Fourier transform with branch cuts

Solution 1:

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).