Asymptotic behaviours from Fourier transforms

Question

I have completely forgotten how one derives the asymptotic behavior in frequency space, given the asymptotic behavior of the function in real space (e.g. time). As an example example, it is often said that when $f(t)\sim t^\alpha$ for $t\to\infty$, then $f(\omega)\sim\omega^{-\alpha-1}$ for $\omega\to 0$. Aside from a dimensional analysis, how do you derive this result a bit more strictly?

Answer 1

When we replace $\omega$ in $$ \tilde f(\omega)=\int dt \exp(i\omega t) t^\alpha $$ by $k\omega$, the behavior of the exponential is exactly the same assuming that $t$ is replaced by $t/k$. But if that happens, $t^\alpha dt$ gets replaced by $t^\alpha \ dt / k^{\alpha+1}$. The whole integral calculating $\tilde f(k\omega)$ will therefore be the integral of the product of the same exponential times the same $t^\alpha d\alpha$ times $k^{-\alpha-1}$, so $\tilde f(\omega)$ has to go like $\omega^{-\alpha-1}$. All these things are just the parametric estimates of the leading terms.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ See Method of Steepest Descent : With $\ds{\int_{-\infty}^{\infty}t^{\alpha}\expo{\ic\omega t}\dd t}$ you get the ${\it saddle\ point}$ $\ds{t_{s} = \alpha\ic/\omega}$ and the ${\it asymptotic\ behavior}$ $$ \bbx{\pars{\root{2\pi}\,\ic^{\alpha - 1}\,\alpha^{\alpha + 1/2}\,\,\expo{-\alpha}}\ {1 \over \omega^{\alpha + 1}}} $$