Consider Green's representation formula \begin{equation} 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, \end{equation} where $\Gamma: \mathbb{R}^n\setminus\{0\}\rightarrow \mathbb{R}$ is the fundamental solution of the Laplace equation, i.e. \begin{equation} \Gamma(x) = -\frac{1}{2}\log|x| \end{equation} for $n=2$ and \begin{equation} \Gamma(x)=-\frac{1}{n(n-2)\alpha(n)|x|^{n-2}} \end{equation} 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 \begin{equation} \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. \end{equation} 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?
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.