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## 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.