Question is at the bottom in red.
Our boundary conditions on \(x\) are
\[
\lim_{x\to\pm\infty}\phi(x, y) = 0.
\]
Let \(\phi(x, y) = \varphi(x)\psi(y)\).
Then \(\frac{\varphi''}{\varphi} = - \frac{\psi''}{\psi} = -k^2\).
\begin{align}
\varphi(x) &\sim\left\{\cos(kx), \sin(kx)\right\}\\
\psi(y) &\sim\left\{\cosh(ky), \sinh(ky)\right\}
\end{align}
From the boundary conditions on \(y\), we see that we need to make a change of
variables.
That is, \(y\to b - y^*\); however, since the choice of variables are
arbitrary, let the change of variable be \(b - y\).
So we have is
\[
\psi(y) \sim\left\{\cosh(k(b - y)), \sinh(k(b - y))\right\}.
\]
Using the boundary condition \(\phi(x, b) = 0\), we have that
\[
\psi(y) \sim\sinh(k(b - y)).
\]
The general solution is then
\begin{alignat}{2}
\phi(x, y) &= \int_0^{\infty}\left[A(k)\cos(kx) + B(k)\sin(kx)\right]
\sinh(k(b - y))dk\\
\phi(x, 0) &= \int_0^{\infty}\left[A(k)\cos(kx) + B(k)\sin(kx)\right]
\sinh(kb)dk && ={} f(x)
\end{alignat}
Let \(A^*(k) = A(k)\sinh(kb)\) and \(B^*(k) = B(k)\sinh(kb)\),
and let's multiple through by \(\cos(k'x)\).
\begin{alignat}{2}
\phi(x, 0) &= \int_{-\infty}^{\infty}\int_0^{\infty}
\left[A^*(k)\cos(kx)\cos(k'x) + B^*(k)\sin(kx)\cos(k'x)\right]dkdx
&& ={} \int_{-\infty}^{\infty}f(x)\cos(k'x)dx\\
\phi(x, 0) &= \int_0^{\infty}\int_{-\infty}^{\infty}
A^*(k)\cos(kx)\cos(k'x)dxdk && ={} \int_{-\infty}^{\infty}f(x)\cos(k'x)dx
\end{alignat}
Note that \(\cos(kx)\cos(k'x) = \frac{1}{2}\left[\cos(x\Delta k) +
\cos[x(k + k')]\right]\).
From class, we know that the integral of \(\cos[x(k + k')]\) is zero.
\begin{align}
\int_{-\infty}^{\infty}\cos(kx)\cos(k'x)dx &=
\int_{0}^{\infty}\cos(x\Delta k)dx\\
&= \mathcal{R}e
\left\{\int_0^{\infty}\exp\left[(i\Delta k - \eta)x\right]dx\right\}\\
&= \mathcal{R}e\left\{\left.
\frac{\exp\left[(i\Delta k - \eta)x\right]}
{i\Delta k - \eta}\right|_0^{\infty}\right\}\\
&= \mathcal{R}e\left\{\frac{1}{-i\Delta k + \eta}\right\}\\
&= \frac{\eta}{\eta^2 + (\Delta k)^2}
\end{align}
In class, we have shown that
\(\frac{\eta}{\eta^2 + (\Delta k)^2} = \pi\delta(k - k')\).
Therefore, we have that
\[
\int_{-\infty}^{\infty}\cos(kx)\cos(k'x)dx = \pi\delta(k - k').
\]
Now we have that
\[
\phi(x, 0) = \pi\int_0^{\infty}A^*(k)\delta(k - k')dk = \int_{-\infty}^{\infty}f(x)\cos(k'x)dx
\]
The integral \(\int_0^{\infty}A^*(k)\delta(k - k')dk = A^*(k')\) so
\begin{align}
A^*(k') &= \frac{1}{\pi}\int_{-\infty}^{\infty}f(x)\cos(k'x)dx\\
A(k) &= \frac{1}{\pi\sinh(kb)}\int_{-\infty}^{\infty}f(x')\cos(kx')dx'
\end{align}
Now we can write the general solution \cref{lapgensoln} as
\[
\phi(x, y) = \frac{1}{\pi}\int_0^{\infty}\int_{-\infty}^{\infty}
\frac{\sinh(k(b - y))}{\sinh(kb)}f(x')\cos(kx)\cos(kx')dx'dk.
\]
Again we have that \(\cos(kx)\cos(kx') = \frac{1}{2}\left[\cos(k(x' - x)) +
\cos(k(x' + x))\right]\) and \(\cos(k(x' + x))\) integrates out to zero.
Let \(x' = \xi\) and \(k = u\).
Then the solution is
\[
\phi(x, y) = \frac{1}{2\pi}\int_0^{\infty}\int_{-\infty}^{\infty}
\frac{\sinh(u(b - y))}{\sinh(ub)}f(\xi)\cos(u(\xi - x))d\xi dk.
\]
I have an extra factor of 1/2. Where am I missing a factor of 2?
