I wonder if the self-convolution of $$ f(\vec{r}) = \frac{e^{-(x^2+y^2)}}{r^2}, $$ (where $\vec{r} = (x,y,z)$, and $r^2 = x^2+y^2+z^2$), $$ (f*f)(\vec{r}') = \iiint_{-\infty}^\infty d^3\vec{r}\, f(\vec{r}) f(\vec{r}'-\vec{r}), $$ can be evaluated or simplified in any way?
Thanks.
Edit:
Using the identity $$ \frac{1}{A} = \int_0^\infty d\nu\, e^{-A\nu} \text{ for } A>0, $$ the integration dimension can be decreased by 1: $$ \begin{align*} \iiint_{-\infty}^\infty d^3\vec{r}\, f(\vec{r}) f(\vec{r}'-\vec{r}) &= \iiint_{-\infty}^\infty d^3\vec{r}\, \frac{e^{-x^2-y^2}}{r^2} \frac{e^{-(x'-x)^2-(y'-y)^2}}{|\vec{r}' - \vec{r}|^2} \\ &= \iint_0^\infty d^2\vec{\nu}\iiint_{-\infty}^\infty d^3\vec{r}\, e^{-x^2-y^2-(x'-x)^2-(y'-y)^2- r^2\nu_1 - |\vec{r}' - \vec{r}|^2 \nu_2} \\ &= \pi^{3/2} \iint_0^\infty d^2\vec{\nu} \frac{\exp \left(-\frac{(\nu_1+1) (\nu_2+1) (x'^2+y'^2)}{\nu_1+\nu_2+2}-\frac{\nu_1 \nu_2 z'^2}{\nu_1+\nu_2}\right)}{(\nu_1+\nu_2+2)\sqrt{\nu_1+\nu_2}}. \end{align*} $$
Can this integral be further reduced?