= Flux of a vector field with an integrable tangential singularity
Consider the <vector field> $\mathbf F=(p(x),q(x),r(z))$ in the open unit <cube>, where $p$ and $r$ extend to continuously differentiable functions on $[0,1]$, and $q$ is continuous on $(0,1]$ with $\int_0^1|q(x)|\,dx<\infty$. Even if $q$ is unbounded as $x\downarrow0$, the <normal component> on an $x$-face has the finite trace $\pm p(x)$. The two $y$-faces have finite <improper integral> fluxes $\pm\int_0^1q(x)\,dx$ and therefore cancel. The remaining face contributions give
$$
\boxed{\text{outward flux}=p(1)-p(0)+r(1)-r(0)}.
$$
To justify this use of the <divergence theorem>, first apply it to $[\varepsilon,1]\times[0,1]^2$, on which the field is regular. Its <divergence> is $p'(x)+r'(z)$, independent of $q$. Its outward flux is $p(1)-p(\varepsilon)+(1-\varepsilon)[r(1)-r(0)]$. Taking $\varepsilon\downarrow0$ yields the displayed answer, and <absolute convergence> makes the two $y$-face limits finite separately. For example, $q(x)=1/[x(1+(\ln x)^2)]$ diverges at zero but
$$
\int_0^1q(x)\,dx=\left[\arctan(\ln x)\right]_{0^+}^{1}=\frac{\pi}{2}.
$$
This argument defines the normal traces and the face integrals explicitly; it does not assert that the whole <vector field> has a finite value on the singular face.
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