Single layer potential

Remarks

The integral is a surface integral. For $x \notin \partial\Omega$ the integrand is smooth in $x$, and $S\varphi$ is harmonic on $\R^n \setminus \partial\Omega$. For $x \in \partial\Omega$ the singularity of $\Phi(x - y)$ is integrable on the $(n-1)$-dimensional surface, so $S\varphi(x)$ is still defined. It is the potential of the measure $\varphi\,\sigma$ carried by $\partial\Omega$.

Example: let $\Omega$ be the unit ball in $\R^3$ and $\varphi \equiv 1$. Then $\Phi(x) = 1/(4\pi\abs{x})$ and the mean of $1/\abs{x - y}$ over the unit sphere in $y$ is $1/\max(\abs{x}, 1)$. Since the sphere has area $4\pi$, $S\varphi(x) = 1$ for $\abs{x} \le 1$ and $S\varphi(x) = 1/\abs{x}$ for $\abs{x} \ge 1$. It is continuous across the sphere, while its radial derivative jumps from $0$ to $-1$.

Non-example: the integral $\int_\Omega \Phi(x - y) f(y)\, dy$ over the open set itself is a volume potential, the Newtonian potential of $f$, and not a single layer potential. It is a potential of a charge spread through a solid, not over a surface.