Laplace Equation using fourier transforms? So the question is:
$$u_{xx} + u_{yy} = 0$$ 
$$-\infty < x < \infty$$
$y>0$ , $u(x,0)=f(x)$
As  $y \to \infty$  $u \to 0$
Express the solution as a convolution.
What i've done:
$$\hat{u}(k,y) = \int_{-\infty}^{\infty} dx \, u(x,y) e^{i k x}$$
so 
$$-k^2 \hat{u} + \frac{\partial^2 \hat{u}}{\partial y^2} = 0$$
which means that
$$\hat{u}(k,y) = A e^{k y} + B e^{-k y}$$
subject to 
$$\hat{u}(k,0) = \hat{f}(k)$$
This leads to 
A + B = $\hat{f}(k)$
But for U to go to $0$ as y goes to infinity A must equal $0$. Hence,
$$\hat{u}(k,y) = \hat{f}(k)e^{-k y}$$
Taking the inverse fourier transform of this i then get 
$$u(x,y) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \hat{f}(k)e^{k(ix- y)}  $$
= $f(ix-y)$
So yeah im not sure whether i've done something wrong or if i need to do something to display this as a convolution. Any help would be appreciated!
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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A 'particular solution'
$\ds{\exp\pars{\ic k_{x}}\exp\pars{-\verts{q}y}}$
$\ds{\pars{~\mbox{with}\ k_{x}, q \in \mathbb{R}~}}$ satisfies
$$
\pars{-k_{x}^{2} + q^{2}}\exp\pars{\ic k_{x}}\exp\pars{-\verts{q}y} = 0
\quad\implies\quad
\bbx{\ds{k_{x} = \pm q}}
$$
So, the solution can be written as
\begin{align}
\mrm{u}\pars{x,y} & = \int_{-\infty}^{\infty}\hat{\mrm{u}}\pars{q}
\expo{\ic qx - \verts{q}y}\,\dd q
\\[5mm]
\mrm{f}\pars{x} = \mrm{u}\pars{x,0} & = \int_{-\infty}^{\infty}\hat{\mrm{u}}\pars{q}
\expo{\ic qx}\,\dd q \implies \hat{\mrm{u}}\pars{q} =
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic qx}\,{\dd x \over 2\pi}
\end{align}

Then,
\begin{align}
\mrm{u}\pars{x,y} & =
\int_{-\infty}^{\infty}\bracks{%
\int_{-\infty}^{\infty}\mrm{f}\pars{x'}\expo{-\ic qx'}\,{\dd x' \over 2\pi}}
\expo{\ic qx - \verts{q}y}\,\dd q
\\[5mm] &=
{1 \over 2\pi}\int_{-\infty}^{\infty}\mrm{f}\pars{x'}
\int_{-\infty}^{\infty}\expo{-\ic q\pars{x - x'} - \verts{q}y}\,\dd q\,\dd x'
\\[5mm] & =
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\mrm{f}\pars{x'}
\int_{0}^{\infty}\expo{-\ic q\pars{x' - x} - qy}\,\dd q\,\dd x' =
{1 \over \pi}\,\Re\int_{-\infty}^{\infty}\mrm{f}\pars{x'}
{1 \over y + \pars{x' - x}\ic}\,\dd x'
\\[5mm] & =
\bbx{\ds{\int_{-\infty}^{\infty}\mrm{f}\pars{x'}
{y/\pi \over \pars{x - x'}^{2} + y^{2}}\,\dd x'}}
\end{align}
