u^{\kappa - 1} E_{\eta , \kappa }^{(\sigma ;c)} \bigl(\lambda u^{\eta } \bigr) \bigr] &= \sum_{m=0}^{\infty } \frac{B_{q}(\sigma + m + 1, c - \sigma )}{B_{q}(\sigma , c - \sigma )} \frac{ (q^{c };q)_{m}}{(q;q)_{m}} \\ &\quad {}\times \frac{{\lambda ^{m}}(1-q^{\eta m+\kappa -1})}{1-q} \frac{u^{\eta m + \kappa -2}}{\Gamma _{q}(\eta m + \kappa )}. \end{aligned} \end{aligned}$$
Since, according to the functional equation (2.9), the right-hand side of the above expression can be written as
$$ \sum_{m=0}^{\infty } \frac{B_{q}(\sigma + m + 1, c - \sigma )}{B_{q}(\sigma , c - \sigma )} \frac{ (q^{c };q)_{m}}{(q;q)_{m}} \frac{\lambda ^{m} u^{\eta m + \kappa -2}}{\Gamma _{q}(\eta m + \kappa - 1)}=u^{ \kappa -2} E_{\eta , \kappa -1}^{(\sigma ;c)} \bigl(\lambda u^{\eta };q \bigr). $$
Conclusively, we obtain
$$ D_{q} \bigl[u^{\kappa - 1} E_{\eta , \kappa }^{(\sigma ;c)} \bigl( \lambda u^{\eta };q \bigr) \bigr] = u^{\kappa - 2} E_{\eta ,\kappa -1}^{(\sigma ;c)} \bigl(\lambda u^{\eta };q \bigr). $$
Iterating the above result \(m-1\) times, we obtain the required result (6.2). □
Theorem 6.3
Let \(\xi , \zeta , \sigma , \kappa \in \mathbb{C}\); \(\Re (\xi ), \Re ( \kappa ), \Re (\sigma )> 0\); \(\zeta \neq 0, -1, -2,\ldots \) , then
$$\begin{aligned}& \int _{0}^{1} u^{\xi - 1}(1 - qu)_{(\zeta - 1)} E_{\eta ,\kappa }^{( \sigma ;c)} \bigl(xu^{\rho };q \bigr) \,d_{q}u \\& \quad = \sum_{m=0}^{\infty } \frac{B_{q}(\sigma + m, c - \sigma )(q^{c};q)_{m}}{B_{q}(\sigma , c - \sigma )(q;q)_{m}} \frac{x^{m} \Gamma _{q}(\xi + \rho m)\Gamma _{q}(\xi )}{\Gamma _{q}(\eta m + \kappa )\Gamma _{q}(\xi + \zeta + \rho m )}. \end{aligned}$$
(6.3)
In particular,
$$ \int _{0}^{1} u^{\xi - 1}(1 - qu)_{(\zeta - 1)} E_{\eta ,\kappa }^{( \sigma ;c)} \bigl(xu^{\rho };q \bigr) \,d_{q}u = \Gamma _{q}(\zeta ) E_{\eta , \kappa +\zeta }^{(\sigma ;c)}(x;q ) . $$
(6.4)
Proof
By using definition (3.2), the left-hand side of equation (6.3) can be written as
$$ \int _{0}^{1} u^{\xi - 1}(1 - qu)_{(\zeta - 1)}\sum_{m=0}^{ \infty } \frac{B_{q}(\sigma + m, c - \sigma )(q^{c};q)_{m}}{B_{q}(\sigma , c - \sigma )(q;q)_{m}} \frac{u^{\rho m} x^{m}}{\Gamma _{q}(\eta m + \kappa )} \,d_{q}u. $$
Interchanging the order of summation and integration and in view of equation (2.10), we obtain the required result (6.3).
In equation (6.3) replacing \(\eta =\rho \), \(\xi =\kappa \), then in view of equation (3.2), we can clearly obtain (6.4). □
Theorem 6.4
(q-Laplace transform)
The q-analogue of the generalized Laplace transform is defined as follows:
$$\begin{aligned} _{q}L_{s} \bigl[E_{\eta ,\kappa }^{(\sigma ;c)} \bigl(xu^{\rho };q \bigr) \bigr] = {}&\frac{1}{s} \sum _{m=0}^{\infty } \frac{B_{q}(\sigma + m, c - \sigma )(q^{c};q)_{m}}{B_{q}(\sigma , c - \sigma )(q;q)_{m}} \frac{ \Gamma _{q}(1 + \rho m)}{\Gamma _{q}(\eta m + \kappa )} \\ &{}\times \biggl(\frac{(1-q)^{\rho } x}{s^{\rho }} \biggr)^{m} \end{aligned}$$
(6.5)
provided that \(\kappa , \sigma , s \in \mathbb{C}\); \(\Re (\beta ), \Re (\kappa ), \Re (s) > 0 \).
Proof
The q-Laplace transform of a suitable function is given by means of the following q-integral [7]:
$$ _{q}L_{s} \bigl\{ f(u) \bigr\} = \frac{1}{(1-q)} \int _{0}^{s^{-1}} E_{q}^{qsu}f(u) \,d_{q}u. $$
(6.6)
The q-extension of the exponential function [6] is given by
$$ E_{q}^{u} = {_{0}\phi _{0}} (-,-; q, -u) = \sum_{m = 0}^{ \infty } \frac{q^{\binom{{m}}{{2}}} u^{m}}{(q;q)_{m}} = (-u;q)_{\infty } $$
(6.7)
and
$$ e_{q}^{u} = {_{1}\phi _{0}}(0,-; q, -u) = \sum_{m = 0}^{ \infty } \frac{u^{m}}{(q;q)_{m}} = \frac{1}{(u;q)_{\infty }},\quad \vert u \vert < 1. $$
(6.8)
By using the above q-exponential series and the q-integral equation (2.12), we can write equation (6.6) as
$$ _{q}L_{s} \bigl\{ f(u) \bigr\} = \frac{(q;q)_{\infty }}{s} \sum_{j=0}^{\infty } \frac{q^{j} f(s^{-1}q^{j})}{(q;q)_{j}}. $$
(6.9)
Using definition (3.2) and the definition of q-Laplace transform, we obtain
$$\begin{aligned}& \begin{aligned} {}_{q}L_{s} \bigl[E_{\eta ,\kappa }^{(\sigma ;c)} \bigl(xu^{\rho };q \bigr) \bigr]={}& \frac{(q;q)_{\infty }}{s}\sum _{j=0}^{\infty }\frac{q^{j}}{(q;q)_{j}} \\ &{}\times \sum_{m=0}^{\infty } \frac{B_{q}(\sigma + m, c - \sigma )}{B_{q}(\sigma , c - \sigma )} \frac{(q^{\sigma };q)_{m}}{(q;q)_{m}} \frac{[u(s^{-1} q^{j})^{\sigma }]^{m}}{ \Gamma _{q}(\eta m + \kappa )}. \end{aligned} \end{aligned}$$
On interchanging the order of summation and writing the j series as \(_{1}\phi _{0}\), which can be summed up as \(\frac{1}{(q^{1 + \rho m};q)_{\infty }}\), and after some simplifications, we obtain the required result (6.5). □