# Integral over fundamental solution in Green's formula

Consider Green's representation formula $$$$u(x) = \int_{\Omega} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y+\oint_{\partial\Omega}\Big(\Gamma(x-y)\nabla u(y)-u(y)\nabla_y(\Gamma(x-y)) \Big)\cdot\nu\,\mathrm{d}S,$$$$ where $$\Gamma: \mathbb{R}^n\setminus\{0\}\rightarrow \mathbb{R}$$ is the fundamental solution of the Laplace equation, i.e. $$$$\Gamma(x) = -\frac{1}{2}\log|x|$$$$ for $$n=2$$ and $$$$\Gamma(x)=-\frac{1}{n(n-2)\alpha(n)|x|^{n-2}}$$$$ for $$n\geq 3$$, $$\Omega$$ is a bounded Lipschitz domain, $$u\in C^2(\bar{\Omega})$$ and $$x\in\Omega$$.

My question is why the first integral in this formula exists at all, as we integrate over the singularity of the fundamental solution $$\Gamma$$ at $$y=x$$. Is this an integral in the classical sense? In the proof, one considers Green's identity on $$\Omega\setminus B_\epsilon(x)$$ for $$\epsilon>0$$. At one step in the proof, we also perform the first integral in the representation formula over $$B_\epsilon(x)$$ in order to show that it vanishes for $$\epsilon\to 0$$. When integrating over $$B_\epsilon(x)$$, it is also possible that $$y=x$$. At the end of the proof, we obtain $$$$\int_{\Omega} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y = \lim_{\epsilon\to 0}\int_{\Omega\setminus B_\epsilon(x)} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y.$$$$ Is this how one defines the integral on the left-hand side? Or am I just confused and the singularity of $$\Gamma$$ does not matter for the integration as it is only a null set?

• Those are all integrable singularities by $p$ test Feb 8 at 21:19
• Could you elaborate a little bit more on that? If I understand correctly, you refer to the case of one-dimensional integrals $\frac{1}{x^p}$. Feb 8 at 21:33

In $$\Bbb{R^n}$$ we have that
$$\int_{B(0,1)} \frac{dx}{|x|^p} = C_n \int_0^1 \frac{r^{n-1}dr}{r^p}$$
where $$C_n$$ is the "surface area" of the unit $$n$$-ball by switching over to $$n$$ dimensional spherical coordinates. The integral converges if $$p-n+1<1$$ by $$p$$ test.