How to integrate two-electron interaction in $\rm He$? (variational method) A probe wavefunction in the variational method is $$\psi(r_1, r_2) =\frac{\alpha^6}{\pi^2}e^{-\alpha(r_1+r_2)}.$$ In $\left<\psi \right|H\left|\psi\right>$ with $$H = \frac{p_1^2+p_2^2}{2m} - \frac{Ze^2}{r_1}-\frac{Ze^2}{r_2}+\frac{e^2}{|\vec{r_1}-\vec{r_2}|}$$
the last term is to be integrated like that: $$\idotsint_{} \frac{\left|\psi\right|^2 e^2}{|\vec{r_1}-\vec{r_2}|}r_1^2\, r_2^2\,\sin{\theta_1}\sin{\theta_2}\, d\theta_1 d\theta_2\,d \phi_1d\phi_2\,dr_1dr_2, $$
which is quite challenging for me. Does anyone know how to integrate it or some workaround method to solve this task?
PS
They may want me just to throw that term away, but I would nevertheless like to know is there a way to integrate such thing
 A: Of course, just after a day I've found the way to do it in two of my books at once
You should rewrite the denominator as follows:
$$\sqrt{r_1^2+r^2_2-2r_1r_2\cos{\theta}},$$
where $\theta$ is the angle between vectors. Then you fix Z axis collinear with $r_1$, for instance, and integrate the $\phi$-part, so you get something like this:
$$2\pi\int_0^\pi\frac{\sin\theta d\theta}{\sqrt{r_1^2+r^2_2-2r_1r_2\cos{\theta}}},$$
which is integrateable easily and gives an eye-pleasing
$$
\left\{
\begin{aligned}
\frac{4pi}{r_1}, r_1>r_2 \\
\frac{4pi}{r_2}, r_2>r_1.
\end{aligned}
\right.
$$
That's not very strict description, but it gives all the clues.
A: There are two ways to carry out this integral.
Method I
Write
$$
\frac{1}{|\vec{r}_1 - \vec{r}_2|} = \begin{cases}
\frac{1}{r_2}\sum_{n=0}^\infty\left(\frac{r_1}{r_2}\right)^nP_n(cos\theta) & \;\text{if}\; 0 < r_1  \le r_2 \\
\frac{1}{r_1}\sum_{n=0}^\infty\left(\frac{r_2}{r_1}\right)^nP_n(cos\theta) & \;\text{if}\; r_2 \le r_1  < \infty
\end{cases}
$$
Let us denote the integral by $I$ so that
$$
I = \frac{\alpha^6}{\pi^2}\iint \frac{e^{-2\alpha(r_1 + r_2)}}{r_{12}}dV_1 dV_2 =
\frac{\alpha^6}{\pi^2}\int e^{-2\alpha r_2}\left\{4\pi\left(\frac{1}{r_2}\int_0^{r_2}r_1^2 e^{-2\alpha r_1} dr_1 + \int_{r_2}^\infty r_1 e^{-2\alpha r_1}dr_1 \right)\right\}dV_2 
$$
All the integrals in the above expression are easy to evaluate.  think there is a typo in the OP's first equation. The coefficient of $\psi$ should be $\alpha^3/\pi$.
Method II If we ignore the constant $\alpha^6/\pi^2$, the rest of the integral can be viewed as the energy of a charge configuration with density $e^{-2\alpha r_1}$ at $\vec{r}_1$ and $e^{-2\alpha r_2}$ at $\vec{r}_2$. We can evaluate the integral by first computing the potential due to one of the distributions and then calculating the energy of the other distribution in the former's field. A spherical shell of radius $r_1$ and charge $4\pi r_1^2 e^{-2\alpha r_1}dr_1$ is
$$
dV(r) =\begin{cases}
4\pi r_1^2e^{-2\alpha r_1}dr_1 r_1^{-1} & \;\text{if}\; r < r_1 \\
4\pi r_1^2e^{-2\alpha r_1}dr_1 r^{-1} & \;\text{if}\; r > r_1.
\end{cases}
$$
Thus,
$$
V(r) = \frac{4\pi}{r}\int_0^r e^{-2\alpha r_1}r_1^2 dr_1 + 4\pi\int_r^\infty e^{-2\alpha r_1} r_1 dr_1.
$$
The energy of the charge configuration is
$$
I = \frac{\alpha^6}{\pi^2}\int V(r_2)e^{-2\alpha r_2}dV_2 = \int e^{-2\alpha r_2}\left\{4\pi\left(\frac{1}{r_2}\int_0^{r_2}r_1^2 e^{-2\alpha r_1} dr_1 + \int_{r_2}^\infty r_2 e^{-2\alpha r_1}dr_1 \right)\right\}dV_2 
$$
The two integrals are identical.
