I want to calculate
$\nabla\cdot(\frac{\hat{r}}{r^2})$
Naively applying the expression for divergence in spherical coordinates gives
$\nabla\cdot(\frac{\hat{r}}{r^2})=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 *\frac{1}{r^2})=\frac{1}{r^2}\frac{\partial}{\partial r}(1)=0$
However, based on the context in which this problem arose, I believe there should be a delta function. You can get this by noticing:
$\nabla^2\frac{1}{r}=\nabla\cdot(\nabla\frac{1}{r})=-\nabla\cdot (\frac{\hat{r}}{r^2})$
Thus
$\nabla\cdot(\frac{\hat{r}}{r^2})=-\nabla^2\frac{1}{r}=4\pi\delta(\vec{r})$
Can anyone point out where the error is? I suspect there is an issue cancelling out the $r^2$ in the first method due to that covering up the zero-in-denominator issue, but I would like to understand precisely what goes wrong.
The divergence is indeed $0$ on $\Bbb{R}^3\setminus\{0\}$, so there's no error there. The only reason the dirac delta appears is because of the desire to "force" the divergence theorem to hold true even when it classically doesn't apply anymore. In other words, we WANT the equation $\int_{\Omega}\text{div}(F)\,dV = \int_{\partial \Omega}F\cdot n\,dA$ to hold true. The surface integral for this radial vector field is ALWAYS $4\pi$ for any surface which "encloses" the origin. Therefore, the only way this can hold true is if we define $\text{div}(F)=4\pi \delta$