= Removing a singular axis in a flux integral
{title2=$\lim_{\varepsilon\downarrow0}\int_{\rho=\varepsilon}\mathbf F\cdot\hat{\mathbf n}\,dS=0$}
A <vector field> singular on an axis need not satisfy the classical <divergence theorem> on the original closed region. Remove a thin tube $\rho<\varepsilon$, apply the theorem to the remaining smooth region, then examine the added cylindrical boundary. If its <flux> tends to zero and the remaining <volume integral> converges, the improper boundary <flux> equals the improper <integral> of the <divergence>. For $\mathbf F=(\log\rho^2,\log\rho^2,z/\rho)$ with $\rho=\sqrt{x^2+y^2}$, the cylindrical <unit normal> has zero $z$ component, so the added <flux> is $O(\varepsilon|\log\varepsilon|)$ on a bounded axial interval. The divergence $(2\cos\phi+2\sin\phi+1)/\rho$ is locally integrable under the <cylindrical coordinates> measure $\rho\,d\rho\,d\phi\,dz$. Hence no finite line-source correction survives, despite the field itself being unbounded.
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