Fourier transform with branch cuts I want to compute the Inverse Fourier transform of the following function (it appears as a certain correlation function in a physical model I am interested it):
$$
\widetilde{f}(\omega) = \frac{2i}{\omega + \sqrt{\omega^2 - 1}}
$$
$$
f(t) = \int_{-\infty}^{\infty} \frac{d\omega}{2\pi} e^{-i \omega t}
\widetilde{f}(\omega) = \,?
$$
Can this be done explicitly? If yes, how does one do the integral and what is the answer?
If it cannot be done explicitly, then how can one derive the fact that $f(t)$ will behave like $e^{\pm i t} t^{-3/2}$ for large $t$?
 A: This can be done explicitly: using that
$$ \frac{1}{\omega+\sqrt{\omega^2-1}} = \omega-\sqrt{\omega^2-1} $$
(rationalise the denominator), we can express the Fourier transform as
$$ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-i\omega t}(\omega-\sqrt{\omega^2-1}) \, d\omega. $$
The first term is (distributionally) equal to
$$ i\delta'(t), $$
by using the derivative rule on the Fourier transform of the delta-function. Obviously this has no large-$t$ asymptotics.
The interesting term is
$$ -\frac{1}{2\pi}\int_{0}^1 i\cos{\omega t} \, \sqrt{1-\omega^2} \, dt - \frac{1}{2\pi}\int_1^{\infty} \cos{\omega t} \, \sqrt{\omega^2-1} \, d\omega, $$
which tables of Fourier transforms (or the integral representations here, or computer algebra software) suggest has the value
$$ -\frac{Y_1(\left| t\right| )}{2 \left| t\right| }-\frac{i J_1(\left| t\right| )}{2 \left| t\right| }, $$
where these are the Bessel functions of order $1$. This is in fact $i H^{(2)}_1(\lvert t \rvert)/(2\lvert t \rvert)$, which has asymptotic
$$ \frac{i H^{(2)}_1(\lvert t \rvert)}{2\lvert t \rvert} \sim -\frac{(1+i)}{2\sqrt{\pi}\lvert t \rvert^{3/2}}e^{-i\lvert t \rvert}, $$
which is, I think, what you are looking for (see here for the asymptotic relation).
