Divergence not defined I’m currently working on the practice problems in Introduction to Electrodynamics by Griffiths. I got confused by the solution to this problem.

What does “ill-defined divergence” even mean? I understand how and when to use delta function, but I don’t understand how divergence is not defined.

I think we can use
$$
\nabla \cdot (\psi \vec{a}) = \vec{a} \cdot \nabla \psi + \psi \nabla \cdot \vec{a}
$$
to see what's happened for $n \lt -2$.
$$\begin{align*}
\nabla \cdot (r^{-3} \hat{r})
&= \left( \frac{1}{r^2} \hat{r} \right)
  \cdot \nabla \frac{1}{r}
  + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r} \\
&= - \frac{1}{r^4}
  + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r}
  && \left( \nabla \frac{1}{r} = - \frac{1}{r^2} \right) \\
&= - \frac{1}{r^4}
  + \frac{4 \pi}{r} \delta^3(\vec{r})
\end{align*}$$
When $r$ toward $0$ (below misuses the delta function, delta function is meaning less outside of integral),
$$\begin{align*}
\nabla \cdot (r^{-3} \hat{r})
&= - \frac{1}{r^4}
  + \frac{4 \pi}{r} \delta^3(\vec{r}) \\
&= -\infty + \infty \cdot \infty
\end{align*}$$
We can not assign a meaningful value to $\nabla \cdot (r^{-3} \hat{r})$, so it is called "ill-defined".
