Integrate 1/|r-r'| in 3 dimensions

Question

I would like to find the volume dependence of the following integrale $$ I(V) = \int_V \text d^3 r \int_V \text d^3 r' \frac 1 {|\vec r - \vec r' |} $$ Here $V$ is just a sphere in $\mathbb R^3$ centered around the origin. I am sure that this is not so difficult with the right tool. But a simple substitution $ \vec r'' = \vec r' -\vec r$ does not help. Can someone help me out?

Answer 1

For $V=B(r)$, with the usual parameterisation this is $|B(1)|^2$ times \begin{align} \int_0^r\int_0^r \frac{x^2y^2}{|x-y|} \ dxdy &= 2\int_{x=0}^r\int_{y>x}\frac{x^2y^2}{y-x}\ dydx\\ &= 2 \int_0^r x^2\int_{s=0}^{r-x}\frac{s^2+2sx+x^2}{s} \ dxds \\ &= 2 \int_0^r x^2\int_{s=0}^{r-x} s+2x+\color{red}{ \frac{x^2}{s}} \ dxds \end{align} The term in red doesn't integrate well at $s=0,x≠0$.