Given the Hankel function of first kind and zero order
$$ H_0^{(1)}(\omega|x-y|), $$
with $|x-y| \geq C >0$, I would like to calculate its Fourier transform, that is
$$ \int_{-\infty}^{\infty} H_0^{(1)}(\omega|x-y|) e^{-i \omega t} \; d\omega. $$
What I thought was the following:
Given
$$ U_y(x,t) = \frac{H(t-|x-y|)}{2\pi\sqrt{t^2-|x-y|^2}}, $$
where $H$ is the Heaviside function at $0$, we know that $U_y(x,t)$ is the fundamental solution of the 2D transient wave equation [1]:
$$ \begin{cases} (\partial_t^2 - \Delta)U_y(x,t) = \delta_{x=y} \; \delta_{t=0} & (x,t) \in \mathbb{R}^2\times \mathbb{R}, \\ U_y(x,t) = 0 & x \in \mathbb{R}^2, t\ll 0. \end{cases} $$
By applying the Fourier transform to the equation, we should get the Helmholtz equation
$$ (\Delta + \omega^2) \hat{U}(x,\omega) = \delta_{x=y}, $$
which has the Hankel function $H_0^{(1)}(\omega|x-y|)$ as the fundamental solution.
Therefore, I am tempted to write
$$ U_y(x,t) = \int_{-\infty}^{\infty} \hat{U}(x,\omega) e^{-i \omega t} \; d\omega, $$
that is
$$ \frac{H(t-|x-y|)}{2\pi\sqrt{t^2-|x-y|^2}} = \int_{-\infty}^{\infty} H_0^{(1)}(\omega|x-y|) e^{-i \omega t} \; d\omega, $$
but I have the feeling that I am missing something, since no books with tables of Fourier transforms mention it.
[1] Mansur, Carrer, Oyarzun, $Time$-$domain$ $BEM$ $techniques$