Let $B_R(0)$ be a ball in $\mathbb R^3 $ and define
$$u(x)=\int_{B_R(0)}\frac{1}{|y-x|}dy$$
Prove that
$$ u(x) = \begin{cases} \frac{2}{3}\pi(3R^2-|x|^2) & \quad \text{for $0 \le|x|\le R$ }\\[8pt] \frac{4}{3}\pi \frac{R^3}{|x|} & \quad \text{for $|x|>R$} \end{cases}$$
My attempt:
For $|x|>R$, $u$ is harmonic and it goes to zero as $|x|$ goes to infinity, so we can use the fundamental solution $\frac{c}{|x|}$, c is some constant. By the continuity of $u(x)$, we get $u(x)=\frac{4}{3}\pi \frac{R^3}{|x|}$.
For $|x| \le R$, I found that the function $w(x)=\frac{2}{3}\pi(3R^2-|x|^2)$ solves $\Delta w=-4\pi$, and $w(x)=\frac{4}{3}\pi R^3$ on $\partial B_R(0)$. Therefore I can use the Green's representatio formula in a ball. So I get something like this, when $n=3$:
$$u(x)=\frac{R^2-|x|^2}{R^2}\int _{\partial B_R(0)}\frac{1}{|x-y|^3}ds(y)+\frac{3}{R^3}\int_{B_R(0)}\frac{1}{|x-y|}-\frac{R|y|}{|y|^2x-R^2y}\,dy,$$
then I don't know what to do next to get the desired equation $u(x)=\int_{B_R(0)}\frac{1}{|y-x|}dy$.
This is really bothering me, your help will be appreciated!:)