The approximate ground state for the two electrons of the He atom is given by $$\psi_0(r_1,r_2) = \frac{8}{\pi a^3}e^{-2(r_1+r_2)/a}$$, where $a$ is the Bohr radius.
I want to calculate the expectation value $\big\langle \frac{1}{|\mathbf r_1-\mathbf r_2|} \big\rangle$, but I really don't see how to proceed. The integrand seems to be correct.
We have
\begin{align} \require{cancel} \bigg\langle \frac{1}{|r_1-r_2|} \bigg\rangle &= \bigg( \frac{8}{\pi a^3} \bigg)^2 \cancel{\int_{r_1=0}^{\infty}\int_{r_2=0}^{\infty} \frac{1}{\sqrt{(r_1-r_2)^2}} e^{-4(r_1+r_2)/a} dr_1 dr_2} \\ &= \ldots \end{align}
Any hints would be appreciated.
The integrals $$\int xe^{-\alpha x}dx = e^{-\alpha x}\bigg( -\frac{x}{\alpha} - \frac{1}{\alpha} \bigg)$$ and $$ \int x^2 e^{-\alpha x} dx = e^{-\alpha x} \bigg( -\frac{x^2}{\alpha}-\frac{2x}{\alpha^2} - \frac{2}{\alpha^3} \bigg) $$ are given as additional information.
As @pawel_winzig pointed out, we need to integrate over all space:
\begin{align} \bigg\langle \frac{1}{|\mathbf r_1-\mathbf r_2|} \bigg\rangle &= \bigg( \frac{8}{\pi a^3} \bigg)^2 \int_{\mathbf{r}_1 \in \mathbb R^3} \int_{\mathbf{r}_2 \in \mathbb R^3} \frac{1}{\sqrt{(\mathbf r_1-\mathbf r_2)^2}} e^{-4(r_1+ r_2)/a} d\mathbf r_1 d \mathbf r_2 \end{align}