What´s wrong in this computation of $\Delta(r^{-1})$ as a distribution? maybe this is an idiot question, but I could not figure out what´s wrong. I know how to compute $\Delta (r^{-1})$ in $\mathbb{R}^{3}$ putting a ball with center in $0$ and then get $\Delta(r^{-1}) = -4 \pi \delta$, however I tried to compute it in other way and I got $\Delta(r^{-1}) =0$.
If $\phi \in \mathcal{D}(\mathbb{R}^3)$, then, choosing $\Omega$ such that $supp(\phi) \subset \Omega$ and $\partial \Omega \cap supp{\phi} = \emptyset$, $\langle\Delta(r^{-1}), \phi \rangle = \langle r^{-1}, \Delta\phi \rangle = \int_\Omega r^{-1} \Delta \phi = \int_\Omega \Delta(r^{-1})\phi - \int_{\partial\Omega} r^{-1}\frac{\partial\phi}{\partial n}  + \int_{\partial\Omega} \phi\frac{\partial\ r^{-1}}{\partial n} = 0 $ since the first integral is zero because $\Delta (r^{-1}) = 0$  in $\mathbb{R}^{3} \setminus \{0\}$, and the last two integrals vanish because $supp(\Delta\phi) \subset supp(\phi)$ and $supp(\phi) \cap \partial\Omega = \emptyset$. What´s wrong here??!!
Thanks in advance.
 A: 
the laplacian will vanish everywhere except at $0$, so it's just a set of null measure. 

Sets of measure zero may be negligible for integration   of functions but they are not negligible for application of the  Fundamental Theorem of Calculus, or of its higher dimensional forms (Green, Stokes, etc). They are not negligible for working with  distributions, either. 
Consider a simpler example: 
$$f(x) = \begin{cases} 1 \quad &\text{if }\ x\ge 0, \\ 0\quad &\text{if }\ x<0\end{cases}$$
Following your logic, one could say that 
$$  f(1)-f(-1) = \int_{-1}^1 f'(x)\,dx = 0 \tag{1}$$
because $f'(x)=0$ except at one point. Clearly, something is wrong here because $f(1)-f(-1)=1$. 
There are two ways to resolve the issue:


*

*Say that the first step is invalid, because the fundamental theorem of calculus does not apply to $f$. 

*Extend the notion of derivative to "distributional  derivative". Then the first step in (1) is correct, but the second is not: $f'$ is the distribution supported at one point, namely the Dirac delta. The integral of $f'$ over $[-1,1]$ is $1$.  


The same options apply to your problem. Either Green's identity is understood in the sense of classical derivatives (and then it's  not applicable to $r^{-1}$ on a domain that includes $0$), or it is understood in the sense of distributions, and then the integral involving $\Delta(r^{-1})$ is nonzero: it's not really an integral, but evaluation of the distribution $\Delta(r^{-1})$ on the test function $\phi$. The latter option is not helpful either (it's basically a tautology), if your goal is to find $\Delta(r^{-1})$.
So, what one usually does for evaluation of $\Delta(r^{-1})$ is to apply the classical Green's identity on a domain that does not include $0$ (not even on the boundary). Namely, on the complement of $B(0,\epsilon)$ in the domain.
