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Use the Fourier transform convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{-ikx}dx$ and inverse factor $1/(2\pi )$. Laplace equation gives ${\hat{u}}_{yy}-k^2\hat{u}=0$. Selecting its decaying Fourier branch gives $\hat{u}(k,y)={\hat{u}}_0(k)e^{-|k|y}$, the standard bounded solution for bounded continuous boundary data. The inverse transform of the multiplier is the Poisson kernel for the upper half-plane:The convolution theorem therefore gives the Poisson integralFor suitably regular boundary data, this kernel is an approximate identity as $y\downarrow 0$, recovering $u_0$.

For $a>0$, contour integration gives ${\hat{u}}_0(k)=(\pi /a)e^{-a|k|}$. Multiplication by $e^{-y|k|}$ and inversion yieldThe stated form assumes $a>0$; if $a<0$, replace it by $|a|$ because the boundary function only depends on $a^2$. The case $a=0$ has singular boundary data and is outside this regular calculation.

The transform argument requires a growth condition. A zero-boundary harmonic function with pointwise vertical decay shows that the printed conditions alone do not ensure uniqueness:vanishes at $y=0$ and tends to zero as $y\to \infty$ for every fixed $x$. It can be added without changing those conditions, but its growth in $x$ excludes it from the bounded solution class. The displayed Poisson integral is the standard bounded solution for regular bounded boundary data; pointwise vertical decay by itself cannot justify a uniqueness claim.

For the rotationally invariant free-space fundamental solution of the Laplace equation, put $s=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}$. Away from the source, the radial Laplacian gives $(s{G}^{\prime }{)}^{\prime }=0$, hence $G=A\log s^2+C$. Integrate its Poisson equation over a disk of radius $r$ about the source. The divergence theorem gives $2\pi r(2A/r)=-1/{ϵ}_0$, soA fixed reference length can be inserted inside the logarithm and absorbed into $C$. The resulting electric field decays like $1/s$.

Strictly, the stated pointwise decay as $y\to \pm \infty$ alone does not justify “most general”: one can also add the harmonic function $\mathrm{Re}{e}^{(x+iy{)}^2}$, whose gradient has that decay for every fixed $x$. This is the same growth issue as a zero-boundary harmonic function with pointwise vertical decay. The displayed answer is the usual isotropic free-space choice; requiring the field to vanish uniformly in every direction at spatial infinity removes these extra harmonic additions. An additive constant remains undetermined.