By using Fourier transforms and a conformal mapping

$$w=\sin \left(\frac{\pi z}{a}\right)$$

with $z=x+i y$ and $w=\xi+i \eta$, and a suitable real constant $a$, show that the solution to

$$\begin{array}{rlrl} \nabla^{2} \phi & =0 & -2 \pi \leqslant x \leqslant 2 \pi, y \geqslant 0 \\ \phi(x, 0) & =f(x) & -2 \pi \leqslant x \leqslant 2 \pi \\ \phi(\pm 2 \pi, y) & =0 & y>0, \\ \phi(x, y) & \rightarrow 0 & y \rightarrow \infty,-2 \pi \leqslant x \leqslant 2 \pi \end{array}$$

is given by

$$\phi(\xi, \eta)=\frac{\eta}{\pi} \int_{-1}^{1} \frac{F\left(\xi^{\prime}\right)}{\eta^{2}+\left(\xi-\xi^{\prime}\right)^{2}} d \xi^{\prime}$$

where $F\left(\xi^{\prime}\right)$ is to be determined.

In the case of $f(x)=\sin \left(\frac{x}{4}\right)$, give $F\left(\xi^{\prime}\right)$ explicitly as a function of $\xi^{\prime}$. [You need not evaluate the integral.]