Let $\phi (x):= -{1 \over {2 \pi}} \ln |x| $ for every $x \in D: R^2$ \ $ \{0\}$. Calculate $\Delta \phi$ in the distributional sense.
From general theory, I know that , having a $\psi \in C_c^{\infty} (E)$, we have: $\langle \Delta T_\phi , \psi\rangle=-\langle \nabla T_\phi ,\nabla \psi\rangle=\langle T_\phi ,\Delta\psi\rangle$.
Here I can identify $T_\phi = \phi(x)$ and so $(T_\phi)' = \phi(x)'$ and $(T_\phi)'' = \phi(x)''$ .
But in this way, applying the standard definition of distributional derivative I have: $\langle \Delta T_\phi , \psi\rangle = \int_D \Delta(-{1 \over {2 \pi}} \ln |x|)\psi dx dx = \int_D ({1 \over {2 \pi x^2}} )\psi dx dx = \int_D (-{1 \over {2 \pi}} \ln |x|) \Delta \psi dx dx $ for the general property of distribution.
But I can't identify $\Delta \phi(x)$ better that $\phi''$ here. Am I missing something?
Set $u = \ln |x|,$ and take some $\varphi \in C_c^\infty(\mathbb{R}^2).$ Then, $\langle \Delta u, \varphi \rangle = \langle u, \Delta \varphi \rangle$. Now note that $u \in L^1_{\text{loc}}(\mathbb{R}^2)$ so we have $$\langle u, \Delta \varphi\rangle = \iint \ln|x| \, \Delta\varphi(x) \, dx = \lim_{\epsilon\to 0} \iint_{|x|>\epsilon} \ln|x| \, \Delta\varphi(x) \, dx.$$
Using integration by parts and using polar coordinates $(r,\theta)$ we get $$ \iint_{|x|>\epsilon} \ln|x| \, \Delta\varphi(x) \, dx = \int_{|x|=\epsilon} \ln|x| \, \nabla\varphi(x) \cdot \hat{r} \, ds - \iint_{|x|>\epsilon} \nabla\ln|x| \cdot \nabla\varphi(x) \, dx \\ = \int_{|x|=\epsilon} \ln|x| \, \nabla\varphi(x) \cdot \hat{r} \, ds - \iint_{|x|>\epsilon} \hat{r}r^{-1} \cdot (\hat{r}\partial_r+\hat{\theta}r^{-1}\partial_\theta)\varphi(x) \, r \, dr \, d\theta \\ = \int_{|x|=\epsilon} \ln\epsilon \, \partial_r\varphi(x) \, \epsilon\,d\theta - \iint_{|x|>\epsilon} \partial_r\varphi(r,\theta) \, dr \, d\theta . $$
The first integral vanishes as $\epsilon \to 0$ and the second integral becomes $$ \iint_{|x|>\epsilon} \partial_r\varphi(r,\theta) \, dr \, d\theta = \int_0^{2\pi} \left( \int_\epsilon^\infty \partial_r\varphi(r,\theta) \, dr \right) \, d\theta \\ = \int_0^{2\pi} \left( -\varphi(\epsilon,\theta) \right) \, d\theta \to \int_0^{2\pi} \left( -\varphi(0,\theta) \right) \, d\theta \\ = -2\pi \, \varphi(0) = \langle -2\pi\delta, \varphi \rangle . $$
Thus, $\Delta u = -2\pi \, \delta.$