Looking for an analog of the proof of $\nabla\cdot\left(\dfrac{\vec{r}}{r^3}\right) = 4\pi\delta^{(3)}(\vec{r})$ for $\nabla\delta^{(3)}$

In a physics problem I came across a function which seems to be $$0$$ over all space except in $$\vec{0}$$ where it is undefined. However the function represents a charge density so I have strong reasons to believe that it takes the form of some kind of dirac-like distribution. My idea was therefore to apply a test function and see what I could get out of it. The function is the following in spherical coordinates (with $$p$$ and $$k$$ two constants) :

$$\varrho(r, \theta, \phi)=\frac{1}{4\pi}\nabla\cdot\underbrace{\left[2\left(\frac{1}{r^2}-\frac{ik}{r}\right)\frac{e^{ikr}}{r}p\cos(\theta)\vec{u}_r+\left(\frac{1}{r^2}-\frac{ik}{r}-k^2\right)\frac{e^{ikr}}{r}p\sin(\theta)\vec{u}_{\theta}\right]}_{\vec{A}(r, \theta, \phi)}$$

It indeed can be shown using the expression of divergence in spherical coordinates that this zero is everywhere except in $$\vec{0}$$ where it is undefined. From a physics perspective it is expected that $$\varrho = \vec{p}\cdot\nabla\delta^{(3)}$$ where $$\vec{p}=p\vec{u}_z=p\cos(\theta)\vec{u}_r-p\sin(\theta)\vec{u}_{\theta}$$. The idea was therefore to apply a test function $$f$$ and proove that somehow : $$\int_{\mathbb{R}^3}f\varrho = p\frac{\partial f}{\partial z}\Big|_{r=0}$$ I thought that the fact that $$\varrho$$ could be expressed as the divergence of another vector field could help me, the idea would be that similarly to showing that $$\nabla\cdot\left(\frac{\vec{r}}{r^3}\right) = 4\pi\delta^{(3)}(\vec{r})$$ I could show that $$\nabla\cdot\vec{A}=4\pi\vec{p}\cdot\nabla\delta^{(3)}$$. I tried using Green's first identity : $$\int_Vf\nabla\cdot\vec{A}\;dV = \int_{\partial V}f\vec{A}\cdot\vec{dS} - \int_V\nabla f\cdot\vec{A}\; dV$$ but I couldn't quite make it work. Could someone help me to proove it please ?

• The electric dipole is treated in my Phys.SE answer here. Commented Mar 7, 2022 at 9:11

The meaning of the equation $$\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)=4\pi~\delta(\boldsymbol r)$$ Is that, for any $$\epsilon > 0$$, $$\int\limits_{\mathbb{B}(0,\epsilon)}\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)\mathrm d^3 \boldsymbol r=4\pi$$ Owing to the definition of the Dirac delta $$\int\limits_{\mathbb{B}(0,\epsilon)}\delta(\boldsymbol r)\mathrm d^3\boldsymbol r=1$$ This is easy to show. Due to the divergence theorem, $$\int\limits_{\mathbb{B}(0,\epsilon)}\operatorname{div}\left(\frac{\boldsymbol r}{r^3}\right)\mathrm d^3 \boldsymbol r=\int\limits_{\partial\mathbb{B}(0,\epsilon)}\frac{\boldsymbol {r}}{r^3}\cdot \hat{\boldsymbol r}~ \mathrm d^2 \boldsymbol r\\ =\int\limits_{\partial\mathbb{B}(0,\epsilon)}\frac{r\hat{\boldsymbol {r}}}{r^3}\cdot \hat{\boldsymbol r}~ \mathrm d^2 \boldsymbol r \\ =\int\limits_0^\pi\int\limits_0^{2\pi}\frac{1}{\epsilon^2}~\epsilon^2\sin\phi ~\mathrm d\theta\mathrm d\phi \\ =4\pi.$$ QED.