A while ago I asked about the Green's function on the surface of a cylinder in $\mathbb{R}^3$, and it turned out to be very easy to compute by conformally mapping the cylinder to the half-plane. Now I'm interested in the Green's function of the Laplacian on the three-dimensional cylinder $(x, y, \cos\theta,\sin\theta)$ in $\mathbb{R}^4$; by symmetry and developability it is enough to find the Green's function $G(x,y,\theta)$ at the origin of the three-dimensional slab $\mathbb{R}^2 \times [-\pi,\pi]$, with periodic boundary conditions in $\theta$.
I thought this would be a simple calculation, but after several attempts I still have not gotten anywhere. One first approach is to use the reflection principle: $\frac{1}{\sqrt{x^2+y^2+\theta^2}}$ is (up to a constant factor) the Green's function for free space, so $$G(x,y,\theta) \propto \sum_{i=-\infty}^{\infty} \frac{1}{\sqrt{x^2+y^2+(\theta+2\pi i)^2}}$$ satisfies the periodicity condition and is harmonic everywhere it is defined; unfortunately $G(0,0,\theta)$ diverges for every $\theta$.
A second attempt was to start from $\Delta G = \delta(x)\delta(y)\delta(\theta)$ and take the Fourier transform in $\theta$; this leads to the modified Bessel equation and $$G(x,y,\theta) \propto \sum_{i=1}^{\infty} \cos(i \theta) K_0\left(i\sqrt{x^2+y^2}\right)$$ where $K_0$ is the modified Bessel function of the second kind. Unfortunately, applying Poisson summation takes me right back to the divergent $G$ above.
Where am I going wrong? How can I compute the Green's function on this domain?
UPDATE: After more carefully doing the above separation of variables I get that the Green's function must be of the form $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} + \sum_{i=1}^{\infty} \left(\cos(i\theta)\left[k_i I_0\left(i \sqrt{x^2+y^2}\right) - \frac{1}{2\pi^2}K_0\left(i \sqrt{x^2+y^2}\right)\right]\right)$$ where the $k_i$ are arbitrary constants; since the $I_0$ terms are harmonic everywhere we might as well set $k_i=0$ to get $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} - \frac{1}{2\pi^2} \sum_{i=1}^{\infty} \cos(i\theta)K_0\left(i \sqrt{x^2+y^2}\right)$$ which does seem to converge, painfully slowly, as $\sqrt{x^2+y^2}\to 0$. I suppose the terms can be combined into the even more dubious $$G(x,y,\theta) = - \frac{1}{2\pi^2} \sum_{i=1}^{\infty} \cos(i\theta)\left[\log \sqrt{x^2+y^2} + K_0\left(i \sqrt{x^2+y^2}\right)\right].$$ But looking at series expansions of the bracketed terms etc. hasn't yet yielded fruit.
I'm still hoping for a nice closed-form formula.
UPDATE 2: As Words points out in the comments, there is no hope for a closed form unless special cases like $\sum_{i=1}^{\infty} K_0(i)$ have a closed form. Even if a closed form is not possible, I would be happy even with a series that converges at a reasonable rate at any point away from the origin (including near $x=y=0$).
But I haven't given up yet on transforming $\sum K_0$ into something more reasonable. For instance, by Poisson summation \begin{align*} \sum_{i=1}^{\infty} K_0(i) &= \sum_{i=1}^{\infty} \begin{cases}\log i + K_0(i), & i\leq 1\\K_0(i), & i > 1\end{cases}\\ &= -\lim_{i\to 0} \frac{\log i+K_0(i)}{2} + \frac{\pi-2}{2} + \sum_{i=1}^{\infty}\left(\frac{\pi}{\sqrt{(2\pi i)^2+1}} - \frac{\textrm{Si}(2\pi i)}{\pi i}\right)\\ &= \frac{\pi+\gamma-\log 2 -2}{2} + \sum_{i=1}^{\infty}\left(\frac{\pi}{\sqrt{(2\pi i)^2+1}} - \frac{\textrm{Si}(2\pi i)}{\pi i}\right). \end{align*} Of course I don't have reason to believe at the moment that a series involving the sine integral is somehow better than one involving the modified Bessel function, but it's not obvious to me that there aren't additional or different transformations possible to tame the series.
Answering my own question with the best result so far in case anybody encounters this problem in the future.
For $\sqrt{x^2+y^2} \gg 0$: $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} - \frac{1}{2\pi^2}\sum_{i=1}^{\infty} \cos(i\theta) K_0\left(i \sqrt{x^2+y^2}\right).$$ For $\sqrt{x^2+y^2} \approx 0$ the above converges extremely slowly (and is undefined at $x^2+y^2=0$) and the following (derived by Poisson summation from the above) can be used instead: $$G(x,y,\theta) = \frac{\log 2-\gamma}{4\pi^2} - \frac{1}{2\pi}\sum_{i=-\infty}^{\infty} \left[\frac{\pi}{2\sqrt{(\theta+2\pi i)^2+x^2+y^2}} - \frac{\operatorname{Si}(\theta+2\pi i)}{\theta+2\pi i}\right].$$
In the above, $K_0$ is the modified Bessel function of the second kind, $\gamma$ is the Euler-Mascheroni constant, and $\operatorname{Si}$ is the sine integral.
I'm setting aside this problem for now, but will give a nice bounty if anybody in the future can improve on the above series (series whose summands do not depend on expensive special functions would be nice, for instance).