What is the error hiding in this trivial integral with spherical polar coordinates and divergence theorem?

Question

I am working in spherical polar coordinates $(r,\theta,\phi)$ where $\theta$ and $\phi$ are the polar and azimuthal angles respectively. I define a vector field $\mathbf{V}=\phi \hat{\boldsymbol{\phi}}$, which has divergence $$ \boldsymbol{\nabla}\cdot\mathbf{V}=\frac{1}{r^2}\frac{\partial}{\partial r}(0)+ \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(0) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}(\phi) = \frac{1}{r\sin\theta} .$$

Also define $u=e^{-r^2}$, with gradient $$ \boldsymbol{\nabla}u=\Bigg(\hat{\mathbf{r}}\frac{\partial}{\partial r} + \frac{1}{r}\hat{\boldsymbol{\theta}}\frac{\partial}{\partial \theta} + \frac{1}{r\sin\theta} \hat{\boldsymbol{\phi}} \frac{\partial}{\partial \phi}\Bigg) e^{-r^2} = -2r e^{-r^2} \hat{\mathbf{r}} . $$

Then I can write $$ \boldsymbol{\nabla}\cdot(u\mathbf{V}) = u\boldsymbol{\nabla}\cdot\mathbf{V} + \boldsymbol{\nabla}u\cdot\mathbf{V} = \frac{e^{-r^2}}{r\sin\theta} $$

I want to integrate the above expression over all $\mathfrak{R}^3$. I apply divergence theorem to the left hand side, evaluated at the surface of a sphere at infinity. So I get

$$ LHS= \int\boldsymbol{\nabla}\cdot(u\mathbf{V}) dV= \int u \mathbf{V}\cdot d\mathbf{S} = \int u \mathbf{V}\cdot \hat{\mathbf{r}} dS = 0,$$

as expected. But for the RHS I get $$ RHS=\int \frac{e^{-r^2}}{r\sin\theta} dV = 2\pi \int_{0}^{\infty} dr\, r e^{-r^2} \int_{0}^{\pi} d\theta = \pi^2 .$$

Obviously this cannot be right! I have pored over this but I cannot figure out what is happening. Thanks in advance.

Answer 1

What the volume integral is really calculating is the following limit. Take the spherical "slice" which goes from $\phi = 0$ to $\phi = \beta$. The volume integral calculation is altered by

$$\beta\int_0^\infty re^{-r^2}dr\int_0^\pi d\theta = \frac{\pi\beta}{2}$$

The surface integral now has three components: The spherical surface itself is $0$ as before, what's new is the addition of the semicircular planes at $\phi = 0$ and $\phi = \beta$. The contribution from the first plane is $0$, but the contribution from the second plane is

$$\int_{\phi=\beta} u\mathbf{V}\cdot d\mathbf{S} = \int_0^\infty \int_0^\pi e^{-r^2}\beta\hat{\phi}\cdot\hat{\phi}(rdrd\theta) = \beta\int_0^\infty re^{-r^2}dr\int_0^\pi d\theta = \frac{\pi\beta}{2}$$

Ordinarily as you take the limit $\beta\to2\pi$, the two planes would close up and since the function takes on the same values but with faces oriented opposite ways, they would cancel. But the vector field has a jump discontinuity at the $xz$ plane, $x>0$, so this does not happen. Instead,

$$\lim_{\beta\to2\pi}\int_{S_{\beta}} u\mathbf{V}\cdot d\mathbf{S} = \pi^2 \neq 0$$

Since the vector field had a discontinuity, it's divergence was hiding a secret Dirac delta on the "prime meridian" of the sphere. The divergence theorem is valid if the delta is considered properly.