 Solution

Use the horizontal Fourier convention $\tilde{f}(\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{R}}f(\mathbf{R})d^{2}R$. For each nonzero $k=|\mathbf{k}|$, the Poisson equation becomesDecay away from the sheet and continuity at it give $\tilde{\Phi }=C{e}^{-k|z|}$. Its derivative jump is $-2kC=4\pi G\tilde{\Sigma }$, hence the off-plane razor-thin Poisson kernel isThe spatially uniform mode has a different vertical solution and an arbitrary additive potential reference; it is not obtained by substituting $k=0$ into this decaying-mode formula.