Solution
= Solution
Part (b) gives
$$
\widetilde g(k)=-i\pi\operatorname{sgn}(k)
(e^{-a|k|}-e^{-3a|k|}).
$$
Since $(e^{-a q}-e^{-3aq})/\sinh(aq)=2e^{-2aq}$,
$$
\widetilde u=-i\pi\operatorname{sgn}(k)
\left(e^{-(2a-y)|k|}-e^{-(2a+y)|k|}\right).
$$
Inverting with part (b) yields
$$
\boxed{
u(x,y)=
\frac{x}{x^2+(2a-y)^2}
-\frac{x}{x^2+(2a+y)^2}}.
$$