I'm trying to analytically find the following expectation


$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right],$$
where $a$ and $b$ are constant values, $\mathcal{Q}$ is the Gaussian Q-function, which is defined as $\mathcal{Q}(x) = \frac{1}{2 \pi}\int\_{x}^{\infty} e^{-u^2/2}du$ and $\gamma$ is a random variable with Gamma distribition, i.e., $f\_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta} $.


By using Mathematica, I've found the following solution:


$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right] = a 2^{-\frac{\kappa }{2}-3} b^{-\frac{\kappa }{2}-\frac{1}{2}} \theta ^{-\kappa -1} \left(2 \sqrt{2} \sqrt{b} \theta \, \_2\tilde{F}\_2\left(\frac{\kappa +1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa +2}{2};\frac{1}{2 b \theta ^2}\right)-\kappa \, \_2\tilde{F}\_2\left(\frac{\kappa +1}{2},\frac{\kappa +2}{2};\frac{3}{2},\frac{\kappa +3}{2};\frac{1}{2 b \theta ^2}\right)\right),$$


however, I'd like to know the steps to find this solution or to find another one.



Finding the expectation of $a \mathcal{Q} \left( \sqrt{b } \gamma \right) $, where $\gamma$ is a Gamma r.v
$\newcommand\Ga\Gamma$
Without loss of generality $a=1$. Let then $Q:=\mathcal Q$, $k:=\kappa>0$, and $t:=\theta\sqrt b>0$, so that $\sqrt b\,\gamma$ has the gamma distribution with parameters $k,t$. Let also $c:=\Ga(k)t^k$. Then, letting $f$ denote the standard normal pdf, we have  

$$c\,EaQ(\sqrt b\,\gamma)=\int\_0^\infty dy\,y^{k-1} e^{-y/t}Q(y) 
=\sum\_{j=0}^\infty\frac{(-1/t)^j}{j!}I\_j,\tag{1}$$
where
$$I\_j:=\int\_0^\infty dy\,y^{k+j-1}Q(y)=\frac1{k+j}\,\int\_0^\infty dy\,y^{k+j}f(y) \\
=\frac{2^{(j+k)/2-1} \Ga((j+k+1)/2)}{(k+j)\sqrt{\pi }};\tag{2}$$
the penultimate equality here is obtained by integration by parts and the last equality is obtained by the substitution $w=y^2/2$. Using now the identity $\Ga(x+1)=x\Ga(x)$ about $j/2$ times to reduce $\Ga((j+k+1)/2)$ to $\Ga((k+1)/2)$ for even $j$ and to $\Ga(k/2)$ for odd $j$, from (1) and (2) we get
$$c\_1\,EQ(\sqrt b\,\gamma) \\
=
\sqrt{2} (k+1) t \Gamma \left(\frac{k+1}{2}\right) \,
 \_2F\_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2};\frac{1}{2},\frac{k}{2}+1;\frac{1}{2
 t^2}\right) \\ 
-k^2 \Gamma \left(\frac{k}{2}\right) \,
 \_2F\_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2}+1;\frac{3}{2},\frac{k}{2}+\frac{3}{2};\frac{1}{2 t^2}\right),$$
where $c\_1:=c\,\sqrt{\pi } k (k+1) t\,2^{(3-k)/2}$.


