Consider the following integral: $$\oint_{C} v\frac{\partial u}{\partial \vec{n}} - u\frac{\partial v}{\partial \vec{n}} \quad ds$$
where $\vec{n}$ is a normal vector (I don't think inward vs outward matters here), and $C$ is just some simple closed contour in the plane.
Under what circumstances will this integral go to $0$ as $C$ expands outward to $\infty$ in the plane? Is "expands outward to $\infty$ in the plane" necessarily a well-defined statement?
I know that the integral will always be zero if the integrand can be written as the gradient of a potential function, but I'm not sure if this can be done.
Also, will this change if either $v$ or $u$ has a singularity somewhere?
If the function $u\triangle v-v\triangle u$ is well defined inside $C$, and $u$ and $v$ satisfy the Laplace equation, then by second's Green identity the integral you have is exactly 0, no matter what type of contour you have.
However if the function $u\triangle v-v\triangle u$ has a singularity and functions $u$ and $v$ satisfy Laplace's equation for all points inside $C$ except in the singularity, you have the following result instead: $$\oint_{C}{\left(v\frac{\partial u}{\partial n}-u\frac{\partial v}{\partial n}\right)\,ds}+\oint_{C_\epsilon}{\left(v\frac{\partial u}{\partial n}-u\frac{\partial v}{\partial n}\right)\,ds}=0$$
Note that the normal vector $\textbf{n}$ in the first integral corresponds to the outward one, while for the circle enclosing the singularity must be pointing inwards.