A Few Explicit Properties Involving Integral Transforms and Fractional Calculus of Srivastava’s Triple Hypergeometric Function Hλ, lC, P,α

1. Introduction and Preliminaries

There is a long history of hypergeometric functions of a single variable in many areas of physics, statistics, and mathematics. The Gauss hypergeometric function is defined as

$${}_{2}F_1\left(\begin{array}{c}{a}_1,a_2\\ b_1\end{array};x\right)=\sum _{i=0}^{\infty }{\displaystyle \frac{{(a_1)}_i{\left(a_2\right)}_i}{{\left(b_1\right)}_i}}{\displaystyle \frac{x^i}{i!}}(a_1,a_2\in \mathsf{C},b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-},\left|x\right|<1),$$

(1.1)

where ${\left(a_1\right)}_i$ denotes the Pochhammer symbol defined by

$${\left(a_1\right)}_i={\displaystyle \frac{\Gamma (a_1+i)}{\Gamma \left(a_1\right)}}=\left\{\begin{array}{c}1,(i=0,a_1\in \mathsf{C}\setminus \left\{0\right\}),\hfill \\ \\ a_1(a_1+1)...(a_1+i-1),(i\in \mathsf{N},a_1\in \mathsf{C})\hfill \end{array}\right\}.$$

(1.2)

Here, the classical Gamma function [13] is defined as

$$\Gamma \left(a_1\right)={\int }_0^{\infty }{e}^{-t}{t}^{a_1-1}dt,(R\left(a_1\right)>0).$$

(1.3)

Extensions of the function (1.1) to include p numerator parameters $a_i$ ($1\le i\le p$) and q denominator parameters $b_i$ ($1\le i\le q$) also find wide application; see [17]. Triple hypergeometric functions have been introduced and studied by Srivastava and Karlsson ([23], Chapter 3), who provide a table of 205 distinct such functions. In [19,20], Srivastava introduced the triple hypergeometric functions $H_A$, $H_B,$ and $H_C$ of the second order. It is known that $H_C$ and $H_B$ are generalizations of Appell’s hypergeometric functions $F_1$ and $F_2$, while $H_A$ is the generalization of both $F_1$ and $F_2$.

The focus of this study will be on Srivastava’s triple hypergeometric function $H_C$, which is given by ([23], , p. 43, 1.5(11) to 1.5(13)) (see also [19] and [22], p. 68)

$$H_C(a_1,a_2,a_3;b_1;x_1,x_2,x_3):=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_1\right)}_{i+k}{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{i+j+k}}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B(a_1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}.$$

(1.4)

In ([6], p.243), the convergence region for the hypergeometric series $H_C(·)$ is provided as $|x_1|<{\alpha }_1$, $|x_2|<{\alpha }_2$, $|x_3|<{\alpha }_3$, where ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$ satisfy the relation

$${\alpha }_1+{\alpha }_2+{\alpha }_3-2\sqrt{(1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)}<2.$$

(1.5)

Now, we find it suitable to introduce a new parameter c into $H_C(·)$ as follows

$$H_C^{\left(c\right)}(a_1,a_2,a_3;b_1;x_1,x_2,x_3):=$$

$$\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B(a_1+c+i+k,b_1+c+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}},$$

(1.6)

which reduces to (1.4) when $c=0$.

Here, $B(x_1,x_2)$ is the classical Beta function defined as [11], (5.12.1)

$$B(x_1,x_2)=\left\{\begin{array}{cc}{\displaystyle {\int }_0^1{t}^{x_1-1}{(1-t)}^{x_2-1}dt,}\hfill & (R\left(x_1\right)>0,R\left(x_2\right)>0)\hfill \\ \\ {\displaystyle \frac{\Gamma \left(x_1\right)\Gamma \left(x_2\right)}{\Gamma (x_1+x_2)},}\hfill & ((x_1,x_2)\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-}).\hfill \end{array}\right.$$

(1.7)

In 1997, Chaudhry et al. [2], Eq.(1.7) introduced a p-extension of $B(x_1,x_2)$ given by

$$B_p(x_1,x_2)={\int }_0^1{t}^{x_1-1}{(1-t)}^{x_2-1}exp\left[{\displaystyle \frac{-p}{t(1-t)}}\right]dt,(R\left(p\right)\ge 0).$$

(1.8)

In 2018, a further extension of the Beta function was given, as shown by Shadab et al. [15]

$$B_p^{\lambda }(x_1,x_2)={\int }_0^1{t}^{x_1-1}{(1-t)}^{x_2-1}{E}_{\lambda }\left[{\displaystyle \frac{-p}{t(1-t)}}\right]dt,(R\left(p\right)\ge 0),$$

(1.9)

which reduces to (1.8) when $\lambda =1$ and $E_{\lambda }\left(x\right)$ denotes the Mittag-Leffler function defined by [10],

$$E_{\lambda }\left(x\right)=\sum _{i=0}^{\infty }{\displaystyle \frac{x^i}{\Gamma (\lambda i+1)}},(x\in \mathsf{C};\lambda >0).$$

In 2020, Oraby et al. [12] gave a generalized Beta function in the form

$$B_{p,\alpha }^{\lambda ,l}(x_1,x_2)={\int }_0^1{t}^{x_1-1}{(1-t)}^{x_2-1}{E}_{\lambda ,\alpha }\left[{\displaystyle \frac{-p}{t^l{(1-t)}^l}}\right]dt$$

(1.10)

$$(R\left(x_1\right)>0,R\left(x_2\right)>0,R\left(p\right)\ge 0,R\left(l\right)>0),$$

which reduces to (1.9) when $\alpha =l=1$.

