The indefinite integral of $\frac{e^{-\frac{(\alpha +\beta )^2}{4 \alpha }}}{2 \sqrt{\pi } \sqrt{\alpha }}$ w.r.t. $\alpha$

Question

Mathematica tells me the indefinite integral is:

$$ \frac{1}{2} e^{-\frac{\sqrt{\beta ^2}}{2}-\frac{\beta }{2}} \left(e^{\sqrt{\beta ^2}} \left(\text{erf}\left(\frac{1}{2} \sqrt{\alpha } \left(\frac{\sqrt{\beta ^2}}{\alpha }+1\right)\right)-1\right)+\text{erf}\left(\frac{\sqrt{\alpha }}{2}-\frac{\sqrt{\beta ^2}}{2 \sqrt{\alpha }}\right)+1\right) $$

I have tried integration by parts, a few substitutions, to no avail. What trick must I use to integrate this?

Answer 1

I suppose that the first step is $\alpha=x^2$ $$I=\int\frac{e^{-\frac{(\alpha +\beta )^2}{4 \alpha }}}{2 \sqrt{\pi } \sqrt{\alpha }}\,d\alpha=\frac{1}{ \sqrt{\pi }}\int e^{-\frac{\left( x^2+\beta\right)^2}{4 x^2}}\,dx$$ and then $$\frac{\left( x^2+\beta\right)^2}{ x^2}=x^2+2 \beta +\frac{\beta ^2}{x^2}$$

Then, I am stuck ! I did not find anythink looking like this integrand in tables of integrals.