Gradient pairing with a boundary-constant function and a harmonic function (${\int }_V\nabla f\cdot \nabla gdV=0$)

If $f$ is constant on the boundary and $g$ is a regular harmonic function throughout a volume, the divergence theorem applied to $f\nabla g$ gives ${\int }_V\nabla f\cdot \nabla g={\int }_{\partial V}f{\partial }_{n}g$. The constant boundary value factors out; a second application of the divergence theorem makes the remaining flux ${\int }_V\Delta g=0$. This requires enough regularity to use Green's first identity over the entire volume.

Although $1/r$ is harmonic away from the origin in three dimensions, it is not a regular harmonic function on a ball containing the origin. Its gradient is $-\hat{r}/r^2$, so the displayed improper volume integral follows by canceling the $r^2$ in the spherical volume element. On the annulus $\epsilon <r<a$, Green's first identity gives outer boundary contribution $-4\pi$ and inner contribution $4\pi \epsilon /a$, recovering the same limit. Its global Laplacian contains a Dirac delta distribution, so the regular-harmonic gradient-pairing theorem is inapplicable.