How to integrate $\int \frac{e^{-x}e^{-y}}{\lvert \vec x - \vec y \rvert} dx dy$ for the helium atom?

Question

What methods are used to integrate

$$\int \int \frac{e^{-x}e^{-y}}{\lvert \vec x - \vec y \rvert} dx dy$$

which comes up in perturbation theory calculations such as for the helium atom (8.2.6). The integration process itself is often left out, skipping to "in the end we get the answer 34.0". Could anyone show/outline the procedure to perform this integration?

Answer 1

OK, pursuant to the discussion in the comments, here's the solution. We will assume, on physical grounds, that $|x-y|\ge\delta>0$ when we get to the end.

Switching to polar coordinates, we have $x=r\cos\theta$, $y=r\sin\theta$, $r>0$, $0\le\theta\le\pi/2$. Then \begin{align*} \int_0^\infty\int_0^\infty \frac{e^{-(x+y)}}{|x-y|}dx\,dy &= \int_0^{\pi/2}\int_0^\infty \frac{e^{-r(\cos\theta+\sin\theta)}}{r|\cos\theta-\sin\theta|} r\,dr\,d\theta \\ &= \int_0^{\pi/2}\int_0^\infty \frac{e^{-r(\cos\theta+\sin\theta)}}{|\cos\theta-\sin\theta|} \,dr\,d\theta \\ &= 2\int_0^{\pi/4}\int_0^\infty \frac{e^{-r(\cos\theta+\sin\theta)}}{\cos\theta-\sin\theta} \,dr\,d\theta. \end{align*}

Doing the inside integral (with $\int_0^\infty e^{-ar}dr = \frac1a$), we end up with $$2\int_0^{\pi/4} \frac1{\cos^2\theta-\sin^2\theta}d\theta = 2\int_0^{\pi/4}\sec(2\theta)\,d\theta = 2\int_0^{\pi/2}\sec u\,du.$$ Since $\sec u\to\infty$ as $u\to\pi/2^-$, we suspect that the integral will diverge. If we bound $u$ away from $\pi/2$ (which is bounding $\theta$ away from $\pi/4$, where $x=y$), then we'll have \begin{multline*} 2\int_0^{\pi/2-\epsilon}\sec u\,du = 2\ln(\sec u+\tan u)\Big|_0^{\pi/2-\epsilon} = 2\ln\left(\tfrac{1+\sin\theta}{\cos\theta}\right)\Big|_0^{\pi/2-\epsilon} \\ = 2\ln\left(\tfrac{1+\cos\epsilon}{\sin\epsilon}\right) \approx 2\ln (2/\epsilon).\end{multline*}

Answer 2

Summarizing from this and that derivation,

$$\int \frac{e^{ar_1}e^{ar_2}}{\lvert \vec r_1 - \vec r_2 \rvert}$$

$$=\int_0^\infty \color{orange}{\int_0^\pi} \color{red}{\int_0^{2\pi}} e^{ar_1}r_1^2 dr_1 \color{orange}{sin\theta_1 d\theta_1} \color{red}{d\phi_1} \int_0^\infty \int_0^\pi \color{red}{\int_0^{2\pi}}\frac{1}{\lvert \vec {r1} - \vec {r2} \rvert}e^{ar_2}r_2^2\sin\theta_2dr_2d\theta_2 \color{red}{d\phi_2}$$ $$=\color{red}{(2\pi)^2}\color{orange}{(2)}\int_0^\infty e^{ar_1}r_1^2 dr_1 \int_0^\infty \int_0^\pi \frac{1}{\lvert \vec {r1} - \vec {r2} \rvert}e^{ar_2}r_2^2\sin\theta_2dr_2d\theta_2$$

since

$$\color{red}{\int_0^{2\pi} d\phi = 2\pi}$$ $$\color{orange}{\int_0^\pi \sin\theta_1 d\theta_1 = 2}$$