Where $E_{\lambda ,l}\left(x\right)$ is the generalized Mittag-Leffler function introduced in 1905 by Wiman [24] defined as

$$E_{\lambda ,\alpha }\left(x\right)=\sum _{i=0}^{\infty }{\displaystyle \frac{x^i}{\Gamma (\lambda i+\alpha )}},(x,\alpha \in \mathsf{C};\lambda >0,R\left(\alpha \right)>0).$$

In 2021, Abubakar [1] introduced and studied a generalized Beta function as

$${}^{\Psi }B_{\tau }^l(x_1,x_2)={}^{\Psi }B_{\tau }^l\left.\left[\begin{array}{c}{({\rho }_m,{\acute{\rho }}_m)}_{1,{\eta }_1}\\ {({\zeta }_n,{\acute{\zeta }}_n)}_{1,{\eta }_2}\end{array}\right|x_1,x_2\right]$$

$$={\int }_0^1{t}^{x_1-1}{(1-t)}^{x_2-1}{}_{{\eta }_1}\Psi _{{\eta }_2}\left(-{\displaystyle \frac{\tau }{t^l{(1-t)}^l}}\right)dt,$$

(1.11)

( $x_1,x_2,\tau ,l,{\rho }_m,{\zeta }_n\in \mathsf{C};$ ${\acute{\rho }}_m,{\acute{\zeta }}_n\in \mathsf{R},$$R\left(x_1\right)>0,R\left(x_2\right)>0,R\left(\tau \right)>0,R\left(l\right)>0$$andm=1,...,{\eta }_1,n=1,...,{\eta }_2$)

here ${}_{{\eta }_1}\Psi _{{\eta }_2}$ is the Fox-Wright function ([7,8]) defined as

$${}_{{\eta }_1}\Psi _{{\eta }_2}\left.\left[\begin{array}{c}{({\rho }_m,{\acute{\rho }}_m)}_{1,{\eta }_1}\\ {({\zeta }_n,{\acute{\zeta }}_n)}_{1,{\eta }_2}\end{array}\right|z\right]={}_{{\eta }_1}\Psi _{{\eta }_2}\left[\begin{array}{c}({\rho }_1,{\acute{\rho }}_1),...,({\rho }_{{\eta }_1},{\acute{\rho }}_{{\eta }_1});\\ ({\zeta }_1,{\acute{\zeta }}_1),...,({\zeta }_{{\eta }_2},{\acute{\zeta }}_{{\eta }_2});\end{array}z\right]$$

$$=\sum _{k=0}^{\infty }{\displaystyle \frac{{\displaystyle \prod _{m=1}^{{\eta }_1}\Gamma ({\rho }_m+{\acute{\rho }}_{m}k)}}{{\displaystyle \prod _{n=1}^{{\eta }_2}\Gamma ({\zeta }_n+{\acute{\zeta }}_{n}k)}}}{\displaystyle \frac{z^k}{k!}},$$

(1.12)

where $z,{\rho }_m,{\zeta }_n\in \mathsf{C}$, and the coefficients ${\acute{\rho }}_m\ge 0$, ${\acute{\zeta }}_n\ge 0$ ($m=1,...,{\eta }_1;n=1,...,{\eta }_2$). If we put

$$\Delta =\sum _{n=1}^{{\eta }_2}{\acute{\zeta }}_n-\sum _{m=1}^{{\eta }_1}{\acute{\rho }}_m,\delta =\prod _{m=1}^{{\eta }_1}{\acute{\rho }}_m^{-{\acute{\rho }}_m}\prod _{n=1}^{{\eta }_2}{\acute{\zeta }}_n^{{\acute{\zeta }}_n},{\mu }^{*}=\sum _{n=1}^{{\eta }_2}{\zeta }_n-\sum _{m=1}^{{\eta }_1}{\rho }_m+\frac{1}{2}({\eta }_1-{\eta }_2),$$

(1.12) converges for $\left|z\right|<\infty$ when $\Delta >-1$, for $\left|z\right|<\delta$ when $\Delta =-1$ and for $\left|z\right|=\delta$ if, in addition, $R\left({\mu }^{*}\right)>\frac{1}{2}$.

For ${\acute{\rho }}_1=...={\acute{\rho }}_{{\eta }_1}={\acute{\zeta }}_1=...={\acute{\zeta }}_{{\eta }_2}=1$ eq. (1.12) reduces to the generalized hypergeometric function ${}_{{\eta }_1}F_{{\eta }_2}$ [26]

$${}_{{\eta }_1}\Psi _{{\eta }_2}\left[\begin{array}{c}({\rho }_1,1),...,({\rho }_{{\eta }_1},1);\\ ({\zeta }_1,1),...,({\zeta }_{{\eta }_2},1);\end{array}z\right]={\displaystyle \frac{{\displaystyle \prod _{m=1}^{{\eta }_1}\Gamma \left({\rho }_m\right)}}{{\displaystyle \prod _{n=1}^{{\eta }_2}\Gamma \left({\zeta }_n\right)}}}{}_{{\eta }_1}F_{{\eta }_2}\left[\begin{array}{c}{\rho }_1,...,{\rho }_{{\eta }_1};\\ {\zeta }_1,...,{\zeta }_{{\eta }_2};\end{array}z\right].$$

(1.13)

The plan of this paper is as follows. The extended function ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is defined in section 2 based on the extended Beta function in (1.10), and some integral representations are presented involving the generalized Mittag-Leffler function and the Gauss hypergeometric function ${}_{2}F_1$. Further, we discuss some derivative formulas and integral trasforms, namely, the Euler-Beta transform, Laplace transform, Mellin transform, and Whittaker transform. Moreover, some recursion formulas are established. Also, our work on Srivastava triple hypergeometric function is motivated by the work given in ([3,4]).

2. The Extended Srivastava Triple Hypergeometric Function ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

Srivastava introduced the triple hypergeometric function $H_C(·)$, together with its integral representations, in ([19,22]). Here, we define the extended Srivastava triple ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ by means of the extended Beta function defined in (1.10)

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}},$$

(2.1)

where, $a_1,a_2,a_3\in \mathsf{C}$ and $b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-}$ and the region of convergence is $|x_1|<{\alpha }_1$, $|x_2|<{\alpha }_2$, $|x_3|<{\alpha }_3$, where ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$ satisfy (1.5).

Now, using the extended beta function given in (1.10) in (2.1), we get ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(.)$ in terms of extended Mittag-Leffler function as follows:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha }\left[-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right]dt.$$

(2.2)

We use of the above eq. (2.2) throughout the paper.

3. A Few Explicit Properties

In this section, we calculate two explicit properties, namely, integral representations and derivative properties.

3.1. Integral Representations

Here, we get the following integral representations for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ function involving the product of Mittag-Leffler function of two parameter and the Gauss hypergeometric function ${}_{2}F_1$.

Theorem 1.

For $R\left(p\right)\ge 0$, $R\left(a_n\right)>0$ ($n=1,2,3$) and $R(b_1-a_1)>0,$ the integral representation of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows :

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)={\displaystyle \frac{\Gamma \left(b_1\right)}{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}}{\int }_0^1{t}^{a_1-1}{(1-t)}^{b_1-a_1-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)$$

$$\times {(1-x_{1}t)}^{-a_2}{(1-x_{3}t)}^{-a_3}{}_{2}F_1\left[\begin{array}{c}{a}_2,a_3\\ b_1-a_1\end{array};{\displaystyle \frac{(1-t)x_2}{(1-x_{1}t)(1-x_{3}t)}}\right]dt,$$

(3.1)

where $|x_n|<1$ ($n=1,2,3$).

