$$I=\int d^{3}{\vec{r_{1}}} \frac{\exp\bigg[-2\alpha\mid\vec{r_{1}}-\vec{R_{2}}\mid^{2}\bigg]}{\mid\vec{r_{1}}-\vec{R_{1}}\mid}$$
Where $\alpha$ is a real positive number and constant. $\vec{R_{1}}$ and $\vec{R_{2}}$ are constant vectors . Integral is over the full space (i.e) $$-\infty<x_{1}<\infty$$ $$-\infty<y_{1}<\infty$$ $$-\infty<z_{1}<\infty$$
Can someone suggest a trick to evaluate this integral ? Is it exactly integratble?
This answer is just to propose a trick to rewrite the integral as this was asked by OP.
Let us observe that $$ \frac{1}{|{\bf r_1}-{\bf R}_1|}=\int\frac{d^3q}{(2\pi)^3}e^{i{\bf q}\cdot({\bf r_1}-{\bf R}_1)}\frac{1}{q^2} $$ and $$ e^{-2\alpha|{\bf r_1}-{\bf R}_2|^2}=\sqrt{\frac{8\alpha}{\pi}}\int_{-\infty}^\infty dk e^{ik|{\bf r_1}-{\bf R}_2|}e^{-\frac{k^2}{8\alpha}}. $$ Then, the integral takes the form $$ \sqrt{\frac{8\alpha}{\pi}}\int\frac{d^3q}{(2\pi)^3}\frac{1}{q^2} \int_{-\infty}^\infty dke^{-\frac{k^2}{8\alpha}} \int d^3r_1 e^{i{\bf q}\cdot({\bf r_1}-{\bf R}_1)}e^{ik|{\bf r_1}-{\bf R}_2|}. $$ As suggested in the comments, this integral can become singular and is diverging for ${\bf r}_1={\bf R}_1$. This reflects the singularity at the origin of the 1/r potential. Far from the singularity, this is amenable to a multipolar expansion.