Divergence of $1/r$ in cylindrical coordinates In classical textbooks, like "Introduction to Electrodynamics" by J.D. Griffiths, it is given that
$$\nabla\cdot\left(\frac{\widehat{r}}{r^2}\right)=4\pi\delta^3(R).$$
To prove this equality, Griffiths first evaluates the divergence and finds it $0$ for $r \neq 0$ and undefined for $r=0$. Then evaluates the integral of the given divergence and finds it $4\pi$. Then, he concludes that the above equality should hold since the integral has a constant value containing origin.
Now, I am asked about the following equality :
$$\nabla \cdot\left(\frac{\widehat{\rho}}{\rho}\right)=2\pi\delta^2(\rho)$$
(note: the equation is claimed to be valid in cylindrical coordinates).
I was unable to work out the integral properly. Any help would be appreciated.
 A: If one integrates the line integral of the normal component of $\frac{\hat \rho}{\rho}$ along the circle $\rho=\epsilon$, then one finds that
$$\oint_{\rho=\epsilon}\left(\frac{\hat \rho}{\rho}\right)\cdot d\vec\ell=\int_0^{2\pi}\left(\frac{\hat \rho}{\epsilon}\right)\cdot \hat \rho\,\epsilon\,d\phi=2\pi$$
Heuristically, we apply naively the divergence theorem (the conditions of the divergence theorem are not satisfied due to the singularity at $\rho=0$) and find for a smooth function $\phi(\vec \rho)$
$$\begin{align}
\lim_{\epsilon\to 0}\int_0^{2\pi}\int_0^\epsilon \nabla\cdot\left(\frac{\hat\rho}{\rho}\right)\phi(\vec\rho)\,dS&\approx \phi(0) \lim_{\epsilon\to 0}\int_0^{2\pi}\int_0^\epsilon \nabla\cdot\left(\frac{\hat\rho}{\rho}\right)\,dS\\\\
&=\phi(0)\lim_{\epsilon\to0}\oint_{\rho=\epsilon}\left(\frac{\hat \rho}{\rho}\right)\cdot d\vec\ell\\\\
&=2\pi\phi(0)
\end{align}$$
Hence, in distribution, we can write
$$\nabla\cdot\left(\frac{\hat \rho}{\rho}\right)\sim 2\pi \delta(\vec\rho)$$

To be precise, we begin with the regularization 
$$\begin{align}
\delta_a(\vec \rho)&=\nabla \cdot \left(\frac{\vec \rho}{\rho^2+a^2}\right)\\\\
&=\frac{2a^2}{(\rho^2+a^2)^2}
\end{align}$$
Then, for any test function $\phi(\vec \rho)$ and domain $S\in \mathbb{R}^2$ we have
$$\lim_{a\to0}\int_S \phi(\vec \rho)\delta_a(\vec\rho)\,dS=\begin{cases}2\pi \phi(0)&,\vec\rho\in S\\\\0&,\vec\rho\notin S\end{cases}\tag1$$
To show that $(1)$ holds, we used an analogous approach of the analysis in THIS ANSWER for the $3$-D Dirac Delta. 
Thus, it is in this sense that 
$$\lim_{a\to0}\delta_a(\vec\rho)=2\pi\delta(\vec\rho)$$