Proof: By changing the order of integration and summation (with uniform convergence of the integral) in (2.2) and using the relation (1.7), we obtain

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)={\displaystyle \frac{\Gamma \left(b_1\right)}{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}}{\int }_0^1{t}^{a_1-1}{(1-t)}^{b_1-a_1-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)$$

$$\times \sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{(b_1-a_1)}_j}}{\displaystyle \frac{{\left(x_{1}t\right)}^i}{i!}}{\displaystyle \frac{{\left(x_2(1-t)\right)}^j}{j!}}{\displaystyle \frac{{\left(x_{3}t\right)}^k}{k!}}dt.$$

(3.2)

Applying the result ${\left(\lambda \right)}_{i+j}={\left(\lambda \right)}_j{(\lambda +j)}_i$ in the treble sum in (3.2), yields

$$\sum _{j=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_j{\left(a_3\right)}_j}{{(b_1-a_1)}_j}}{\displaystyle \frac{{\left(x_2(1-t)\right)}^j}{j!}}\sum _{i=0}^{\infty }{(a_2+j)}_i{\displaystyle \frac{{\left(x_{1}t\right)}^i}{i!}}\sum _{k=0}^{\infty }{(a_3+j)}_k{\displaystyle \frac{{\left(x_{3}t\right)}^k}{k!}}.$$

Now, emplyoing the binomial theorem

$$\sum _{m=0}^{\infty }{\displaystyle \frac{{\left(\lambda \right)}_m{z}^m}{m!}}={(1-z)}^{-\lambda },(\lambda \in \mathsf{C},|z|<1)$$

in the sum to evaluate the sums over i and k, we get

$$={(1-x_{1}t)}^{-a_2}{(1-x_{3}t)}^{-a_3}\sum _{j=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_j{\left(a_3\right)}_j}{{(b_1-a_1)}_j}}{\left[{\displaystyle \frac{x_2(1-t)}{(1-x_{1}t)(1-x_{3}t)}}\right]}^j{\displaystyle \frac{1}{j!}},$$

identify the sum over j as (1.1), we get the representaion (3.1).

Remark 1. Here, we discuss some special cases by making suitable transformations of the integration variable. If we put

$$\begin{array}{ccc}\hfill \left(i\right)t& =& {\displaystyle \frac{w}{1+w}},{\displaystyle \frac{dt}{dw}}={\displaystyle \frac{1}{{(1+w)}^2}},\hfill \\ \hfill \left(ii\right)t& =& {\displaystyle \frac{({\theta }_2-{\theta }_3)(w-{\theta }_1)}{({\theta }_2-{\theta }_1)(w-{\theta }_3)}},{\displaystyle \frac{dt}{dw}}={\displaystyle \frac{({\theta }_2-{\theta }_3)({\theta }_1-{\theta }_3)}{({\theta }_2-{\theta }_1){(w-{\theta }_3)}^2}},\hfill \\ \hfill \left(iii\right)t& =& {sin}^{2}w,{\displaystyle \frac{dt}{dw}}=2sinwcosw,\hfill \\ \hfill \left(iv\right)t& =& {\displaystyle \frac{(1+\rho )w}{1+\rho w}},{\displaystyle \frac{dt}{dw}}={\displaystyle \frac{(1+\rho )}{{(1+\rho w)}^2}},\hfill \end{array}$$

we get the integral representations as follows:

$$\left(i\right){\displaystyle \frac{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}{\Gamma \left(b_1\right)}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$={\int }_0^{\infty }{w}^{a_1-1}{(1+w)}^{a_2+a_3-b_1}{\Lambda }_1^{-a_2}{\Lambda }_2^{-a_3}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{{\delta }_1^l{\delta }_2^l}}\right){}_{2}F_1\left(\begin{array}{c}{a}_2,a_3\\ b_1-a_1\end{array};{\delta }_3{x}_2\right)dw,$$

(3.3)

where

$${\delta }_1={\displaystyle \frac{w}{1+w}},{\delta }_2={\displaystyle \frac{1}{1+w}},{\delta }_3={\displaystyle \frac{1+w}{{\Lambda }_1{\Lambda }_2}},{\Lambda }_1=1+(1-x_1)w,{\Lambda }_2=1+(1-x_3)w;$$

$$\left(ii\right){\displaystyle \frac{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}{\Gamma \left(b_1\right)}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)={\displaystyle \frac{{({\theta }_2-{\theta }_3)}^{a_1}{({\theta }_1-{\theta }_3)}^{b_1-a_1}}{{({\theta }_2-{\theta }_1)}^{b_1-a_2-a_3-1}}}$$

$$\times {\int }_{{\theta }_1}^{{\theta }_2}{\displaystyle \frac{{(w-{\theta }_1)}^{a_1-1}{({\theta }_2-w)}^{b_1-a_1-1}}{{(w-{\theta }_3)}^{b_1-a_2-a_3}}}{\Lambda }_1^{-a_2}{\Lambda }_2^{-a_3}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{{\delta }_1^l{\delta }_2^l}}\right){}_{2}F_1\left(\begin{array}{c}{a}_2,a_3\\ b_1-a_1\end{array};{\delta }_3{x}_2\right)dw,$$

(3.4)

where, with ${\theta }_3<{\theta }_1<{\theta }_2$,

$${\delta }_1={\displaystyle \frac{({\theta }_2-{\theta }_3)(w-{\theta }_1)}{({\theta }_2-{\theta }_1)(w-{\theta }_3)}},{\delta }_2={\displaystyle \frac{({\theta }_1-{\theta }_3)({\theta }_2-w)}{({\theta }_2-{\theta }_1)(w-{\theta }_3)}},{\delta }_3={\displaystyle \frac{({\theta }_1-{\theta }_3)({\theta }_2-{\theta }_1)({\theta }_2-w)(w-{\theta }_3)}{{\Lambda }_1{\Lambda }_2}},$$

$${\Lambda }_1=({\theta }_2-{\theta }_1)(w-{\theta }_3)-x_1({\theta }_2-{\theta }_3)(w-{\theta }_1),{\Lambda }_2=({\theta }_2-{\theta }_1)(w-{\theta }_3)-x_3({\theta }_2-{\theta }_3)(w-{\theta }_1);$$

$$\left(iii\right){\displaystyle \frac{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}{\Gamma \left(b_1\right)}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$=2{\int }_0^{\pi /2}{\left({sin}^{2}w\right)}^{a_1-{\displaystyle \frac{1}{2}}}{\left({cos}^{2}w\right)}^{b_1-a_1-{\displaystyle \frac{1}{2}}}{\Lambda }_1^{-a_2}{\Lambda }_2^{-a_3}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{{\delta }_1^l{\delta }_2^l}}\right){}_{2}F_1\left(\begin{array}{c}{a}_2,a_3\\ b_1-a_1\end{array};{\delta }_3{x}_2\right)dw,$$

