Let $\vec r=x\hat i+y\hat j+z\hat k$ a position vector and $\delta^3(\vec{r})=\delta (x)\delta (y)\delta (z)$ is the tree-dimensional Dirac's delta function, undertood in the sense of distributions (it is zero at all points except at the origin, where it equals infinite). I'm looking for a proof of the following identity,
$$\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}=4\pi\delta^3(\vec{r})$$
How could it be derived?
Since $\phi\vec{\nabla}\cdot\vec{A}=\vec{\nabla}\cdot(\phi\vec{A})-\vec{\nabla}\phi\cdot\vec{A}$, we want to prove scalar test functions $f$ satisfy$$\int_\Omega\frac{\vec{r}}{r^3}\cdot\vec{\nabla} f(\vec{r})d^3\vec{r}=-4\pi f(\vec{0}),$$where $\Omega$ denotes $3$-dimensional space, which we'll parameterize with spherical polar coordinates so the integral is$$4\pi\int_0^\infty\frac{\hat{r}}{r^2}\cdot\vec{\nabla} f(\vec{r})r^2dr=4\pi\int_0^\infty\frac{\partial f}{\partial r}dr=-4\pi f(\vec{0}).$$