= Solution
Use the <Fourier transform> convention $\widehat F(k)=\int_{\mathbb R}F(x)e^{-ikx}dx$, with inverse factor $1/(2\pi)$. Splitting the inverse integral at zero gives
$$
G(x)=\frac1{2\pi}\left(\int_0^\infty e^{-(a-ix)k}dk+\int_0^\infty e^{-(a+ix)k}dk\right)
=\boxed{\frac{a}{\pi(a^2+x^2)}}.
$$
This is the <Poisson kernel for the upper half-plane> at height $a$.
Take the <Fourier transform> in $x$ of the <Laplace equation>. The transformed boundary-value problem is
$$
\partial_y^2\widehat\psi-k^2\widehat\psi=0,\qquad \widehat\psi(k,0)=1,\qquad \widehat\psi(k,1)=e^{-|k|}.
$$
For $k\ne0$ the general solution is a <linear combination> of $e^{|k|y}$ and $e^{-|k|y}$; the two conditions set their coefficients to zero and one respectively. At $k=0$, the solution is the constant one. Inverting therefore gives
$$
\boxed{\psi(x,y)=\frac{y}{\pi(x^2+y^2)},\qquad 0<y<1.}
$$
The upper boundary holds pointwise. At the lower boundary the <Dirac delta function> must be interpreted distributionally: the kernel has integral one and concentrates at zero, so for every continuous compactly supported <test function> $h$, $\int\psi(x,y)h(x)dx\to h(0)$. This also specifies the solution in the assumed class admitting the stated <Fourier transforms>.
Back to article page