(3.5)

where

$${\delta }_1={sin}^{2}w,{\delta }_2={cos}^{2}w,{\delta }_3={\displaystyle \frac{{cos}^{2}w}{{\Lambda }_1{\Lambda }_2}},{\Lambda }_1=1-x_1{sin}^{2}w,{\Lambda }_2=1-x_3{sin}^{2}w;$$

and

$$\left(iv\right){\displaystyle \frac{\Gamma \left(a_1\right)\Gamma (b_1-a_1)}{\Gamma \left(b_1\right)}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$={(1+\rho )}^{a_1}{\int }_0^1{\displaystyle \frac{w^{a_1-1}{(1-w)}^{b_1-a_1-1}}{{(1+\rho w)}^{b_1-a_2-a_3}}}{\Lambda }_1^{-a_2}{\Lambda }_2^{-a_3}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{{\delta }_1^l{\delta }_2^l}}\right){}_{2}F_1\left(\begin{array}{c}{a}_2,a_3\\ b_1-a_1\end{array};{\delta }_3{x}_2\right)dw,$$

(3.6)

where, with $\rho >-1$,

$${\delta }_1={\displaystyle \frac{(1+\rho )w}{1+\rho w}},{\delta }_2={\displaystyle \frac{1-w}{1+\rho w}},{\delta }_3={\displaystyle \frac{(1-w)(1+\rho w)}{{\Lambda }_1{\Lambda }_2}},$$

$${\Lambda }_1=1+\rho w-(1+\rho )x_{1}w,{\Lambda }_2=1+\rho w-(1+\rho )x_{3}w.$$

3.2. Derivative Properties

In this part, we discuss some interesting derivative properties of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ .

Theorem 2.

The following derivative formula holds for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

$${\left({\displaystyle \frac{d}{dp}}\right)}^m\left[p^{\alpha -1}{\mathbf{H}}_{C,p^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]$$

$$=p^{\alpha -m-1}{\mathbf{H}}_{C,p^{\lambda },\alpha -m}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3),$$

(3.7)

where $\alpha ,a_n\in \mathsf{C}$, $b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-}$; $m\in \mathsf{N}$ and $R(\alpha -m)>0,R\left(\alpha \right)>0,\lambda >0,R\left(a_n\right)>0;n=1,2,3$.

Proof: Using (2.2) and employing term-wise differentiation m times in L.H.S of (3.7), we obtain

$${\left({\displaystyle \frac{d}{dp}}\right)}^m\left[p^{\alpha -1}{\mathbf{H}}_{C,p^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]$$

$$={\left({\displaystyle \frac{d}{dp}}\right)}^m\left[p^{\alpha -1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}\right.$$

$$\left.\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha }\left\{-{\displaystyle \frac{p^{\lambda }}{t^l{(1-t)}^l}}\right\}dt\right]$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}$$

$$\times \sum _{n=0}^{\infty }{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\left({\displaystyle \frac{d}{dp}}\right)}^m\left[p^{\lambda n+\alpha -1}\right]dt$$

$$=p^{\alpha -m-1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}$$

$$\times {(1-t)}^{b_1+j-a_1-1}\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{p^{\lambda }}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha -m)}}dt$$

$$=p^{\alpha -m-1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}$$

$$\times {(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha -m}\left(-{\displaystyle \frac{p^{\lambda }}{t^l{(1-t)}^l}}\right)dt$$

$$=p^{\alpha -m-1}{\mathbf{H}}_{C,p^{\lambda },\alpha -m}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3).$$

Theorem 3.

We have the following differentiation formula as follows:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$=\alpha {\mathbf{H}}_{C,p,\alpha +1}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)+\lambda p{\displaystyle \frac{d}{dp}}{\mathbf{H}}_{C,p,\alpha +1}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3),$$

(3.8)

where $\alpha \in \mathsf{C},$$b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-},$ and $\lambda >0,R\left(\alpha \right)>0$.

Proof: Using (2.2) in the R.H.S of (3.8), yields

$$\alpha {\mathbf{H}}_{C,p,\alpha +1}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)+\lambda p{\displaystyle \frac{d}{dp}}{\mathbf{H}}_{C,p,\alpha +1}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$=\alpha {\mathbf{H}}_{C,p,\alpha +1}^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)+\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}$$

$$\times {(1-t)}^{b_1+j-a_1-1}\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{\lambda n{p}^n}{\Gamma (\lambda n+\alpha +1)}}dt$$

(3.9)

Now employing the relation $\Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right)$ in (3.9), we get

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}$$

$$\times \sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}dt$$

$$={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3).$$

4. Integral Transform of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

Here, we explore some integral transforms, namely, the Euler-Beta transform, Laplace transform, Mellin transform, and Whittaker transform of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$.

4.1. Euler-Beta Transform

The Euler-Beta transform of the function $f\left(p\right)$ is defined as [18]

$$\mathbf{B}\left\{f\left(p\right);a,b\right\}={\int }_0^1{p}^{a-1}{(1-p)}^{b-1}f\left(p\right)dp.$$

(4.1)

Theorem 4.

