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The inverse-square radial vector field has zero ordinary divergence at every nonzero radius but flux $4\pi$ through every centred sphere. It is locally integrable in three dimensions. For a smooth compactly supported test function $\eta$, integrate by parts outside a ball of radius $ϵ$. The inward-oriented inner boundary gives $-\int \mathbf{F}\cdot \nabla \eta ={\int }_{S^2}\eta (ϵ\hat{\mathbf{r}})d\Omega$, which tends to $4\pi \eta (0)$. This proves the displayed distributional divergence, with the three-dimensional Dirac delta function representing the point source.

For $r=\sqrt{x^2+y^2+z^2}>0$, the radial vector field has components $F_i=x_{i}f(r)$. The product rule givesThus for $f(r)=r^{n-1}$,When $n=-2$, $\mathbf{F}=\hat{\mathbf{r}}/r^2$ and the ordinary divergence is zero at every nonzero radius. However, the field is singular at the origin. Its outward flux through a radius-$R$ sphere is $(1/R^2)(4\pi R^2)=4\pi$, so one cannot apply the smooth-field divergence theorem across the origin and conclude that this flux vanishes.

In the language of distributions, the complete distributional divergence identity is $\boxed{\nabla \cdot (\hat{\mathbf{r}}/r^2)=4\pi {\delta }^{(3)}(\mathbf{r})}$. Here ${\delta }^{(3)}$ is the three-dimensional Dirac delta function. Without that generalized interpretation, the correct statement is simply zero divergence away from the origin and undefined ordinary divergence at the singular point. Removing a small ball around the origin restores the ordinary theorem: the inner boundary has inward radial orientation and cancels the outer flux.