Fundamental solution of the Laplace's equation in $\mathbb{R}^n$ is given by:
$$\phi (x)=\begin{cases} -\frac{1}{2\pi}<\log|x| & n=2 \\ \frac{1}{n(n-1)\operatorname{Vol}(B(0,1))}|x|^{2-n} & n\geq 3 \end{cases}$$
I'm supposed to prove that $\Delta \phi = \delta _0$, where $\delta _0$ is Dirac's delta distribution. This should be easy but, for example, for $n=2$. I'm getting $$\Delta \phi = -\frac{1}{2\pi |x|}.$$ This does "explode" in $x=0$, but I don't think that's enough to call it $\delta _0$.
Did I do the calculus incorrectly or is my understanding of what $\Delta \phi = \delta _0$ means wrong?