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Take the source at the origin and write ${\rho }^2=y^2+z^2$. Repeated reflection in $x=\pm 1$ places the sources at $(2n,0,0)$ with strengths $(-1{)}^n$, $n\in \mathbb{Z}$. There are countably infinitely many images; $n=0$ is the original source. A convergent prescription groups adjacent opposite charges:Each bracket is $O(|m{|}^{-2})$ on compact sets away from the source, so the paired series converges absolutely. The individual unsigned point-charge terms would not be absolutely summable. On $x=1$, reflection pairs the index $n$ with $1-n$; on $x=-1$, it pairs $n$ with $-1-n$. Their distances agree and signs oppose. Symmetric finite cutoffs leave only a vanishing far-tail contribution, proving both zero boundary values. All nonzero-index sources lie outside the slab. This is the centered-source version of the Dirichlet Green function between parallel planes, selected to decay at large transverse distance.

Reflecting repeatedly in the two parallel planes creates an infinite method of images. For ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$, introduceThe Dirichlet Green function between parallel planes isEach bracket is $O(n^{-2})$, so the paired series converges absolutely away from the sources. At $x=-1$, reflection pairs the positive image indexed by $n$ with the negative image indexed by $-n$. At $x=1$ it pairs that positive image with the negative image indexed by $1-n$. The distances agree in each pairing, making both boundary values zero. This can be checked with symmetric finite partial sums first; the unmatched end terms tend to zero, so no subtraction of separately divergent series is needed. Only ${\mathbf{r}}_0^{+}={\mathbf{r}}^{\prime }$ lies within the slab, so ${\Delta }_{\mathbf{r}}{G}_{\mathrm{slab}}=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. This is the usual Green function decaying at large transverse separation; such a condition selects the physical solution in the unbounded region.