Hello I want to compute
$\int_0^\infty \sqrt xe^{-x/2} (\operatorname{erfc}(\sqrt{x/2}))^{m}dx $ where $\operatorname{erfc}(x)=1-\operatorname{erf}(x)$ with $\operatorname{erf}(x)$ the standard error function and $m$ is a fixed non-negative integer.
So far I tried substituting in a smart way but I could not find one. The same happens with pratial integration. My problem there is the $m$ in the exponent.
As already said in comments, very little hope for a closed form for $$I_m=\int \sqrt{x}\,e^{-x/2}\Bigg[ \text{erfc}\left(\sqrt{\frac{x}{2}}\right)\Bigg]^{m-1} \,dx$$ which could look nicer using $x=2y^2$ $$I_m=4 \sqrt{2}\int y^2\,e^{-y^2} \Big[\text{erfc}(y)\Big]^{m-1}\,dy$$ Using $$\text{erfc}(y)=1+\frac 2 {\sqrt \pi}\sum_{n=0}^\infty (-1)^{n+1}\frac{y^{2 n+1}}{(2 n+1)\, n!}$$ and using the binomial expansion of $\Big[\text{erfc}(y)\Big]^{m-1}$, we should face integrals $$J_k=\int y^{2+k}\,e^{-y^2}\,dy=-\frac{1}{2} \Gamma \left(\frac{k+3}{2},y^2\right)\qquad \text{with} \qquad k \text{ from }0 \text{ to } \infty$$
I tried for $m=4$ and integrated from $x=0$ to $x=2$ truncating the expansion to $O(y^{21})$.
The decimal representation of the result is $0.0746951$ while the numerical integration would give $I_4=0.0756388$ (relative error of $1.26$%).