Solve Laplace equation in the upper half plane I need to solve 
\begin{eqnarray}
u_{xx} + u_{yy} = 0 \quad \quad y>0 \quad -\infty < x< \infty
\end{eqnarray}
With boundary condition 
\begin{eqnarray}
\frac{\partial u(x,0)}{\partial y} = h(x) \mbox{ and } |u(x,y)|<\infty
\end{eqnarray}
I call $U(w,y)$ the Fourier transorm of $u(x,t)$. I call $H(w)$ the Fourier transform of $h(x)$. Substituting $U(w,y)$ in the partial differential equation I get:
\begin{eqnarray}
\frac{\partial^2 U}{\partial y^2} = w^2U
\end{eqnarray}
This differential equation has solution
\begin{eqnarray}
U(w,y) = C(w)e^{-|w|\cdot y}
\end{eqnarray}
Differentiating gives
\begin{eqnarray}
U_y = -|w| C(w) e^{-|w|\cdot y}
\end{eqnarray}
Plugging in the boundary condition $U_y(w,0) = H(w)$ gives
\begin{eqnarray}
U(w,y) = -\frac{H(w)}{|w|}e^{-|w|y}
\end{eqnarray}
Is this step correct? And how do I need to continue if it is? I need to transform back, such that
\begin{eqnarray}
u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} -\frac{H(w)}{|w|}e^{-|w|y} e^{iwx}dw
\end{eqnarray}
How can I solve this integral? Does convolution apply?
 A: consider the  Fourier transform of $u$ $u_{xx}$ ,$u_{yy}$ and $h(x)$ $$u(x,y)=\sum_{n=1}^{\infty}u_n(y)\sin(\frac{n\pi  x}{l})$$$$u_{xx}=\sum_{n=1}^{\infty}w_n(y)\sin(\frac{n\pi  x}{l})$$$$u_{yy}=\sum_{n=1}^{\infty}v_n(y)\sin(\frac{n\pi x}{l})$$ $$h(x)=\sum_{n=1}^{\infty}h_n(y)\sin(\frac{n\pi  x}{l})$$such that $$u_n(y)=\frac{2}{l}\int_{0}^{l}u(x,y)\sin(\frac{n\pi x}{l})$$$$v_n(y)=\frac{2}{l}\int_{0}^{l}u_{xx}\sin(\frac{n\pi x}{l})$$$$w_n(y)=\frac{2}{l}\int_{0}^{l}u_{yy}\sin(\frac{n\pi x}{l})$$$$h_n(y)=\frac{2}{l}\int_{0}^{l}h(x)\sin(\frac{n\pi x}{l})$$ and we conclude following statments  $$1.\begin{eqnarray}
\frac{\partial^2 U_n(y)}{\partial y^2} = v_n(y)
\end{eqnarray}$$$$2.v_n(y)+w_n(y)=0$$$3.$ and integrate $w_n(y)=\frac{2}{l}\int_{0}^{l}u_{yy}\sin(\frac{n\pi x}{l})$ by part and you take $w_n(y)$respect to $ u_n(y)$ therefore by use of $1,2,3$ you can  easily find  $ u_n(y)$ then you can find$ u(x,y ) $
A: You can do that last integral if you replace $H(w)$ with its definition as the Fourier transform of $H(x)$. You should then be able to rearrange the (double) integral into the form
$$ \int dx'\, h(x') G(x,y,x') $$
where $G(x,y,x')$ is some function that should be able to find in closed form (it is also the solution to your problem in the case $h(x)=\delta(x - x')$).
