Use the full Fourier Transform in x to solve Use the full Fourier Transform in x to solve:
$$ u_{xx} + u_{yy} =0 \, \,\,\, -\infty<x<\infty\,,\,\, 0<y<1$$
$$u(x,0) = \left\{
  \begin{array}{l l}
    0 & \quad \text{if $x<0$}\\
    e^{-x} & \quad \text{if $0\leq x$}
  \end{array} \right.$$
$$u(x,1)=0\,,\,\,\,-\infty < x < \infty$$
 A: Begin by writing the FT of $u$ as
$$\hat{u}(k,y) = \int_{-\infty}^{\infty} dx \: u(x,y) \, e^{i k x}$$
Then the PDE can be rewritten as
$$\frac{\partial^2}{\partial y^2} \hat{u}(k,y) = k^2 \hat{u}(k,y)$$
The general solution to this equation is
$$\hat{u}(k,y) = A e^{k y} + B e^{-k y}$$
Now use the boundary conditions to find $A$ and $B$.
$$\hat{u}(k,0) = A+B =\int_{0}^{\infty} dx \: e^{-x} \, e^{i k x} = \frac{1}{1-i k}$$
$$\hat{u}(k,1) = A e^k + B e^{-k} = 0$$
I leave it to the reader to do the algebra.  The end result is that we may write the solution $u(x,y)$ by taking the inverse FT of $\hat{u}(k,y)$:
$$u(x,y) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{u}(k,y) \, e^{-i k x}$$
or, 
$$\begin{align}u(x,y) &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{dk}{1-i k} \frac{\sinh{[(1-y) k]}}{\sinh{k}} e^{-i k x}\\ &= \frac{1}{4 \pi} \int_{0}^{\infty} dx' \: \frac{\sin{\pi y}}{\cosh{\pi (x-x')} - \cos{\pi y}} e^{-x'} \end{align}$$
For the latter integral, I used the convolution theorem.  It is unclear whether this integral has a closed form.
