Gauss' law in differential form for a point charge I'm trying to understand how the integral form is derived from the differential form of Gauss' law.
I have several issues:  
1) The law states that $ \nabla\cdot E=\frac{1}{\epsilon 0}\rho$, but when I calculate it directly I get that $ \nabla\cdot E=0$ (at least for $ r\neq0$).  
2) Now $ \iiint\limits_\nu \nabla\cdot E d\tau $ should be zero no matter what the value of the divergence is at 0, since the divergence is zero everywhere but 0 (in contrast to the law which states it is non-zero).  
3) 
a. The proof itself goes on to use the divergence theorem to state that for any volume $\nu$, $ \iiint\limits_\nu \nabla\cdot E d\tau = \iint\limits_{\partial\nu} E d a $, however the divergence theorem requires E to be continuously differentiable everywhere in $\nu$ (it is not differentiable at 0, let alone continuously differentiable there).  
b. The function cannot be corrected in any way at 0 since the derivative goes to infinity around 0.  
c. The point 0 cannot be removed from the integrated volume because the divergence theorem requires that the volume of integration be compact.  
d. In light of the former I don't see how the divergence theorem can be used here.
 A: We have that the electric field from a "point charge" $q$ located at the origin is 
$$\vec E=\frac{q}{4\pi \epsilon_0}\frac{\vec r}{|\vec r|^3} \tag 1$$
Clearly from $(1)$, we have $\nabla \cdot \vec E=0$ for $\vec r\ne 0$.  We note that the divergence is undefined (using classical analysis) at the origin.
Now, let $V$ be any region, whose "outer boundary" contains the origin, that excludes the "small" spherical region $0<\epsilon <r$ (i.e., $V$ has a hole to exclude the origin).  Denote the "outer boundary" of $V$ to be $S$ and the boundary of the "small" spherical region to be $S_{\delta}$. Clearly, we have from the Divergence Theorem
$$\begin{align}
\int_V \nabla\cdot \vec E\,dV&=\oint_{S} \vec E\cdot \hat n\,dS-\oint_{S_{\delta}} \vec E\cdot \hat n\,dS\\\\
&=0 \tag 2
\end{align}$$
since $\nabla \cdot \vec E=0$ throughout $V$.  Equation $(2)$ implies that 
$$\oint_{S} \vec E\cdot \hat n\,dS=\oint_{S_{\delta}} \vec E\cdot \hat n\,dS \tag 3$$
for any $S$ surrounding the origin.  We can evaluate the integral on the right-hand side of $(3)$ using $(1)$.  Proceeding we have
$$\begin{align}
\oint_{S_{\epsilon}} \vec E\cdot \hat n\,dS &=\int_0^{2\pi}\int_0^\pi \left(\frac{q}{4\pi \epsilon_0} \frac{\hat r}{\delta^2}\right)\,\cdot \hat r \,\delta^2\,d\theta\,d\phi\\\\
&=\frac{q}{\epsilon_0} \tag 4
\end{align}$$
which yields the integral-form of Gauss's Law for a point charge.

NOTE:
Analysis can be facilitated using Generalized Functions as follows.  In THIS ANSWER, I discuss regularizing the Electric Field of a point charge to assign meaning to the Dirac Delta for use in the Divergence Theorem. This provides a rigorous way forward where Dirac Delta is interpreted in terms of the limit of the regularized function $\vec \psi$ given by 
$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 1$$
Taking the divergence of $(1)$ reveals that 
$$\nabla \cdot \vec \psi(\vec r; a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$
Now, in the Aforementioned Answer, I showed that for any sufficiently smooth test function $\phi$, we have that
$$\lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV=
\begin{cases}
0&, \text{V does not include the origin}\\\\
4\pi \phi(0)&,\text{V includes the origin}
\end{cases}$$
and it is in this sense that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$
Heuristically, let the volume charge density of a point charge be given by $q\delta (\vec r)$, where $\delta(\vec r)$ is the Dirac Delta.  Then, the point-form of Gauss's Law is 
$$\nabla \cdot \vec E=\frac{q}{\epsilon_0}\delta (\vec r)$$
and we have from the divergence theorem for Generalized Functions
$$\int_V \nabla \cdot \vec E\,dV=\frac{q}{\epsilon_0}=\oint_S\vec E\cdot \hat n\,dS$$
where $\delta(\vec r)$ interpreted in terms of the limit of the regularized function.  And we are done!

Other answers I've posted on the subject of the Dirac Delta are HERE, HERE, and HERE.  This latter post provides a good primer on Distributions.
