The Fourier transform $\tilde{f}(k)$ of a function $f(x)$ and its inverse are given by

$$\tilde{f}(k)=\int_{-\infty}^{\infty} f(x) e^{-i k x} d x, \quad f(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \tilde{f}(k) e^{i k x} d k$$

(a) Calculate the Fourier transform of the function $f(x)$ defined by:

$$f(x)= \begin{cases}1 & \text { for } 0<x<1 \\ -1 & \text { for }-1<x<0 \\ 0 & \text { otherwise }\end{cases}$$

(b) Show that the inverse Fourier transform of $\tilde{g}(k)=e^{-\lambda|k|}$, for $\lambda$ a positive real constant, is given by

$$g(x)=\frac{\lambda}{\pi\left(x^{2}+\lambda^{2}\right)}$$

(c) Consider the problem in the quarter plane $0 \leqslant x, 0 \leqslant y$ :

$$\begin{aligned} \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}} &=0 ; \\ u(x, 0) &= \begin{cases}1 & \text { for } 0<x<1, \\ 0 & \text { otherwise; }\end{cases} \\ u(0, y)=\lim _{x \rightarrow \infty} u(x, y)=\lim _{y \rightarrow \infty} u(x, y) &=0 . \end{aligned}$$

Use the answers from parts (a) and (b) to show that

$$u(x, y)=\frac{4 x y}{\pi} \int_{0}^{1} \frac{v d v}{\left[(x-v)^{2}+y^{2}\right]\left[(x+v)^{2}+y^{2}\right]}$$

(d) Hence solve the problem in the quarter plane $0 \leqslant x, 0 \leqslant y$ :

$$\begin{aligned} \frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}} &=0 ; \\ w(x, 0) &= \begin{cases}1 & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases} \\ w(0, y) &= \begin{cases}1 & \text { for } 0<y<1 \\ 0 & \text { otherwise }\end{cases} \\ \lim _{x \rightarrow \infty} w(x, y)=\lim _{y \rightarrow \infty} w(x, y) &=0 \end{aligned}$$

[You may quote without proof any property of Fourier transforms.]