Need help in solving an equation involving volume, single and double layer potentials Let be $V \subset \mathbb{R}^n $, $ 3\leq n$ an open set, where you can apply Gauß's Theorem. 
To show is, that for all $ U \in C^{(1)} ( \bar{V} ) \cap C^{(2)} (V) $ with  bounded 2nd derivatives the following equation for $y \in V $:
$$ (n-2) \omega_{n-1} U(y) = \int_{ \partial V} \left[\frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x)-U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}}\right] d\sigma(x)- \int_V \frac{ \Delta U(x)}{ |x-y|^{n-2}} dx $$
where 


*

*$ \nu_x $ is the outer normal unit vector on $x\in \partial V$ and

*$w_{n-1} := \frac{n \pi^{n/2}}{ \Gamma( \frac{n}{2} +1) }$
Well, I know that $W(x):= |x-y|^{-(n-2)} $ is not defined in $x=y$.
Therefore, instead of integrating over $V$ , first integrate over $V_{\epsilon}:= V$ \ $ K_{\epsilon}(y) $ and use the limes $\epsilon \rightarrow 0+ $
Thats pretty much it. Do you guys maybe know what that is for an Equation? I didn't know a proper title for the question, sorry about that. I find it quite hard to solve.
Any help is therefore very appreciated !!
 A: Here is a rough sketch on how I think this equation can be shown.
As you stated, we should work on the domain $V_\epsilon$ instead of $V$ first.
The other important idea is to use Green's seocond identity.
So we apply Green's second identity to the domain $V_\epsilon$.
If we also use that $\Delta W(x)=0$ for $x\neq y$ then the equation becomes
$$
(n-2) \omega_{n-1} U(y) = \int_{ \partial K_\epsilon} \left[\frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x)-U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}}\right] d\sigma(x).
$$
Note that the terms involving integrals over $V$ or $\partial V$ disappeared.
Now we will work with the right-hand side and consider the convergence $\epsilon\to0$.
First we consider the first term of the integral.
Because the second derivatives of $U$ are bounded it means that the first derivatives of $U(x)$
are bounded by a constant $C$ if $x$ is close to $y$.
We have
$$
\left|\int_{ \partial K_\epsilon} frac{1}{|x-y|^{n-2}} \frac{ \partial U}{ \partial \nu} (x) d\sigma(x)\right|
\leq
C \int_{ \partial K_\epsilon} \frac{1}{|x-y|^{n-2}} d\sigma(x)
= C \epsilon^{2-n} \omega_{n-1} \epsilon^{n-1}\to 0,
$$
where we used that the surface area of a ball with radius $\epsilon>0$ is given by $\omega_{n-1}\epsilon^{n-1}$.
So it remains to consider the other term.
First, we can calculate that
$$
\frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}}
=  -(n-2) \frac{1}{|x-y|^{n-1}}.
$$
Then we have
$$
\int_{ \partial K_\epsilon} U(x) \frac{ \partial }{\partial \nu_x} \frac{1}{|x-y|^{n-2}} d\sigma(x)
= 
-(n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(x) d\sigma(x).
$$
It remains to consider the difference to the left-hand side of the original equation.
$$
\left|
(n-2)\omega_{n-1} U(y)
-(n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(x) d\sigma(x)
\right|
=\left|
(n-2) \epsilon^{-(n-1)} \int_{ \partial K_\epsilon} U(y)-U(x) d\sigma(x)
\right|
\leq \sup_{x\in \partial K_\epsilon} |U(y)-U(x)| (n-2)\epsilon^{-(n-1)}\omega_{n-1} \epsilon^{n-1}
\to0
$$
where the $\sup$-term converges to $0$ because $U$ is continuous.
