The function $u(x, y)$ satisfies Laplace's equation in the half-space $y \geqslant 0$, together with boundary conditions

$$\begin{gathered} u(x, y) \rightarrow 0 \text { as } y \rightarrow \infty \text { for all } x \\ u(x, 0)=u_{0}(x), \text { where } x u_{0}(x) \rightarrow 0 \text { as }|x| \rightarrow \infty \end{gathered}$$

Using Fourier transforms, show that

$$u(x, y)=\int_{-\infty}^{\infty} u_{0}(t) v(x-t, y) d t$$

where

$$v(x, y)=\frac{y}{\pi\left(x^{2}+y^{2}\right)}$$

Suppose that $u_{0}(x)=\left(x^{2}+a^{2}\right)^{-1}$. Using contour integration and the convolution theorem, or otherwise, show that

$$u(x, y)=\frac{y+a}{a\left[x^{2}+(y+a)^{2}\right]}$$

[You may assume the convolution theorem of Fourier transforms, i.e. that if $\tilde{f}(k), \tilde{g}(k)$ are the Fourier transforms of two functions $f(x), g(x)$, then $\tilde{f}(k) \tilde{g}(k)$ is the Fourier transform of $\int_{-\infty}^{\infty} f(t) g(x-t) d t$.]