I am wanting to find the charge density of an infinitely thin [Hertzian dipole](https://en.wikipedia.org/wiki/Dipole_antenna#Hertzian_dipole), but am struggling evaluating the Dirac delta functions gradient.

$$\vec{J} = I\_{0}\cos(\omega t) \delta^3(r) \hat k.$$

Using: $\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t}$

We know $\nabla \cdot ({\phi \vec{v}}) = \phi \nabla \cdot \vec{v} + \nabla \phi \cdot \vec{v}.$

Which reduces to $\nabla \phi \cdot \vec{v}.$

Which ultimately means

$$\nabla \cdot \vec{J} = I\_{0}\cos(\omega t) \delta(x) \delta(y) \frac{\partial}{\partial z} \delta(z).$$

Beyond this point I am not really sure if what I'm doing is really correct

By definition $lim z -> 0$  

$$\delta (z) = \frac{1}{|a|\sqrt{\pi}}e^{-(\frac{z}{a})^2}.$$

So $\frac{\partial}{\partial z} \delta(z) = \frac{1}{|a|\sqrt{\pi}}e^{-(\frac{z}{a})^2} \cdot -2(\frac{z}{a}) \cdot \frac{1}{a}$

= $\delta(z) \cdot -2(\frac{z}{a^2}) $

So,

$$\nabla \cdot \vec{J} = I\_{0}\cos(\omega t) \delta(x) \delta(y) \delta(z) \cdot -2(\frac{z}{a^2}) $$

$$\nabla \cdot \vec{J} = I\_{0}\cos(\omega t) \delta^3(r) \cdot -2(\frac{z}{a^2}).$$

Integrating gets:

$$\rho = I\_{0}\sin(\omega t) \delta^3(r) \cdot 2(\frac{z}{a^2 \omega}).$$

So, plugging this expression into math lab and taking the limit yields an infinite charge density,

I would assume that if I integrate it, it yields a finite charge, but I haven't really been able to find anything on getting the charge density from the current density online, so I would appreciate it if anyone could tell me any mistakes that I've made? **and if not, help me re write this in a form such that I am able to prove that it's integral is finite.**

(Here $I\_{0}$ is not current, and is some finite value)