Let $\mathcal{L}_I(s)$ be the Laplace transform of $I$ which is given by
$\mathcal{L}_I(s)=\left(\frac{2}{r_d^2-r^2}\int_r^{r_d}\mathbb{E}_H\left[r\exp\left(-sHr^{-\eta}\right)\right]dr\right)^k$
$\mathcal{L}_I(s)=\left(\frac{2}{r_d^2-r^2}\int_r^{r_d}\mathcal{L}_H(sr^{-\eta})dr\right)^k=\left(\frac{2}{r_d^2-r^2}\int_r^{r_d}\frac{r}{(1+sr^{-\eta})}dr\right)^k$
Here, $\mathbb{E}_H$ defines the expectation over variable $H$.
It is said that when $H$ is exponentially distributed with mean 1, (i.we., $\exp(-x)$), $\mathcal{L}_I(s)$ can be expressed as
$\mathcal{L}_I(s)=\left(\frac{\Phi(\eta,s,r_d)-\Phi(\eta,s,r)}{r_d^2-r^2}\right)^k$
where,
$\Phi(\eta,s,x)=x^2-x^2\hspace{1mm} _2F_1\left(1,\hspace{1mm}2/\eta,\hspace{1mm}1+2/\eta,\hspace{1mm} -x^{\eta}/s\right)$
Here, $_2F_1$ is the Gauss Hypergeometric function.
How to prove it?