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I'm trying to evaluate the expression:

$\mathbb{E}_h\{h^{0.5}\gamma(0.5, shc)\}$

where, $\gamma(.)$ is a lower incomplete gamma function given by $\gamma(a, z) = \int_{0}^{z}exp(-t)t^{a-1}dt$ and $h$ is an exponential random variable. A couple of steps in my evaluation are as follows:

$\mathbb{E}_h\{h^{0.5}\gamma(0.5, shc)\}$

= $\int_{0}^{\infty}e^{-h}h^{0.5}\int_{0}^{shc}exp(-t)t^{-0.5}dt\,dh$. Here, the inner integral gives me $\sqrt{\pi}\,erf(\sqrt{shc})$. But, I'm unable to find a way to proceed forward. Any pointers? Thanks.

Update:

I wanted to make add more details to make things clearer. This expression appears in a monograph and apparently this result simplifies as follows:

$\mathbb{E}_h\{h^{0.5}\gamma(0.5, shc)\} = \frac{\pi}{2}-arctan\left(\frac{1}{\sqrt{sc}}\right)+\frac{\sqrt{sc}}{sc+1}$ .

However, I'm unable to see how.

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  • $\begingroup$ To start with, Mathematica gives an expression in terms of a hypergeometric $_2 F_2$ $$\frac{1}{30} \left(-\frac{20 \Gamma \left(\frac{5}{4}\right) \, _2F_2\left(\frac{3}{4},\frac{5}{4};\frac{1}{2},\frac{7}{4};\frac{1}{4 c^2 s^2}\right)}{\left(c^2 s^2\right)^{3/4}}+\frac{12 \Gamma \left(\frac{7}{4}\right) \, _2F_2\left(\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{9}{4};\frac{1}{4 c^2 s^2}\right)}{\left(c^2 s^2\right)^{5/4}}+15 \pi \right)$$..do you need a derivation or just the result? $\endgroup$ – Pierpaolo Vivo Jan 18 '16 at 21:58
  • $\begingroup$ Thanks for your comment. I added more info. to make things clearer. I am trying to do the derivation. $\endgroup$ – ariadnus Jan 18 '16 at 22:07
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Change variables in $\int_{0}^{\infty}e^{-h}h^{0.5}\int_{0}^{shc}\exp(-t)t^{-0.5}dt\,dh$ as $t = shc z$, resulting in $$ sc (sc)^{-1/2}\int_{0}^{\infty}e^{-h}h\int_{0}^{1}\exp(-shc z)\frac{1}{\sqrt{z}}dz\,dh $$ and exchanging the order of integration $$ \sqrt{sc}\int_0^1 dz\frac{1}{\sqrt{z}} \int_0^\infty dh e^{-h (1+s c z)}h=\sqrt{sc}\int_0^1 dz\frac{1}{\sqrt{z}} \frac{1}{(1+csz)^2}=\sqrt{c s} \left(\frac{1}{c s+1}+\frac{\tan ^{-1}\left(\sqrt{c s}\right)}{\sqrt{c s}}\right) $$

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