I want to calculate $\Delta \ln{||x||}$ for $x=(x_1, x_2)$.
$$\langle \Delta \ln{||x||}, \phi \rangle= \langle \ln{||x||}, \Delta{\phi}\rangle=\int_{\mathbb{R}^2} \ln{||x||} \Delta{\phi(x)} dx= \lim_{\epsilon \to 0} \int_{\epsilon< ||x||<R} \ln{||x||} \Delta{\phi(x)} dx$$
How could we apply the formula $\int_V (f \Delta{\phi}-\phi \Delta f) dV=\int_{\partial{V}} \left( f \frac{\partial{\phi}}{\partial{n}}-\phi \frac{\partial{f}}{\partial{n}} \right)dS$ in order to find the value of $\Delta{\ln{||x||}}$ ?
Taking partials $$ \begin{align} \frac{\partial^2}{\partial x^2}\frac12\log(x^2+y^2) &=\frac{\partial}{\partial x}\frac{x}{(x^2+y^2)}\\ &=\frac{y^2-x^2}{(x^2+y^2)^2} \end{align} $$ symmetry gives $$ \frac{\partial^2}{\partial y^2}\frac12\log(x^2+y^2) =\frac{x^2-y^2}{(x^2+y^2)^2} $$ Therefore, away from $(0,0)$ $$ \Delta\log|x|=0 $$ Thus, $\Delta\log|x|$ is a multiple of the Dirac delta. To figure out the coefficient, we compute $$ \begin{align} \int_{B(0,r)}\Delta\log|x|\,\mathrm{d}x &=\int_{\partial B(0,r)}\nabla\log|x|\,\cdot n\,\mathrm{d}\sigma\\ &=\int_{\partial B(0,r)}\frac{x}{r^2}\cdot\frac xr\,\mathrm{d}\sigma\\ &=\frac1r\,2\pi r\\[9pt] &=2\pi \end{align} $$ So $$ \bbox[5px,border:2px solid #C0A000]{\Delta\log|x|=2\pi\delta(x)} $$