Confusion about divergence of point particle So in the Griffiths textbook he introduces the dirac delta function by attempting to calculate the divergence of $\frac{\hat r}{r^2}$. When performing the divergence calculation, we get $0$ everywhere, but when taking the Gauss integral we get $4\pi$. This is supposedly because of the singularity at $r=0$. 
However, where exactly are we going wrong, and which answer is right?
 People often write the divergence as $4\pi d(s)$, but how do we even know that the Gauss integral is meaningful with the singularity? Finally, does this only happen with $\frac{1}{r^2}$ or does it happen for all $\frac{1}{r^n}$, $n>0$? 
I am confused by what exactly is going wrong: which answer of divergence is "correct" and why?
 A: The correct answer is $4\pi$ ( More specifically $4\pi\delta ^3 (\textbf{r})$)
The divergence differential is not radius independent and thus at $r=0$, we encounter a "division by zero" leading to a solution that is incorrect. Although it is true that the divergence is $0$ everywhere, it is not so at $r=0$. 
On the other hand, we see that the divergence integral is independent of the radius of the sphere, and thus it gives the value of the divergence correctly as $4\pi$. 
Combining these two facts, we see that the divergence is a function which has its value 0 everywhere, except at the very centre where it suddenly spikes to a constant value of $4\pi$. And that's the definition of a Dirac Delta Function ( a function that is $0$ everywhere, but suddenly blips up to some value at some point, and then is $0$ again everywhere ).
And that's how you get the value of $4\pi\delta ^3 (\textbf{r})$ .
And for $n>2$, the divergence function (both the differential as well as the integral)  becomes radius dependent, making the "division by zero" more explicit, creating all sorts of problems (There is no escape now). (You can even prove that Gauss' Theorem is invalid for $n\ne 2$ )
Cheers!!