So

$$=\color{red}{(2\pi)^2}\color{orange}{(2)}\int_0^\infty r_1^2 e^{ar_1} dr_1 (\color{green}{\int_0^{r_1}} \int_0^\pi \color{green}{\frac{1}{\vec {r1} - \vec {r2}}}r_2^2e^{ar_2}\sin\theta_2\color{green}{dr_2}d\theta_2 + \color{green}{\int_{r_1}^\infty} \int_0^\pi \color{green}{\frac{1}{\lvert \vec {r1} - \vec {r2} \rvert}}r_2^2e^{ar_2}\sin\theta_2\color{green}{dr_2}d\theta_2)$$

$$=\color{red}{(2\pi)^2}\color{orange}{(2)}\int_0^\infty r_1^2 e^{ar_1} dr_1 (\int_0^{r_1} \frac{1}{r_1} \color{green}{\int_0^\pi{\sin\theta_2\sum_{l=0}^\infty (\frac{r_2}{r_1})^nP_l(\cos\theta_2)d\theta_2}}r_2^2e^{ar_2}dr_2 + \int_{r_1}^\infty \frac{1}{r_2} \color{green}{\int_0^\pi\sin\theta_2\sum_{l=0}^\infty (\frac{r_1}{r_2})^nP_l(\cos\theta_2)d\theta_2}r_2^2e^{ar_2}dr_2)$$

where $P_l$ are the Legendre polynomials,

$$=\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}\int_0^\infty r_1^2 e^{ar_1} dr_1 \left(\int_0^{r_1} \frac{1}{r_1} r_2^2e^{ar_2}dr_2 + \int_{r_1}^\infty \frac{1}{r_2} r_2^2e^{ar_2}dr_2\right)$$

since

$$ \color{green}{\int_0^\pi{\sin\theta_2\sum_{l=0}^\infty (\frac{r_2}{r_1})^nP_l(\cos\theta_2)d\theta_2} = 2}$$ $$ \color{green}{\int_0^\pi\sin\theta_2\sum_{l=0}^\infty (\frac{r_1}{r_2})^nP_l(\cos\theta_2)d\theta_2 = 2}$$

Changing variables as

$$ ar_1 = \bar r_1$$ $$ ar_2 = \bar r_2$$

We obtain

$$=\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}\int_0^\infty \frac{\bar r_1^2}{\color{brown}{a^2}} e^{\bar r_1} \frac{1}{\color{brown}{a}}d\bar r_1 \left(\int_0^{r_1} \frac{\color{brown}{a}}{\bar r_1} \frac{\bar r_2^2}{\color{brown}{a^2}}e^{\bar r_2}\frac{1}{\color{brown}{a}}d\bar r_2 + \int_{r_1}^\infty \frac{\bar r_2}{\color{brown}{a}}e^{\bar r_2}\frac{1}{\color{brown}{a}}d\bar r_2\right)$$

$$=\frac{\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}}{\color{brown}{a^5}}\int_0^\infty \bar r_1^2 e^{\bar r_1} d\bar r_1 \left({\frac{1}{\bar r_1}\color{purple}{\int_0^{r_1} \bar r_2^2e^{\bar r_2}d\bar r_2}} + \color{blue}{\int_{r_1}^\infty \bar r_2e^{\bar r_2}d\bar r_2}\right)$$

$$=\frac{\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}}{\color{brown}{a^5}}\int_0^\infty \bar r_1^2 e^{\bar r_1} \left({\frac{1}{\bar r_1}\color{purple}{(-\bar r_1^2 e^{-\bar r_1} - 2\bar r_1 e^{-\bar r_1} - 2e^{-\bar r_1} + 2)}} + \color{blue}{\bar r_1e^{-\bar r_1} + e^{-\bar r_1}}\right)d\bar r_1$$

$$=\frac{\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}}{\color{brown}{a^5}}\int_0^\infty \left(-\color{red}{\bar r_1^2e^{-2\bar r_1}} - \color{orange}{2 \bar r_1e^{-2\bar r_1}} + \color{green}{2\bar r_1e^{-\bar r_1}} \right)d\bar r_1$$

But

$$ \color{red}{\int_0^\infty\bar r_1^2e^{-2\bar r_1} d\bar r_1= \frac{1}{4}} $$ $$ \color{orange}{\int_0^\infty\bar r_1e^{-2\bar r_1} d\bar r_1= \frac{1}{4}}$$ $$ \color{green}{\int_0^\infty\bar r_1e^{-\bar r_1} d\bar r_1= 1}$$

Therefore

$$=\frac{\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}}{\color{brown}{a^5}}\left(-\frac{1}{4} - 2\left(\frac{1}{4}\right) + 2(1) \right)$$ $$=\frac{\color{red}{(2\pi)^2}\color{orange}{(2)}\color{green}{(2)}}{\color{brown}{a^5}}\color{purple}{\left(\frac{5}{4}\right)}$$ $$=\frac{20\pi^2}{a^5}$$