Let the Fourier transform of $f(t)$ be defined as $F(\omega) = \int_{-\infty}^\infty dt f(t) e^{i\omega t}$ for values of $\omega$ where the integral exists. What are the precise conditions on $F(\omega)$ so that $f(t)$ has the large time asymptotics, $t\to\infty$, as
$$f(t) = C \frac{ e^{-at} }{ t^\nu} \left( 1 + O\left( \frac{1}{t^\mu} \right) \right)$$
where $\mu > 0$? Once this type of asymptotics is present, how does one recover $a, C$ and $\nu$ from $F(\omega)$?
Examples:
1.
$F(\omega) = \frac{2\alpha}{\omega^2+\alpha^2}$, with $\alpha>0$, in this case we know that $f(t) = e^{-\alpha|t|}$ So this corresponds to $a = \alpha$, $\nu=0$, $C=1$. In this case the pole of $F(\omega)$ results in the exponential fall-off.
2.
$F(\omega) = \frac{1}{\sqrt{1-\omega^2}} \;\; {\rm if} \;\; -1 < \omega < 1,\;\; {\rm otherwise} \;\; 0$, which gives $f(t) = \pi J_0(t)$ a Bessel function. Asymptotically we have $J_0(t) = \sin(t+\pi/4) \sqrt{\frac{2}{\pi t}} + \ldots$. Now $\nu \neq 0$ and indeed we have a branch cut in $F(\omega)$ at $\omega = \pm 1$.
3.
$F(\omega) = \frac{1}{\sqrt{1+\omega^2}}$, which gives $f(t) = 2K_0(|t|)$, another Bessel function. Asymptotically $f(t) = 2 e^{-t} \sqrt{\frac{2}{\pi t}} + \ldots$, so again we have $\nu = 1/2$ from a branch cut at $\omega = \pm i$.
Basically: what is the precise correspondence between singularities of $F(\omega)$ and the asymptotic form of $f(t)$? Is it true that poles lead to $\nu=0$ and branch cuts are needed for $\nu \neq 0$?
