Properties of the Double Layer Potential Consider the double layer potential
$$
 W_{\nu}(x) = \int_{\partial\Omega} \nu(y) \frac{\partial}{\partial n_y}\left( \frac{1}{|x - y|} \right) d \sigma_y
$$
for a bounded region $\Omega \in \mathbb R^3$ with a constant function $\nu(y) \equiv \nu_0$.
a) Show that $W_{\nu}(x) = 0$ for $x \notin \overline \Omega$.
b) Compute $W_{\nu}^i(x) := \lim_{\xi \to x, \xi \in \Omega} W_{\nu}(\xi)$ for $x \in \partial \Omega$.
c) Show, with b), that $W_{\nu}(x) = -4\pi \nu_0$ for all $x \in \Omega$.
Any hints how to proceed?
 A: $$
    \int_{\partial\Omega}\frac{\partial}{\partial n_y}\frac{1}{|x-y|}\,dS(y)=\int_{\partial\Omega} \nabla_{y}\frac{1}{|x-y|}\cdot\,d\overline{S}(y).
$$
If $x \notin\Omega$, then the right side becomes, by the Divergence Theorem,
$$
        \int_{\Omega}\nabla_{y}^{2}\frac{1}{|x-y|^{2}}dV(y) = 0.
$$
If $x \in \Omega$, let $\Omega_{\epsilon}$ be the region obtained by deleting a sphere of radius $\epsilon$ centered at $x$ from $\Omega$. Then the boundary of $\Omega_{\epsilon}$ is $\partial\Omega\cup \partial B_{\epsilon}(x)$, but the outward normal on $B_{\epsilon}(x)$ points toward the inside point $x$. Because of the inward pointing normal on $\partial B_{\epsilon}$,
$$
       \int_{\partial B_{\epsilon}(x)}\nabla_{y}\frac{1}{|x-y|}\cdot d\overline{S}(y) 
    = 4\pi, \;\;\; \epsilon > 0.
$$
Now the first line of this post may be written as
$$
         -4\pi + \int_{\partial\Omega_{\epsilon}}\nabla_{y}\frac{1}{|x-y|}\cdot d\overline{S}(y)=-4\pi.
$$
This type of technique is farily general because a typical volume integral in the subject can typically be written as a limit as $\epsilon \rightarrow 0$ of the volume integral where a sphere of radius $\epsilon$ centered at $x$ is omitted.
