Divergence of a Vector Field - Surprising Result I'm following the text Introduction to Electrodynamics by Griffiths, and I came across the following in an in-text problem:

Sketch the vector function v = $\frac{\boldsymbol{\hat{\mathbf{r}}}}{r^2}$, and compute its divergence. The answer may surprise you... can you explain it?

Well, the answer did surprise me, for the sketch of the function is indeed indicating a diverging field (like field lines from a point positive charge), yet the math claims the divergence to be zero. What's going wrong? 
This is the solution I have, from a manual, which also doesn't make sense to me:

The answer is that $\nabla·v = 0$ everywhere except at the origin, but at the
  origin our calculation is no good, since $r = 0$, and the expression for $v$ blows up. In fact, $\nabla·v$ is infinite at that one point, and zero elsewhere.

Could someone please help me understand the situation? Any help would be appreciated, thanks!
P.S. I understand that this has been asked earlier on Physics SE, but I didn't understand the answers. The one with most upvotes said:
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. (Answer by @genneth)
What's the delta distribution in conversation?
 A: The "problem" is there because one assumes that the charge is a point at $r=0$.  
To see how one might round the "problem" in a non-rigerous way suppose instead one assumes a uniform charge density $\rho$.  
In a sphere of volume $V$, the total charge is $\displaystyle\int_{\rm V} \rho\,dV$.  
The electric field due to a point charge $q$ is $\vec E = - \dfrac {1}{4\pi \epsilon_0}\dfrac {q}{r^2} \hat r = k \dfrac {q}{r^2} \hat r $ 
In our case the charge is not a point charge but distributed over a volume $V$ and so $\vec E = k \dfrac {\int_{\rm V} \rho\,dV}{r^2} \hat r $.
The electric flux through the surface of a sphere of radius $R$ is $\displaystyle \int_{\rm S}\vec E \cdot d\vec s = k \dfrac {\int_{\rm V} \rho\,dV}{R^2}\,4\pi R^2=\int _{\rm V}4 \pi k \rho\, dV= \int _{\rm V}\nabla\cdot \vec E \,dV$.  
So the divergence of the electric field is $4 \pi k\rho = \dfrac {\rho}{\epsilon _0}$ in a more familiar form.  
Now consider what a point charge implies.
As the radius of the sphere decreases and tends towards zero then the charge density must tend towards infinity.  
To get around this "problem" a function is "loosely" defined as having the property that at $r=0$ the area under the function is $1$ and for all other values of $r$ the area under the function is zero.
It is called a delta function $\delta (r)$ and has infinite height and zero width but with a finite area of $1$ at $r=0$..  
So now the divergence of the electric field from a point charge $q$ is given by $\nabla \cdot \vec E= \dfrac{q}{\epsilon_0} \,\delta (r)$.  
At $r=0$ the divergence of the electric field is $\dfrac{q}{\epsilon_0}$ and the divergence is zero everywhere else as you have found.
A: We want to find $\nabla \cdot \vec{v} $, where $\vec{v} = \hat{r}/r^2 $
First, let's write $\hat{r}$ in terms of cartesian coordinates:
$$ \hat{r} = \vec{r}/|r| $$
$$ = \frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}} $$
Sub this value back into $\vec{v}$:
$$ \vec{v} = \frac{x\hat{x}+y\hat{y}+z\hat{z}}{(x^2+y^2+z^2)^{3/2}} $$
$\nabla \cdot \vec{v} $ is the sum of the $\hat{x}$ component differentiated by x, the $\hat{y}$ component differentiated by y and the $\hat{z}$ component differentiated by z. This is computationally intensive, but easily found using software like Mathematica.
$$\nabla \cdot \vec{v} = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} + \frac{3}{(x^2+y^2+z^2)^{3/2}} = 0$$
Of course, we made this calculation with the assumption that $(x,y,z) > 0$, otherwise we'll have zero on the denominator, and thus an infinite divergence.
Edit: To answer your question about what the dirac-delta distribution is: It's a function that's zero everywhere except for the origin where it's infinite. 
