# Laplacian of an integral on manifold equals zero

Let $$\Omega \subseteq \mathbb R^3$$ be a connected, open, and bounded set, and let $$\Gamma = \partial \Omega$$. Let $$\sigma \in C^1(\mathbb R^3)$$, $$\sigma \geq 0$$. Consider the integral $$\phi(x) = \int_\Gamma \frac{\sigma(y) dS(y)}{|x-y|}.$$ Show that $$\Delta\phi=0$$ for all $$x \in \Omega$$.

The only thing that came to my mind is that $$\Delta f = f''_{rr} + f'_r \cdot \frac{2}{r}$$ if $$f = f(r)$$ where $$r = \sqrt{x^2+y^2+z^2}$$, but that does not seem to be the case as there is an arbitrary function $$\sigma$$. The Divergence theorem which comes at handy when dealing with closed manifolds also doesn't seem to help.

• Where this question came from? Maybe, the theory contained in the book can be useful to solve the problem. Mar 5, 2021 at 22:51
• @DiegoMath that is from our calculus of variations course (the lecturer doesn't use a specific book for it). I honestly can't imagine how can variations help me as it looks like an exercise in anything but calculus of variations
– dnes
Mar 6, 2021 at 10:37
• Now I see here that the integral is differentiable by the parameter y, so straight calculation of $\Delta \phi$ yields the necessary result. Silly me.
– dnes
Jul 27, 2021 at 11:49