Divergence of $\frac{\hat{r}}{r^2}$ In David J. Griffiths's Introduction to Electrodynamics, the author gave the following problem in an exercise.

Sketch the vector function 
  $$ \vec{v} ~=~  \frac{\hat{r}}{r^2}, $$ 
  and compute its divergence, where 
  $$\hat{r}~:=~ \frac{\vec{r}}{r} , \qquad r~:=~|\vec{r}|.$$ 
  The answer may surprise you. Can you explain it?

I found the divergence of this function as
      $$ 
    \frac{1}{x^2+y^2+z^2}
  $$ 
Please tell me what is the surprising thing here.
 A: Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.
A: I have the same book, so I take it you are referring to Problem 1.16, which wants to find the divergence of $\frac{\hat{r}}{r^2}$.
If you look at the front of the book. There is an equation chart, following spherical coordinates, you get $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left (r^2 v_r\right) + \text{ extra terms}$.
Since the function $\vec{v}$ here has no $v_\theta$ and $v_\phi$ terms the extra terms are zero.
Hence $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2 \frac{1}{r^2}\right) = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(1\right) = 0$.
At least this is how I interpret the surprising element of the question. 
A: For me another surprising thing about this question was that the divergence was not negative, seeing as the flow decreases as we move radially outwards. I found an excellent explanation of this here: 
http://mathinsight.org/divergence_subtleties
A: You may wish to check if the divergence is finite everywhere.
