compute the integral $\int_{\Omega}\frac{1}{\mid x-x_0\mid} dx$ Does someone know an easy way to compute the integral
$\frac{1}{R^24\pi}\int_{\Omega}\frac{1}{\mid x-x_0\mid} dx=\frac{1}{d}$, where $\Omega$ is the surface of a sphere of radius $R$, $x_0$ is a point on the surface or outside the sphere and $d$ is the distance between $x_0$ and the center of the sphere?
Thank you very much for any help!

We may assume that ${\bf x_0}=(0,0,d)$ with $d\geq R>0$, and the center of the sphere is ${\bf a}=(0,0,0)$.  Then by using spherical coordinates
$$(R\sin(\phi)\cos(\theta),R\sin(\phi)\sin(\theta),R\cos(\phi))$$ with $(\theta,\phi)\in [0,2\pi]\times [0,\pi]$, we obtain
\begin{align*}\frac{1}{4\pi R^2} \int_{\partial B({\bf a}, R)} \frac{1}{|{\bf x_0 - x}|} \, d\sigma_{\bf x}
&=\frac{1}{4\pi R^2} \int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi}\frac{R^2\sin(\phi)}{\sqrt{R^2\sin(\phi)^2+(R\cos(\phi)-d)^2}}\,d\phi\\
&=\frac{1}{2} \int_{\phi=0}^{\pi}\frac{\sin(\phi)}{
\sqrt{R^2+d^2-2Rd\cos(\phi)}}\,d\phi\\
&=\frac{1}{2Rd}\left[
\sqrt{d^2+R^2-2Rd\cos(\phi)}
\right]_0^{\pi}
=\frac{|d-R|-|d+R|}{2Rd}=\frac{1}{d}.
\end{align*}
P.S. By the way if $0\leq d< R$, then following the same procedure we find that the integral is $1/R$ (it is constant).
