= Neumann image lattice for a point source in a strip
{c}
{title2=$V=\frac{q}{2\pi}\left[\log r_0+\sum_{n\geq1}\log\frac{r_nr_{-n}}{(2nb)^2}\right]+C$}
For $b>0$, the <Poisson equation> $\Delta V=q\delta^{(2)}(x,y)$ in $-b<x<b$ with homogeneous <Neumann boundary conditions> on both edges admits the even-in-$y$ <method of images> solution in the title, where $r_n^2=(x-2nb)^2+y^2$. All image strengths are $q$. Pairing the images at $\pm2nb$ and subtracting their constant logarithms makes each summand $O(n^{-2})$, giving local <uniform convergence> away from the source. The unrenormalized sum of logarithms does not converge.
Differentiating gives
$$
\partial_yV=\frac{q}{2\pi}\sum_{n\in\mathbb Z}\frac{y}{(x-2nb)^2+y^2}.
$$
The corresponding $x$-derivatives cancel in reflected pairs on $x=\pm b$. For $y>0$, integration across the strip yields a telescoping <arctangent> sum:
$$
\int_{-b}^b\partial_yV\,dx=\frac{q}{2\pi}\lim_{N\to\infty}2\arctan\frac{(2N+1)b}{y}=\frac q2.
$$
One closed form, up to an additive constant, is $V=(q/4\pi)\log[\cosh(\pi y/b)-\cos(\pi x/b)]$. Its near-source <logarithmic singularity> has coefficient $q/(2\pi)$. Conditions at large $|y|$ or a symmetry class must be specified for a uniqueness claim; side <Neumann boundary conditions> alone allow additional homogeneous <harmonic functions>.
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