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I am having trouble with calculating the following integral:

\begin{equation} I = \int_{0}^{\infty}x\exp({-\beta x})\large{G}_{2,2}^{1,2}\left( x \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right) \large{G}_{1,2}^{2,0}\left( 2\alpha \sqrt{ab}x \left| \begin{array}{cc} 1/2 \\ v,-v \end{array} \right. \right)dx \end{equation}

where $\beta$, $\alpha$, $a$, $b$ are non-zero positive constants, and $v \in \{0,1\}$. Any ideas or references would be very helpful. Thank you!

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Finally, here is the solution.

Step 1: Expressing the product of two Meijer G function using an identity from http://functions.wolfram.com/07.34.16.0003.01

\begin{align} G_{2,2}^{1,2}\left(x\Bigg\vert\begin{matrix}1,1\cr 1,0\end{matrix}\right)G_{1,2}^{2,0}\left(2\alpha\sqrt{ab}x\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\right) = G_{0,0:2,2:1,2}^{0,0:1,2:2,0}\left(\begin{matrix}-\cr -\end{matrix}\Bigg\vert\begin{matrix}1,1\cr 1,0\end{matrix}\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\Bigg\vert x,2\alpha\sqrt{ab}x\right) \end{align}

Step 2: Expressing above result using extended generalized bivariate Meijer G-function (EGBMGF) from "Sharma, B.L. and Abiodun, R.F.A.: ‘Generating function for generalized function of two variables’ Proc. American Mathematical Society, Oct. 1974, 46, (1), pp. 69-72" \begin{align} G_{0,0:2,2:1,2}^{0,0:1,2:2,0}\left(\begin{matrix}-\cr -\end{matrix}\Bigg\vert\begin{matrix}1,1\cr 1,0\end{matrix}\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\Bigg\vert x,2\alpha\sqrt{ab}x\right)=&S\bigg[x,2\alpha\sqrt{ab}x\bigg\vert\bigg[\begin{matrix} 0,0\cr0,0 \end{matrix}\bigg]\begin{matrix} -\cr- \end{matrix}\bigg\vert\bigg(\begin{matrix} 1,2\cr2,2 \end{matrix}\bigg)\begin{matrix} 1,1\cr1,0 \end{matrix}\bigg\vert\bigg(\begin{matrix} 2,0\cr1,2 \end{matrix}\bigg)\begin{matrix} 1/2\cr v,-v \end{matrix}\bigg] \end{align} Step 3: Expressing the above results from "Shah, M.: ’On generalization of some results and their applications’ Collectanea Mathematica, 1973, 24, (3), pp. 249-266" \begin{align} S\bigg[x,2\alpha\sqrt{ab}x\bigg\vert\bigg[\begin{matrix} 0,0\cr0,0 \end{matrix}\bigg]\begin{matrix} -\cr- \end{matrix}\bigg\vert\bigg(\begin{matrix} 1,2\cr2,2 \end{matrix}\bigg)\begin{matrix} 1,1\cr1,0 \end{matrix}\bigg\vert\bigg(\begin{matrix} 2,0\cr1,2 \end{matrix}\bigg)\begin{matrix} 1/2\cr v,-v \end{matrix}\bigg]=&S\left[\begin{matrix}\label{shaheq} \left[\begin{matrix} 0,0\cr0,0 \end{matrix}\right]\cr\left(\begin{matrix} 2,1\cr0,1 \end{matrix}\right)\cr\left(\begin{matrix} 0,2\cr1,0 \end{matrix}\right) \end{matrix}\left\vert\begin{matrix} -;-\cr(1,1);(1,0)\cr(1/2);(v,-v) \end{matrix}\right\vert\begin{matrix} x\cr2\alpha\sqrt{ab}x \end{matrix} \right] \end{align}

Step 4: Using an identity from "Shah, M.: ’On generalization of some results and their applications’ Collectanea Mathematica, 1973, 24, (3), pp. 249-266"

\begin{align} I&=\int_{0}^{\infty}x\exp(-\beta x)S\left[\begin{matrix} \left[\begin{matrix} 0,0\cr0,0 \end{matrix}\right]\cr\left(\begin{matrix} 2,1\cr0,1 \end{matrix}\right)\cr\left(\begin{matrix} 0,2\cr1,0 \end{matrix}\right) \end{matrix}\left\vert\begin{matrix} -;-\cr(1,1);(1,0)\cr(1/2);(v,-v) \end{matrix}\right\vert\begin{matrix} x\cr2\alpha\sqrt{ab}x \end{matrix} \right]dx\\ &=\frac{1}{\beta^2}S\left[\begin{matrix} \left[\begin{matrix} 1,0\cr1,0 \end{matrix}\right]\cr\left(\begin{matrix} 2,1\cr0,1 \end{matrix}\right)\cr\left(\begin{matrix} 0,2\cr1,0 \end{matrix}\right) \end{matrix}\left\vert\begin{matrix} 2;-\cr(1,1);(1,0)\cr(1/2);(v,-v) \end{matrix}\right\vert\begin{matrix} \frac{1}{\beta}\cr\frac{2\alpha\sqrt{ab}}{\beta} \end{matrix} \right]\\ &=\frac{1}{\beta^2}G_{1,0:2,2:1,2}^{1,0:1,2:2,0}\Bigg(2\Bigg\vert\begin{matrix} 1,1\cr 1,0\end{matrix}\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\Bigg\vert\frac{1}{\beta},\frac{2\alpha\sqrt{ab}}{\beta}\Bigg) \end{align}

That completes the proof.

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