The Euler-Beta transform of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows:

$$\mathbf{B}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);a,b\right\}=\Gamma \left(b\right)\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}$$

$$\times {\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{}^{\Psi }B_1^l\left.\left[\begin{array}{c}(1,1),(a,1)\\ (\alpha ,\lambda ),(a+b,1)\end{array}\right|(a_1+i+k),(b_1+j-a_1)\right],$$

(4.2)

where, $\alpha ,a,b,l,(a_1+i+k),(b_1+j-a_1)\in \mathsf{C};$$\lambda >0,R\left(\alpha \right)>0,R\left(a\right)>0,R\left(b\right)>0,R(a_1+i+k)>0,R(b_1+j-a_1)>0.$

Proof: Using (4.1) in (2.2), we obtain

$$\mathbf{B}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);a,b\right\}$$

$$={\int }_0^1{p}^{a-1}{(1-p)}^{b-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times \left({\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}\left\{{\int }_0^1{p}^{a-1}{(1-p)}^{b-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\right\}\right)dt.$$

(4.3)

By evaluating the integral as

$$\begin{array}{ccc}\hfill & & {\int }_0^1{p}^{a-1}{(1-p)}^{b-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\hfill \\ \hfill & & =\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}{\int }_0^1{p}^{a+n-1}{(1-p)}^{b-1}dp\hfill \\ \hfill & & =\Gamma \left(b\right)\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{\Gamma (a+n)\Gamma (n+1)}{\Gamma (\lambda n+\alpha )\Gamma (a+b+n)n!}},\hfill \end{array}$$

employing the Fox-Wright function (1.12) in the above sum and using in (4.3), we get,

$$\mathbf{B}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);a,b\right\}=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times \Gamma \left(b\right)\left({\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{}_2\Psi _2\left[\begin{array}{c}(1,1),(a,1)\\ (\alpha ,\lambda ),(a+b,1)\end{array};-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right]\right)dt.$$

Upon evaluating the integral using the generalized Beta function defined in (1.11), we get the required result (4.2).

4.2. Laplace Transform

The Laplace transform of a function $f\left(p\right)$ is defined as [18]

$$\mathbf{L}\left\{f\left(p\right)\right\}={\int }_0^{\infty }{e}^{-sp}f\left(p\right)dp$$

(4.4)

Theorem 5.

The Laplace transform of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows:

$${\int }_0^{\infty }{p}^{a-1}{e}^{-sp}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp={\displaystyle \frac{1}{s^a}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}$$

$$\times {\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{{\displaystyle \frac{x_3^k}{k!}}}^{\Psi }{B}_{\left({\displaystyle \frac{1}{s}}\right)}^l\left.\left[\begin{array}{c}(1,1),(a,1)\\ (\alpha ,\lambda )\end{array}\right|(a_1+i+k),(b_1+j-a_1)\right],$$

(4.5)

where, $\lambda ,\alpha ,a,(a_1+i+k),(b_1+j-a_1)\in \mathsf{C};$$R\left(\lambda \right)>0,R\left(\alpha \right)>0,R\left(a\right)>0,R\left(s\right)>0,R(a_1+i+k)>0,R(b_1+j-a_1)>0$ and $|-{\displaystyle \frac{1}{s{t}^l{(1-t)}^l}}|<1$.

Proof: Using (2.2) in the integral on the left-hand side of (4.5), we get

$${\int }_0^{\infty }{p}^{a-1}{e}^{-sp}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times \left({\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}\left\{{\int }_0^{\infty }{e}^{-sp}{p}^{a-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\right\}\right)dt.$$

(4.6)

By evaluating the integral as

$$\begin{array}{ccc}\hfill & & {\int }_0^{\infty }{e}^{-sp}{p}^{a-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\hfill \\ \hfill & & =\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}{\int }_0^{\infty }{e}^{-sp}{p}^{a+n-1}dp\hfill \\ \hfill & & ={\displaystyle \frac{1}{s^a}}\sum _{n=0}^{\infty }{\displaystyle \frac{\Gamma (a+n)\Gamma (n+1)}{\Gamma (\lambda n+\alpha )n!}}{\left(-{\displaystyle \frac{1}{s{t}^l{(1-t)}^l}}\right)}^n,\hfill \end{array}$$

employing the Fox-Wright function (1.12) in the above sum and using in (4.6), we get,

$${\int }_0^{\infty }{p}^{a-1}{e}^{-sp}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times {\displaystyle \frac{1}{s^a}}\left({\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{}_2\Psi _1\left[\begin{array}{c}(1,1),(a,1)\\ (\alpha ,\lambda )\end{array};-{\displaystyle \frac{1}{s{t}^l{(1-t)}^l}}\right]\right)dt.$$

Upon evaluating the integral using the generalized beta function defined in (1.11), we reach the result (4.5).

4.3. Mellin Transform

The Mellin transform of the function $f\left(p\right)$ is defined as [18],

$$f^{*}\left(s\right)=\mathbf{M}\left\{f\left(p\right);s\right\}={\int }_0^{\infty }{p}^{s-1}f\left(p\right)dp,R\left(s\right)>0,$$

(4.7)

and the inverse Mellin transform is given by

$$f\left(p\right)={\mathbf{M}}^{-\mathbf{1}}\left\{f^{*}\left(s\right);p\right\}={\displaystyle \frac{1}{2\pi i}}{\int }_{c-i\infty }^{c+i\infty }{p}^{-s}{f}^{*}\left(s\right)ds,c\in \mathsf{R}.$$

(4.8)

Theorem 6.

The Mellin transform of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows:

$$\mathbf{M}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);s\right\}={\int }_0^{\infty }{p}^{s-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp$$

$$={\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{H}_C^{\left(ls\right)}\left(a_1,a_2,a_3;b_1;x_1,x_2,x_3\right),$$

(4.9)

where $\alpha ,s\in \mathsf{C}$; $R\left(s\right)>\alpha >0$, $R\left(\alpha \right)>0,\lambda >0,$$b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-}$ and $H_C^{\left(ls\right)}(·)$ is defined in (1.6).

Proof: Using (2.2) in the integral on the left-hand side of (4.9), we get

$$\mathbf{M}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);s\right\}=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}\left\{{\int }_0^{\infty }{p}^{s-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\right\}dt.$$

(4.10)

Now, if we put $\gamma =q=1$ and $\xi =-p/t^l{(1-t)}^l$ in the following result ( [16], Th. 4.1)

$$E_{\lambda ,\alpha }^{\gamma ,q}\left(\xi \right)d\xi ={\displaystyle \frac{1}{2\pi i\Gamma \left(\gamma \right)}}{\int }_L{\displaystyle \frac{\Gamma \left(s\right)\Gamma (\gamma -qs)}{\Gamma (\alpha -\lambda s)}}{(-\xi )}^{-s}ds,$$

where L is the contour of integration that begins at $c-i\infty$ and ends at $c+i\infty$, $c\in \mathsf{R},$

we get,

$$E_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)={\displaystyle \frac{1}{2\pi i}}{\int }_L{\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{\left({\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)}^{-s}ds$$

$$={\displaystyle \frac{1}{2\pi i}}{\int }_L{f}^{*}\left(s\right)p^{-s}ds,$$

where

$$f^{*}\left(s\right)={\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{\left[t^l{(1-t)}^l\right]}^s.$$

(4.11)

Using (4.7), (4.8), and (4.11), which leads to

$${\int }_0^{\infty }{p}^{s-1}{E}_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp={\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{\left[t^l{(1-t)}^l\right]}^s$$

(4.12)

Finally, put (4.12) in (4.10), yields

$$\mathbf{M}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);s\right\}=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}$$

$$\times {\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{\int }_0^1{t}^{a_1+i+k+ls-1}{(1-t)}^{b_1+j+ls-a_1-1}dt.$$

Evaluate the integral using the classical Beta function, yields

$$\mathbf{M}\left\{{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3);s\right\}$$

$$={\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B(a_1+ls+i+k,b_1+ls+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}.$$

Identifying the above sum as $H_C^{\left(ls\right)}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$ defined in (1.6), we get the result (4.9).

