Proof of Gradshteyn & Ryzhik, ed. 8, #7.522.1

Question

Looking for a proof to formula 7.522.1 in Gradshteyn & Ryzhik (8th edition) which states \begin{equation} \int_{0}^{\infty}e^{-\lambda x}x^{\gamma-1}{_{2}F_1}(\alpha,\beta;\delta;-x)\,\mathrm{d}x=\frac{\Gamma(\delta)\lambda^{-\gamma}}{\Gamma(\alpha)\Gamma(\beta)}E(\alpha\,;\beta:\gamma\,;\delta:\lambda), \end{equation} for $\Re \lambda>0$ and $\Re \gamma>0$ where $E(p\,;\alpha_{r}:q\,;\varrho_{s}:x)$ is the MacRobert E-Function.

Update 4/13/18

This gets it started: \begin{equation} \int_{0}^{\infty}e^{-\lambda x}x^{\gamma-1}{_{2}F_1}(\alpha,\beta;\delta;-x)\,\mathrm{d}x =\frac{\Gamma(\delta)}{\Gamma(\alpha)\Gamma(\beta)} \int_{0}^{\infty}x^{\gamma-1}e^{-\lambda x}G^{1,2}_{2,2}\left(x\,\bigg|{1-\alpha,1-\beta \atop 0,1-\delta}\right)\,\mathrm{d}x. \end{equation} This is in the form of formula 7.813.1 in Gradshteyn & Ryzhik (8th edition). If all the conditions are met (and there are a lot of them) then \begin{equation} \int_{0}^{\infty}e^{-\lambda x}x^{\gamma-1}{_{2}F_1}(\alpha,\beta;\delta;-x)\,\mathrm{d}x =\frac{\Gamma(\delta)\lambda^{-\gamma}}{\Gamma(\alpha)\Gamma(\beta)} G^{1,3}_{3,2}\left(\frac{1}{\lambda}\,\bigg|{1-\gamma,1-\alpha,1-\beta \atop 0,1-\delta}\right). \end{equation}

Answer 1

You might have copied 7.522.1 incorrectly. In the 7th edition of G&R, the rhs is $$\frac {\Gamma(\delta) \lambda^{-\gamma}} {\Gamma(\alpha) \Gamma(\beta)} E(\alpha, \beta, \gamma : \delta : \lambda).$$ Then, using the formula for the relation of the E-function with the G-function and the reflection formula for the G-function, $$E(\alpha, \beta, \gamma : \delta : \lambda) = G_{2,3}^{3,1} \left( \lambda \middle| {1, \delta \atop \alpha, \beta, \gamma} \right) = G_{3,2}^{1,3} \left( \frac 1 \lambda \middle | {1 - \alpha, 1 - \beta, 1 - \gamma \atop 0, 1 - \delta} \right),$$ in agreement with 7.813.1.

This is a special case of the general formula for the integral of $x^p$ times a product of two linear G-functions.