2D Green's Function - Bessel function equivalence

[RUber]

Homework Statement

This is not a homework problem per se, but I have been working on it for a few days, and cannot make the logical connection, so here it is:
-- The problem is to show that
$\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{-\sqrt{\xi ^2 + \alpha^2 } |y-y'| + i \xi (x-x') }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi= \frac i4 H_0^{(1)}(i \alpha R) = \frac 1{2\pi} K_0(\alpha R)$
Where $R=\sqrt{(x-x')^2 + (y-y')^2 }$.
$\alpha$ is the constant wavenumber of a time-harmonic wave.
This is part of the nitty-gritty explanation of the 2D free space Green's function for waves.
Because the answer is a Bessel function, I expect there to be a change to cylindrical coordinates, which I attempt in section 3.

Homework Equations

I have a claim that the equation above is equivalent to
$i \int_{C(\phi)} e ^{ i k r \cos \beta } d\beta$
with $C(\phi)$ defined by:
$C(\phi) = \left\{ \begin{array}{l l} \displaystyle x=-|\phi| & y \text{ from } i \infty \text{ to } 0 \\ y=0 & x \text{ from } -|\phi| \text{ to } \pi - |\phi|\\ x= \pi - |\phi| & y \text{ from } 0 \text{ to } -i \infty \end{array} \right.$

Referring to Gradshteyn and Ryzhik, \cite{Gradshteyn2000}, this functional form is equivalent to $\rmi \pi H_0^1(kr)$.

The Attempt at a Solution

Attempting to change to polar coordinates centered at (x',y') using $x = R\cos\theta, y = R\sin\theta$ gives:
$\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{-\sqrt{\xi ^2 + \alpha^2 } R\sin\theta + i \xi R\cos\theta }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi$ for $\theta \in [0, \pi]$
and
$\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{\sqrt{\xi ^2 + \alpha^2 } R\sin\theta + i \xi R\cos\theta }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi$ for $\theta \in [ -\pi,0]$.
The reference I have simply says that with an appropriate subsitition, the equivalence can be seen.
I cannot see what change of variables to make from here. I wonder if perhaps there is a more intuitive form of the Bessel K function that might help me see the connection.
Thank you to anyone who might be able to point me in the right direction.