Corollary 1: The inverse Mellin formula for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)={\mathbf{M}}^{-1}\left\{f^{*}\left(s\right);p\right\}$$

$$={\displaystyle \frac{1}{2\pi i}}{\int }_{c-i\infty }^{c+i\infty }{p}^{-s}{\displaystyle \frac{\Gamma \left(s\right)\Gamma (1-s)}{\Gamma (\alpha -\lambda s)}}{H}_C^{\left(ls\right)}\left(a_1,a_2,a_3;b_1;x_1,x_2,x_3\right)ds,$$

(4.13)

where $c>\alpha$

4.4. Whittaker Transform

Theorem 7.

The Whittaker transform ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ is as follows:

$${\int }_0^{\infty }{p}^{a-1}{e}^{-{\displaystyle \frac{1}{2}}bp}{\mathbf{W}}_{\eta ,\mu }\left(bp\right){\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp$$

$$={\displaystyle \frac{1}{b^a}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$${\times }^{\Psi }{B}_{\left({\displaystyle \frac{1}{b}}\right)}^l\left.\left[\begin{array}{c}(1,1),({\displaystyle \frac{1}{2}}\pm \mu +a,1)\\ (\alpha ,\lambda ){,}_{(}1-\eta +a,1)\end{array}\right|(a_1+i+k),(b_1+j-a_1)\right]$$

(4.14)

where, $\alpha ,a,(a_1+i+k),(b_1+j-a_1)\in \mathsf{C}$; $\lambda >0,R\left(\alpha \right)>0,R\left(a\right)>0,R(a_1+i+k)>0,R(b_1+j-a_1)>0$.

Proof: Using (4.14) in (2.2), we get

$${\int }_0^{\infty }{p}^{a-1}{e}^{-{\displaystyle \frac{1}{2}}bp}{\mathbf{W}}_{\eta ,\mu }\left(bp\right){\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times \left({\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}\left\{{\int }_0^{\infty }{p}^{a-1}{e}^{-{\displaystyle \frac{1}{2}}bp}{\mathbf{W}}_{\eta ,\mu }\left(bp\right)E_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\right\}\right)dt.$$

(4.15)

Now consider the right-most integral of (4.15) as I and put $bp=v$, yields

$$\begin{array}{ccc}\hfill I& =& {\int }_0^{\infty }{p}^{a-1}{e}^{-{\displaystyle \frac{1}{2}}bp}{\mathbf{W}}_{\eta ,\mu }\left(bp\right)E_{\lambda ,\alpha }\left(-{\displaystyle \frac{p}{t^l{(1-t)}^l}}\right)dp\hfill \\ \hfill & =& {\int }_0^{\infty }{\left({\displaystyle \frac{v}{b}}\right)}^{a-1}{e}^{{\displaystyle \frac{-v}{2}}}{\mathbf{W}}_{\eta ,\mu }\left(v\right)\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{v}{b{t}^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}{\displaystyle \frac{1}{b}}dv\hfill \end{array}$$

$$={\displaystyle \frac{1}{b^a}}\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{b{t}^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}{\int }_0^{\infty }{v}^{n+a-1}{e}^{-{\displaystyle \frac{v}{2}}}{\mathbf{W}}_{\eta ,\mu }\left(v\right)dv.$$

(4.16)

Now employing the result

$${\int }_0^{\infty }{v}^{a-1}{e}^{-{\displaystyle \frac{v}{2}}}{\mathbf{W}}_{\eta ,\mu }\left(v\right)dv={\displaystyle \frac{\Gamma (\frac{1}{2}+\mu +a)\Gamma (\frac{1}{2}-\mu +a)}{\Gamma (1-\eta +a)}},\left(R\left(\mu \pm a\right)>{\displaystyle \frac{-1}{2}}\right),$$

where ${\mathbf{W}}_{\eta ,\mu }\left(v\right)$ is the Whittaker function [25], in (4.16), we get

$$I={\displaystyle \frac{1}{b^a}}\sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{b{t}^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{\Gamma (\frac{1}{2}+\mu +a+n)\Gamma (\frac{1}{2}-\mu +a+n)\Gamma (n+1)}{\Gamma (\lambda n+\alpha )\Gamma (1-\eta +a+n)n!}}$$

Using (1.12), we get

$$I={\displaystyle \frac{1}{b^a}}{}_3\Psi _2\left[\begin{array}{c}(1,1),({\displaystyle \frac{1}{2}}+\mu +a,1),({\displaystyle \frac{1}{2}}-\mu +a,1)\\ (\alpha ,\lambda ),(1-\eta +a,1)\end{array};-{\displaystyle \frac{1}{b{t}^l{(1-t)}^l}}\right].$$

(4.17)

Applying (4.17) in (4.15), we get

$${\int }_0^{\infty }{p}^{a-1}{e}^{-{\displaystyle \frac{1}{2}}bp}{\mathbf{W}}_{\eta ,\mu }\left(bp\right){\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)dp=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{\displaystyle \frac{1}{b^a}}{}_3\Psi _2\left[\begin{array}{c}(1,1),({\displaystyle \frac{1}{2}}\pm \mu +a,1)\\ (\alpha ,\lambda ),(1-\eta +a,1)\end{array};-{\displaystyle \frac{1}{b{t}^l{(1-t)}^l}}\right]dt.$$

Upon evaluating the integral in terms of the generalized Beta function defined in (1.11), we obtain the desired result (4.14).

