Stuck while trying to show fundamental solution $\frac{1}{2\pi}\ln|x|$ satisfies $\Delta \phi(x) = \delta(x)$ in 2d?

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I am trying to show that the fundamental solution to the Laplacian in 2D satisfies

$$\Delta \phi(x) = \delta(x)$$

where $x = (x_1, x_2) \in \mathbb{R}^2$.

So the fundamental solution in 2D is $\frac{1}{2\pi}\ln|x|$ and we have

$$\Delta \frac{1}{2\pi}\ln|x| = \delta(x)$$

$$\int_B \Delta \frac{1}{2\pi}\ln|x| dx = \int_B\delta(x)dx$$

where $B$ is the unit ball.

$$\frac{1}{2\pi}\int_B \Delta \ln|x| dx = 1$$

$$\int_B \Delta \ln|x| dx = 2\pi$$

Using the divergence theorem we transform this to a surface integral

$$\int_{\partial B} \frac{\partial \ln|x|}{\partial \nu(x)} dx = 2\pi$$

where $\nu$ is the normal derivative. I can't see where to go from here?