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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?

sunenj73
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    This would be better suited to math.SE – lemon May 20 '16 at 09:30
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    Perhaps, but I am not so interested in a mathematician's derivation (involving e.g. epsilons and deltas etc), more a physicist's derivation, if you know what I mean. – sunenj73 May 20 '16 at 09:51
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    This is called the Hardy-Littlewood Tauberian Theorem see https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_tauberian_theorem , and is not a simple result. – hyportnex May 20 '16 at 11:55
  • Thanks, hyportnex, that's helpful. I was hoping for a qualitative argument, a la $f(\omega)=\int e^{i\omega t}f(t)dt=\int e^{i y}f(y/\omega)dy/\omega $ which would allow me to use the asymptotic behavior of $f(t)$ as $y/\omega\to\infty$ when $\omega\to 0$. But I don't feel comfortable about the contribution/behavior from the region close to $y=0$ – sunenj73 May 20 '16 at 12:55

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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 ${\it asymptotic\ behavior}$ $$ \bbx{\pars{\root{2\pi}\,\ic^{\alpha - 1}\,\alpha^{\alpha + 1/2}\,\,\expo{-\alpha}}\ {1 \over \omega^{\alpha + 1}}} $$

Felix Marin
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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.

Luboš Motl
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  • This is kind of like my comment above, but still has the problem close to 0, because $f(t)$ is not everywhere like $t^\alpha$. I guess we need to argue that the contribution ($\sim \int_0^T f(t) e^{i\omega t}dt$) is subdominant/vanishes? Otherwise we argued that $\tilde f(\omega)\sim a+b \omega^{-\alpha-1}$ – sunenj73 May 20 '16 at 13:23
  • Dear @sunenj73 - apologies, when I was writing the answer, your comment with formulae was unreadable due to a TeX error. ... The contribution you mention is surely subdominant but doesn't vanish. My assumption was that one scales $T$ as well, and chooses it so that the integral above $T$ dominates over or at most parametrically matches the integral below $T$. – Luboš Motl May 20 '16 at 13:37
  • I see that it is subdominant for $\alpha > -1$ -- so does the result only hold for this case? – sunenj73 May 20 '16 at 15:53
  • Yes, or to say the least, there surely has to be a similar inequality. I am sure that the original statement doesn't hold for negative $\alpha$, for example, does it? For a negative $\alpha$, one only says that the function is nicely decreasing to zero at infinity but this doesn't mean that the Fourier transform decreases near $\omega=0$ - for example, we may always add a Gaussian to $f(t)$ and the Gaussian decreases superquickly at infinity so doesn't change the asymptotics but its Fourier transform, another Gaussian, isn't zero at zero. ;-) – Luboš Motl May 20 '16 at 16:41
  • Right :) I guess in such a case, we can determine the low $\omega$ behavior from expanding the $e^{i\omega t}$ within the integral. – sunenj73 May 20 '16 at 19:35
  • It's unfortunate that my answer was migrated after it was written because it was specifically meant to be an answer appropriate for people doing physics. An appropriate mathematics answer would look different (even mine). – Luboš Motl May 22 '16 at 04:22
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    I agree, it belongs in physics. – sunenj73 May 23 '16 at 10:28
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    Which by the way is also supported by the fact that some very related questions in this forum are left unanswered, while mine got an answer in physics… – sunenj73 May 23 '16 at 11:34