5. Fractional Calculus of ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

Here, we investigate some results associated with the right-sided Riemann-Liouville fractional integral operator $I_{a+}^{\xi }$ and the derivative operator $D_{a+}^{\xi },$ which are defined respectively as follows [9,14]:

$$\left(I_{a+}^{\xi }f\right)\left(z\right)={\displaystyle \frac{1}{\Gamma \left(\xi \right)}}{\int }_a^z{\displaystyle \frac{f\left(p\right)}{{(z-p)}^{1-\xi }}}dp,(R\left(\xi \right)>0,\xi \in \mathsf{C}),$$

(5.1)

and

$$\left(D_{a+}^{\xi }f\right)\left(z\right)={\left({\displaystyle \frac{d}{dz}}\right)}^m\left(I_{a+}^{m-\xi }f\right)\left(z\right),(m=\left[R\left(\xi \right)\right]+1;R\left(\xi \right)>0,\alpha \in \mathsf{C}),$$

(5.2)

where $\left[t\right]$ is the greatest integer.

Hilfer [5] defined a generalized Riemann-Liouville fractional derivative operator $D_{a+}^{\xi ,\gamma }$ of order $0<\xi <1$ and type $0\le \gamma \le 1$ with respect to z as follows:

$$\left(D_{a+}^{\xi ,\gamma }f\right)\left(z\right)=\left(I_{a+}^{\gamma (1-\xi )}{\displaystyle \frac{d}{dz}}\right)\left(I_{a+}^{(1-\gamma )(1-\xi )}f\right)\left(z\right),(m=\left[R\left(\xi \right)\right]+1;R\left(\xi \right)>0,\xi \in \mathsf{C}).$$

(5.3)

Theorem 8.

Let $a\in {\mathsf{R}}_{+};$$\alpha ,\xi \in \mathsf{C};$$\lambda >0,R\left(\alpha \right)>0,R\left(\xi \right)>0,$ and $b_1\in \mathsf{C}\setminus {\mathsf{Z}}_0^{-}$, then for $z>a$ the following relations hold for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

$$\left(i\right)\left(I_{a+}^{\xi }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$={(z-a)}^{\alpha +\xi -1}{\mathbf{H}}_{C,{(z-a)}^{\lambda },\alpha +\xi }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

(5.4)

$$\left(ii\right)\left(D_{a+}^{\xi }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$={(z-a)}^{\alpha -\xi -1}{\mathbf{H}}_{C,{(z-a)}^{\lambda },\alpha -\xi }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

(5.5)

and

$$\left(iii\right)\left(D_{a+}^{\xi ,\gamma }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$={(z-a)}^{\alpha -\xi -1}{\mathbf{H}}_{C,{(z-a)}^{\lambda },\alpha -\xi }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3).$$

(5.6)

Proof. Use (2.2) in (5.1), and with the help of the relation [14]

$$\left(I_{a+}^{\xi }\left[{(p-a)}^{\mu -1}\right]\right)\left(z\right)={\displaystyle \frac{\Gamma \left(\mu \right)}{\Gamma (\mu +\xi )}}{(z-a)}^{\xi +\mu -1},$$

for $z>a$ yields,

$$\left(I_{a+}^{\xi }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$=\left(I_{a+}^{\xi }\left[{(p-a)}^{\alpha -1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}\right.\right.$$

$$\left.\left.\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha }\left\{-{\displaystyle \frac{{(p-a)}^{\lambda }}{t^l{(1-t)}^l}}\right\}dt\right]\right)\left(z\right)$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}$$

$$\times \sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}\left(I_{a+}^{\xi }\left[{(p-a)}^{\lambda n+\alpha -1}\right]dt\right)\left(z\right)$$

$$={(z-a)}^{\alpha +\xi -1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha +\xi }\left\{-{\displaystyle \frac{{(z-a)}^{\lambda }}{t^l{(1-t)}^l}}\right\}dt$$

$$={(z-a)}^{\alpha +\xi -1}{\mathbf{H}}_{C,{(z-a)}^{\lambda },\alpha +\xi }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3).$$

Next, employing (2.2) in (5.2), we get

$$\left(D_{a+}^{\xi }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$={\left({\displaystyle \frac{d}{dz}}\right)}^m\left(I_{a+}^{m-\xi }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$={\left({\displaystyle \frac{d}{dz}}\right)}^m\left[{(z-a)}^{\alpha +m-\xi -1}{\mathbf{H}}_{C,{(z-a)}^{\lambda },\alpha +m-\xi }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right].$$

Using (3.7), we get the desired result (5.5).

Finally, with the help of eqs. (2.2) and (5.3), we obtain

$$\left(D_{a+}^{\xi ,\gamma }\left[{(p-a)}^{\alpha -1}{\mathbf{H}}_{C,{(p-a)}^{\lambda },\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)\right]\right)\left(z\right)$$

$$=\left(D_{a+}^{\xi ,\gamma }\left[{(p-a)}^{\alpha -1}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}\right.\right.$$

$$\left.\left.\times {\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}{E}_{\lambda ,\alpha }\left\{-{\displaystyle \frac{{(p-a)}^{\lambda }}{t^l{(1-t)}^l}}\right\}dt\right]\right)\left(z\right)$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_{j}B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}{\int }_0^1{t}^{a_1+i+k-1}{(1-t)}^{b_1+j-a_1-1}$$

$$\times \sum _{n=0}^{\infty }{\left(-{\displaystyle \frac{1}{t^l{(1-t)}^l}}\right)}^n{\displaystyle \frac{1}{\Gamma (\lambda n+\alpha )}}\left(D_{a+}^{\xi ,\gamma }\left[{(p-a)}^{\lambda n+\alpha -1}\right]dt\right)\left(z\right).$$

(5.7)

Applying the relation of Srivastava and Tomovski ([21], p.203, eq.(2.18))

$$\left(D_{a+}^{\xi ,\gamma }\left[{(p-a)}^{\beta -1}\right]\right)\left(z\right)={\displaystyle \frac{\Gamma \left(\beta \right)}{\Gamma (\beta -\xi )}}{(z-a)}^{\beta -\xi -1}$$

$$(z>a;0<\xi <1;0\le \gamma \le 1;R(\beta )>0),$$

in (5.7), we get the relation (5.6).

6. Recursion Formulas for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$

In this section, we investigate two recursion formulas for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$. The first formula gives recursions with respect to the numerator parameters $a_2$ and $a_3$, and the second a recursion with respect to the denominator parameter $b_1$.

Theorem 9.

