How to evaluate the following integral may be related to the incomplete gamma function? \begin{align}
& \int_{0}^{\infty}y^{\nu-1}e^{-y}\int_{0}^{\infty}e^{-x}(\alpha x+y)^{\beta}dxdy\\
=& \int_{0}^{\infty}y^{\nu-1}e^{-y}(\alpha^{\beta}e^{\frac{y}{\alpha}}\Gamma(\beta+1,\frac{y}{\alpha}))dy,\quad (3)
\end{align}
where $\Gamma(\cdot,\cdot)$ represents the upper incomplete gamma function, i.e., $\Gamma(s,x)=\int_{x}^{\infty}t^{s-1}e^{-t}dt$.
If I can find a compact form of the above integral (3) ?
Thanks a lot, Liu.
 A: 
Given real parameters $\left(\beta,\nu,a\right)\in\mathbb{R}^{3}$, suppose
$$0<\beta+\nu+1\land0<\nu\land\left[0<a\lor\left(a=0\land0<\beta+\nu\right)\right].$$
These are the most general conditions for which the following integral is convergent and real-valued:
$$\mathcal{I}{\left(\beta,\nu,a\right)}=\int_{0}^{\infty}\mathrm{d}y\,y^{\nu-1}e^{-y}\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}.$$
Changing the order of integration and rescaling the integrals in the right way, the double integral in question turns out to be separable. We obtain
$$\begin{align}
\mathcal{I}{\left(\beta,\nu,a\right)}
&=\int_{0}^{\infty}\mathrm{d}y\,y^{\nu-1}e^{-y}\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}y^{\nu-1}e^{-y}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\left(ax+y\right)^{\beta}y^{\nu-1}e^{-x-y}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}t\,x\,\left(ax+xt\right)^{\beta}(xt)^{\nu-1}e^{-x-xt};~~~\small{\left[y=xt\right]}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}t\,\left(a+t\right)^{\beta}x^{\beta+\nu}t^{\nu-1}e^{-x(1+t)}\\
&=\int_{0}^{\infty}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}x\,t^{\nu-1}\left(a+t\right)^{\beta}x^{\beta+\nu}e^{-(1+t)x}\\
&=\int_{0}^{\infty}\mathrm{d}t\,\frac{t^{\nu-1}\left(a+t\right)^{\beta}}{\left(1+t\right)^{\beta+\nu+1}}\int_{0}^{\infty}\mathrm{d}u\,u^{\beta+\nu}e^{-u};~~~\small{\left[x=\frac{u}{1+t}\right]}\\
&=\Gamma{\left(\beta+\nu+1\right)}\int_{0}^{\infty}\mathrm{d}t\,\frac{t^{\nu-1}\left(a+t\right)^{\beta}}{\left(1+t\right)^{\beta+\nu+1}};~~~\small{\left[-1<\beta+\nu\right]}\\
&=\nu^{-1}\Gamma{\left(\beta+\nu+1\right)}\,{_2F_1}{\left(-\beta,1;\nu+1;1-a\right)},\\
\end{align}$$
where in the last line above we've made use of this integral representation variant for the Gauss hypergeometric function.

