Fourier Transform of Laplace's equation in the first quadrant Here's the question I'm trying to solve:

We have Laplace's equation in the first quadrant: $U_{xx}+U_{yy} = 0$, with boundary value conditions $U_x(0,y) = 0$ and $U(x,0)=F(x)$. The goal is to find $U(x,y)$.

I want to apply the Fourier Sine (or Cosine) transformation for the Laplace equation:

$S(F(x)) = \int_0^\infty F(x)\sin(\alpha x)dx$,  $C(F(x)) = \int_0^\infty F(x)\cos(\alpha x)dx$

If I apply the sine transformation, we will have
$$
S(U(x,y)) = u(\alpha,y), -\alpha^2u+\alpha U(0,y)+u_{yy}=0
$$
Then I got really confused. I'm not pretty sure how to apply the boundary conditions. Should I use the Cosine transformation here? How I can find $U(x,y)$? Thanks so much for the help!
 A: A better choice is to use the cosine transform with respect to the $x$ coordinate. You can show that then the function $u_c(a,y)=\int_{0}^\infty U(x,y)\cos a xdx$ obeys the equation
$$(u_c)_{yy}-a^2u=-U_x(0,y)=0$$
Here we have assumed that $\lim_{x\to\infty}F(x)=0$. The general solution to the equation above contains rising exponentials that explode at infinity. To construct a well-behaved solution we only keep the term $\propto e^{-ay}$ and applying the boundary condition $u_{c}(a,0)=f_c(a)$ we get
$$u_{c}(a,y)=f_c(a)e^{-ay}$$
The cosine transform is it's own inverse here, given that we're interested only in $x>0$ and inverting in this way recovers the even extension of $U$ but it nevertheless should agree with the full solution at $x>0$. We find that
$$U(x,y)=\frac{2}{\pi}\int_0^\infty  dx f_c(a)e^{-ay}\cos ax$$
which after substituting $f_c(a)=\int_0^\infty F(t)\cos at dt$ and performing the integral over $a$ yields $(z=x+iy)$:
$$\pi U(x,y)=z\int_{0}^\infty dt\frac{F(t)}{t^2+z^2}+\bar{z}\int_{0}^\infty dt\frac{F(t)}{t^2+\bar{z}^2}$$
and more explicitly in terms of $x,y$:
$$U(x,y)=\int_{0}^\infty \frac{dt}{2\pi}~F(t)\frac{x(t^2+x^2+y^2)}{(x^2-y^2+t^2)^2+4x^2y^2}$$
PS It would be interesting to study the family of solutions that can be constructed for a special super-exponentially decaying $f_c(a)$ (for example $f_c(a)\propto e^{-a^2}$), in which case we have more freedom to choose the solutions but seemingly not enough boundary conditions to fix a unique one.
