Why is this true in two dimensions? $$\nabla^2\bigg(\ln(r)\bigg)=2\pi\delta^{(2)}(\mathbf{r}),$$ where $\delta^{(2)}$ denotes the two-dimensional $\delta$-function and $r=\sqrt{x^2+y^2}$ in Cartesian coordinates.
I understand that both function will look the same. But I do not know how to prove this rigorously.
Hint: Let $\phi \in C^\infty_{c}(\mathbb{R}^2)$. You need to show \begin{align} \int_{\mathbb{R}^2} \log|x| \Delta \varphi(x)\ dx = 2\pi \varphi(0). \end{align}
This can be done by splitting the left-hand side into \begin{align} \int_{B(0, \epsilon)} \log|x| \Delta \varphi(x)\ dx + \int_{\mathbb{R}^2\backslash B(0, \epsilon)} \log |x| \Delta \varphi(x)\ dx = I_1+I_2. \end{align} For $I_2$, use integration by parts to put the Laplacian on to $\log |x|$ (don't forget the boundary term). For $I_1$, show it vanishes as $\epsilon\rightarrow 0$. Remember the boundary term? Show that converges to the desired quantity as $\epsilon \rightarrow 0$.
Sorry to use the Answer panel, but I am not allowed to comment. Please, see the answer to this related question for more details on the calculations:
Distributional Laplacian of logarithm and the Dirac delta distribution
