Fundamental solution of Laplace's equation For $n=2$ the fundamental solution of the Laplace's equation is given by
$P=-\frac{\log(|x|)}{2\pi}$.
But how do I show that
$\int_{\mathbb{R}^2}\Delta P \, f \,dx=f(0)$
for all $f\in C^\infty_c(\mathbb{R}^2)$.
My Idea is
\begin{align}
& \int_{\mathbb{R}^2}\Delta Pf \, dx =\lim\limits_{\varepsilon\to 0} \int_{\mathbb{R}^2\setminus B_\varepsilon} \Delta Pf \, dx \\[8pt]
= {} &-\lim_{\varepsilon\to 0} \left(\int_{\partial B_\varepsilon}\nabla Pf\cdot n \, dS+\int_{\mathbb{R}^2\setminus B_\varepsilon}\nabla P\cdot \nabla f \, dx \right) \\[8pt]
= {} &-\lim_{\varepsilon\to 0} \left( \int_0^{2\pi}\frac{1}{\varepsilon^2}f \, d\varphi+\int_{\mathbb{R}^2\setminus B_\varepsilon}\nabla P\cdot \nabla f \, dx \right)
\end{align}
 A: You're idea is on the right track.  Here is a way forward if $f$ is differentiable and of compact support.
It can be shown that we can differentiate once (but not twice) under the integral to find
$$\begin{align}
\Delta \int_{\mathbb{R}^2}P(\vec x, \vec x')f(\vec x')\,d\vec x'&=\nabla \cdot \int_{\mathbb{R}^2}\nabla P(\vec x, \vec x')f(\vec x')\,d\vec x'\\\\
&=-\nabla \cdot \int_{\mathbb{R}^2}\nabla' P(\vec x, \vec x')f(\vec x')\,d\vec x'\\\\
&=\nabla \cdot \int_{\mathbb{R}^2} P(\vec x, \vec x')\nabla'f(\vec x')\,d\vec x'\\\\
&=- \int_{\mathbb{R}^2} \nabla ' P(\vec x, \vec x')\cdot \nabla'f(\vec x')\,d\vec x'\\\\
&=-\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}^2\setminus B_\varepsilon} \nabla ' P(\vec x, \vec x')\cdot \nabla'f(\vec x')\,d\vec x'\tag1
\end{align}$$
Now, we integrate by parts the integral on the right-hand side of $(1)$ to obtain
$$\begin{align}
\Delta \int_{\mathbb{R}^2}P(\vec x, \vec x')f(\vec x')\,d\vec x'&=\lim_{\varepsilon\to 0^+} \oint_{ B_\varepsilon} \hat n'\cdot \nabla ' P(\vec x, \vec x')f(\vec x')\,dS'\\\\
&=\frac1{2\pi }\lim_{\varepsilon\to 0^+} \int_0^{2\pi}\frac1\varepsilon f(\vec x'=\vec x+\varepsilon \hat n')\,\varepsilon\,d\phi'\\\\
&=f(\vec x)
\end{align}$$
as was to be shown!  And we are done.
A: You have
$$
\int_{\mathbb{R}^2}\Delta Pf\,dx=\int_{\mathbb{R}^2\setminus B_\varepsilon}\Delta Pf\,dx+\int_{B_\varepsilon}\Delta Pf\,dx.
$$
Since $\Delta P=0$ outside of $B_\varepsilon$, the first integral is zero. Therefore, integrating by parts,
$$
\int_{\mathbb{R}^2}f\Delta P \, dx=\int_{B_\varepsilon}f\Delta P \, dx=-\int_{B_\varepsilon}\nabla P\nabla f \, dx+\int_{\partial B_\varepsilon}f\nabla P\cdot n \, dS_\varepsilon=
$$
$$
=-\frac{1}{2\pi}\int_{B_\varepsilon}\frac{x}{\|x\|^2}\cdot \nabla f\,dx-\frac{1}{2\pi}\int_{\partial B_\varepsilon}f\frac{x}{\|x\|^2}\cdot n dS_\varepsilon
$$
Now, the first integral approaches zero as $\varepsilon\to 0$ because $\nabla f$ is bounded and $x/\|x\|^2\sim 1/\varepsilon$ while $Area (B_\varepsilon)\sim\varepsilon^2$. As for the second, $x\cdot n=-\|x\|$ and $x\cdot n/\|x\|^2=-1/\varepsilon$. Therefore,
$$
-\frac{1}{2\pi}\int_{\partial B_\varepsilon}f\frac{x}{\|x\|^2}\cdot n dS_\varepsilon=\frac{1}{2\pi\varepsilon}\int_{\partial B_\varepsilon}fdS_\varepsilon=f(\xi)
$$
for some $\xi\in\partial B_\varepsilon$. In the limit $\varepsilon\to 0$, $f(\xi)\to f(0)$, as desired.
