I'm not sure if I'm correct in my logic here, but I wish to solve the boundary value problem:
$$\Delta u(x,y)=0, \ \ \ \ \ x,y\geq0, \ \ \ \ \ u(x,y)\text{ is bounded}$$ Subject to boundary conditions: $$u_x(0,y)=0$$ $$u(x,0)=\frac{1}{1+x^2}$$
I want to use the strategy of solving the general problem described by:
$$\Delta u(x,y)=0, \ \ \ \ \ x,y\geq0, \ \ \ \ \ u(x,y)\text{ is bounded}$$ Subject to boundary conditions: $$u_x(0,y)=0$$ $$u(x,0)=f(x)$$
So far, I've taken the Fourier transform of the equation, yielding
$$-\xi^2\hat{u}(\xi,y)+\hat{u}_{yy}(\xi,y)= 0$$
In other words $$\hat{u}_{yy}(\xi,y)=\xi^2\hat{u}(\xi,y)$$
Which has general solution $$\hat{u}(\xi,y)=a(\xi)\exp(\xi y)+b(\xi)\exp(-\xi y)$$
Then because the function is bounded, we can set (I'm not sure if this step is valid)
$$ c(\xi):=\begin{cases} \ a(\xi), & \xi < 0,\\ b(\xi), & \xi>0. \end{cases} $$ which makes the general solution $\hat{u}(\xi,y)=c(\xi)\exp(-|\xi|y)$
Since $\exp(0)=1$, we have $\hat{u}(\xi,0)=c(\xi) \implies \hat{u}(\xi,y)=\hat{u}(\xi,0)\exp(-|\xi|y)$
Taking the Fourier transform of the first Boundary condition yields 0=0, so this isn't helpful. (Frankly the fact that it was included in the question makes me think I'm doing something wrong because I'm neglecting it here.)
Applying the Fourier transform to the second boundary condition gives $$\hat{u}(\xi,0)=\mathcal{F}\left\{f(x)\right\}$$ hence $$\hat{u}(\xi,y)=\mathcal{F}\left\{f(x)\right\}\exp(-|\xi|y)$$
Then by the Fourier inversion theorem:
$$u(x,y)= \frac{1}{2\pi}\int_{-\infty}^\infty \mathcal{F}\left\{f(x)\right\}\exp(-|\xi|y)\exp(-i\xi y)d\xi$$ $$=\frac{1}{2\pi}\int_{-\infty}^\infty \mathcal{F}\left\{f(x)\right\}\exp(-|\xi|y-i\xi y)d\xi$$
So what's with that boundary condition not being used? Also is the step I'm unsure of the validity of valid?
In order to comply with the boundary condition $u_x(0,y)=0$, you should use the cosine Fourier transform, $$ u(x,y)=\int_0^{\infty}U(k,y)\cos(kx)\,dk. \tag{1} $$ Plugging $(1)$ into the Laplace equation, we reduce it to the ODE $$ \left(\frac{\partial^2}{\partial y^2}-k^2\right)U(k,y)=0\qquad(k\geq 0). \tag{2} $$ The solution to $(2)$ that is bounded for $y\geq 0$ is $U(k,y)=A(k)e^{-ky}$, so we end up with $$ u(x,y)=\int_0^{\infty}A(k)e^{-ky}\cos(kx)\,dk, \tag{3} $$ where $A(k)$ is determined by the other boundary condition, $u(x,0)=f(x)$: $$ \int_0^{\infty}A(k)\cos(kx)\,dk = f(x) \implies A(k)=\frac{2}{\pi}\int_0^{\infty}f(x)\cos(kx)\,dx. \tag{4} $$