We have the following recursions for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ with respect to the numerator parameters $a_2$ and $a_3$ as follows:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$+{\displaystyle \frac{x_1{a}_1}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2+1,a_3;b_1+1;x_1,x_2,x_3)+{\displaystyle \frac{x_2{a}_3}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3+1;b_1+1;x_1,x_2,x_3).$$

(6.1)

and

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3+1;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$+{\displaystyle \frac{x_2{a}_2}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3+1;b_1+1;x_1,x_2,x_3)+{\displaystyle \frac{x_3{a}_1}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3+1;b_1+1;x_1,x_2,x_3).$$

(6.2)

Proof. From (2.1) and using the identity ${(a_3+1)}_{j+k}={\left(a_3\right)}_{j+k}(1+j/a_3+k/a_3)$, we obtain

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3+1;b_1;x_1,x_2,x_3)$$

$$=\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{(a_3+1)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$+{\displaystyle \frac{x_2}{a_3}}\sum _{i=0}^{\infty }\sum _{j=1}^{\infty }\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x^i}{i!}}{\displaystyle \frac{x_2^{j-1}}{(j-1)!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$+{\displaystyle \frac{x_3}{a_3}}\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{\left(a_3\right)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x_1^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^{k-1}}{(k-1)!}}.$$

(6.3)

Consider the first sum in (6.3), which we denote by $S_1$. Put $j\to j+1$ and apply the identity ${\left(a\right)}_{j+1}=a{(a+1)}_j$, we find

$$S_1={\displaystyle \frac{x_2}{a_3}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j+1}{\left(a_3\right)}_{j+k+1}}{{\left(b_1\right)}_{j+1}}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+j+1-a_1)}{B(a_1,b_1+j+1-a_1)}}{\displaystyle \frac{x^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$={\displaystyle \frac{x_2{a}_2}{b_1}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{(a_2+1)}_{i+j}{(a_3+1)}_{j+k}}{{(b_1+1)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+i+k,b_1+1+j-a_1)}{B(a_1,b_1+j+1-a_1)}}{\displaystyle \frac{x^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}.$$

$$={\displaystyle \frac{x_2{a}_2}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3+1;b_1+1;x_1,x_2,x_3).$$

(6.4)

Similarly, take the second sum in (6.3) as $S_2$ with $k\to k+1$, we obtain

$$S_2=x_3\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{(a_3+1)}_{j+k}}{{\left(b_1\right)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+1+i+k,b_1+j-a_1)}{B(a_1,b_1+j-a_1)}}{\displaystyle \frac{x^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}.$$

Now, applying the result

$$B(p+1,q)={\displaystyle \frac{p}{p+q}}B(p,q),$$

yields

$$S_2={\displaystyle \frac{x_3{a}_1}{b_1}}\sum _{i,j,k=0}^{\infty }{\displaystyle \frac{{\left(a_2\right)}_{i+j}{(a_3+1)}_{j+k}}{{(b_1+1)}_j}}{\displaystyle \frac{B_{p,\alpha }^{\lambda ,l}(a_1+1+i+k,b_1+j-a_1)}{B(a_1+1,b_1+j-a_1)}}{\displaystyle \frac{x^i}{i!}}{\displaystyle \frac{x_2^j}{j!}}{\displaystyle \frac{x_3^k}{k!}}$$

$$={\displaystyle \frac{x_3{a}_1}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3+1;b_1+1;x_1,x_2,x_3).$$

(6.5)

Combining (6.4) and (6.5) with (6.3), we get the required result (6.2). Similarly, we can establish the proof of the result (6.1).

Corollary 2: From (6.1) and (6.2), the following recursions for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ hold:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+M,a_3;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$+{\displaystyle \frac{x_1{a}_1}{b_1}}\sum _{k=1}^M{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2+k,a_3;b_1+1;x_1,x_2,x_3)+{\displaystyle \frac{x_2{a}_3}{b_1}}\sum _{k=1}^M{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+k,a_3+1;b_1+1;x_1,x_2,x_3)$$

(6.6)

and

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3+M;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$+{\displaystyle \frac{x_2{a}_2}{b_1}}\sum _{k=1}^M{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3+k;b_1+1;x_1,x_2,x_3)+{\displaystyle \frac{x_3{a}_1}{b_1}}\sum _{k=1}^M{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3+k;b_1+1;x_1,x_2,x_3),$$

(6.7)

for positive integer M.

Theorem 10.

The function ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ satisfies the following recursion formulas:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2-1,a_3;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$-{\displaystyle \frac{x_1{a}_1}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3;b_1+1;x_1,x_2,x_3)-{\displaystyle \frac{x_2{a}_3}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3+1;b_1+1;x_1,x_2,x_3).$$

(6.8)

and

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3-1;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$-{\displaystyle \frac{x_2{a}_2}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3;b_1+1;x_1,x_2,x_3)-{\displaystyle \frac{x_3{a}_1}{b_1}}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3;b_1+1;x_1,x_2,x_3).$$

(6.9)

Proof. Replace $a_2$ and $a_3$ with $a_2-1$ and $a_3-1$ in (6.1) and (6.2), we get the results (6.8) and (6.9) respectively.

Corollary 3: From (6.1) and (6.2), the following recursions for ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ hold:

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2-M,a_3;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$-{\displaystyle \frac{x_1{a}_1}{b_1}}\sum _{k=0}^{M-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2-k,a_3;b_1+1;x_1,x_2,x_3)-{\displaystyle \frac{x_2{a}_3}{b_1}}\sum _{k=0}^{M-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2-k,a_3+1;b_1+1;x_1,x_2,x_3)$$

(6.10)

and

$${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3-M;b_1;x_1,x_2,x_3)={\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2,a_3;b_1;x_1,x_2,x_3)$$

$$-{\displaystyle \frac{x_2{a}_2}{b_1}}\sum _{k=0}^{M-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1,a_2+1,a_3-k;b_1+1;x_1,x_2,x_3)-{\displaystyle \frac{x_3{a}_1}{b_1}}\sum _{k=0}^{M-1}{\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(a_1+1,a_2,a_3-k;b_1+1;x_1,x_2,x_3),$$

(6.11)

for positive integer M.

7. Conclusion

In this paper, we have introduced the extended Srivastava’s triple hypergeometric function defined by ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$ in (2.1). We have given some integral representations of this function that involve the generalized Mittag-Leffler function and Gauss hypergeometric function. We have also explored some derivative properties and integral transforms of the function ${\mathbf{H}}_{C,p,\alpha }^{\lambda ,l}(·)$, namely the Euler-Beta transform, Laplace Transform, Mellin transform, and Whittaker transform and some recursion relations.

This article does not contain any studies with human participants or animals performed by any of the author.