Calculation double Integral over Ball (optical size) I hope that someone can help me with the following problem.
I have to show that
$$\int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy=4\pi^2~,$$
with $B_1(0)\subset\mathbb{R}^3$.
I have no idea how to calculate those integrals, the common(for me) tricks won't help.
 A: The integral is connected with the mean inverse squared distance between two points within a unit ball.
Following the method used by Christian Blatter in this post we have that,
$$\begin{align*}
\int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy
&=|B_1(0)|^2
\int_0^1\int_0^1\int_0^\pi \frac{f_R(r) f_S(s)f_\Theta(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\
&=\left(\frac{4\pi}{3}\right)^2
\int_0^1\int_0^1\int_0^\pi \frac{3r^2\cdot 3s^2\cdot1/2\sin(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\
&=8\pi^2
\int_0^1\int_0^1r^2 s^2\left(\int_0^\pi \frac{\sin(\theta) \ d\theta}{r^2+s^2-2rs\cos\theta}\right)\ ds\ dr\\
&=4\pi^2
\int_0^1r\left(\int_0^1 s\ln\left(\frac{(r+s)^2}{(r-s)^2}\right)\ ds\right)\ dr\\
&=4\pi^2
\int_0^1r\left(\ln\left(\frac{1+r}{1-r}\right)(1-r^2)+2r\right)\ dr\\
&=4\pi^2.
\end{align*}$$
and we are done.
A: I have a solution for 
$$
\frac{1}{2} \int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|}dxdy=\frac{4 \pi^2}{3} \frac{4}{5} 
$$
By Newton theorem, you have for every spherical symmetric function g (so g(y)=f(|y|):
$$
\int \frac{g(y)}{|x-y|}d y = \int_0^\infty \frac{f(r)4\pi r^2}{max\{|x|,r\}} d r
$$
With this in mind and the spherically transformation, you get 
$$
\frac{1}{2} \int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|}dxdy = \frac{1}{2} \int_{B_1(0)}\int_0^1 \frac{4\pi r^2}{max\{|x|,r\}} d r \ d x \\
= \frac{1}{2} \int_0^1 4\pi t^2 \int_0^1 \frac{4\pi r^2}{max\{t,r\}} d r \ d t = \frac{1}{2} 4^2 \pi^2\int_0^1 t^2 \bigg(\int_0^t r^2 \frac{1}{t} dr + \int_t^1 r \ dr\bigg) dt=\frac{4 \pi^2}{3} \frac{4}{5} 
$$
A: Let us recall the Laplace's multipole expansion formula into Legendre polynomials:
$$\frac{1}{|\vec{r}'-\vec{r}|}=\frac{1}{r}\sum_{n=0}^\infty \left(\frac{r'}{r}\right)^n P_n(Y)\qquad ;r\geq r',$$
where I have denoted $Y:=\hat{r}\bullet\hat{r}'$. Recall as well their orthogonality relation
$$\oint P_n(Y)P_m(Y) d\Omega = \frac{4\pi}{2n+1} \delta_{nm}.$$
Denote our integral $L$, by symmetry, we can write
$$I = \int\int \frac{dV'dV}{|\vec{r}'-\vec{r}|^2} = 2\int\int \theta(r-r')\frac{dV'dV}{|\vec{r}'-\vec{r}|^2}.$$
Expanding $1/|\cdot|$ into series (then squared), we get
$$I=2\sum_{nm}\int_0^1\int_0^r \frac{1}{r^2} \left(\frac{r'}{r}\right)^n \left(\frac{r'}{r}\right)^m P_n(Y) P_m(Y) r'^2 r^2 d\Omega'd\Omega dr' dr.$$
By orthogonality, we get immediatelly
$$I=32\pi^2\sum_{n=0}^\infty\int_0^1\int_0^r \frac{1}{r^2}\left(\frac{r'}{r}\right)^{2n} \frac{1}{2n+1} r'^2 r^2 dr' dr=8\pi^2\sum_{n=0}^\infty\frac{1}{(2n+1)(2n+3)}.$$
By partial fraction decomposition and due to telescopicity of the resulting sum, we get
$$I=4\pi^2\sum_{n=0}^\infty\frac{1}{2n+1}-\frac{1}{2n+3}=4\pi^2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots\right)=4\pi^2$$
Note: that the partial result in terms of only $r'$ and $r$ with $\sum_n$ can be summed up into the integral given by @Robert Z
A: I assume that $B_1(0)$ is the ball of radius $1$ about the origin.
Since this space has nice spherical symmetry, it will probably be
easier to convert to spherical coordinates. Let our change of coordinate
map be
$$ H(\rho,\theta,\phi)
   = (\rho\cos\theta\cos\phi,
      \rho\sin\theta\cos\phi,
      \rho\sin\phi), \qquad
      \rho \in [0,1], \theta \in [0,2\pi], \phi \in [-\pi/2,\pi/2] $$
so we wish to integrate
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over |r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2}
   \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$
Now absolute value is usually defined as $|x|=\sqrt{x^2}$, so we have
$$ \begin{align}
|r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2
   &= r^2\cos^2\theta\cos^2\phi
      + r^2\sin^2\theta\cos^2\phi-2r^2\cos^2\phi\sin\theta\cos\theta \\\
   &= r^2\cos^2\phi(1- 2\sin\theta\cos\theta)
\end{align}$$
Then our integral is
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over r^2\cos^2\phi(1- 2\sin\theta\cos\theta)}
   \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$
easily simplifying to
$$ \int_{-\pi \over 2}^{\pi \over 2}
   \int_0^{2\pi}\!\!\!\int_0^1
   {1 \over \cos\phi}{1 \over (1- 2\sin\theta\cos\theta)}
   \thinspace drd\theta d\phi $$
Clearly $r$ integrates out. By the standard tricks,
$$ \int {1 \over \cos x}\thinspace dx = \log(\tan x + \sec x) $$
and
$$ \int {1 \over 1-2\sin x \cos x} = {\sin x \over \cos x - \sin x} $$
So the antidifferentiation results in
$$ \begin{align}
   &\log(\tan\phi-\sec\phi)\bigg|_{-\pi \over 2}^{\pi \over 2}
   \times {\sin \theta \over \cos \theta - \sin\theta}\bigg|_0^{2\pi} \\
\end{align} $$
which is not defined. Hence the integral does not exist.
