Timeline for Asymptotic decay rate of an oscillatory integral
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2020 at 18:42 | comment | added | Willie Wong | math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf gets to it very quickly (as an exercise with hints). | |
| Feb 25, 2020 at 16:35 | comment | added | Ludwig | Since I'm not familiar with Paley-Wiener theorem, could you suggest a good reference on this? Thanks. | |
| Feb 25, 2020 at 16:23 | comment | added | Willie Wong | In terms of rates: yes. Paley-Wiener is if-and-only-if: is if you have a strictly faster exponential decay rate, it will contradict the fact that your integrand has poles when \cos(x) = \cos(y) = k. | |
| Feb 25, 2020 at 14:51 | comment | added | Ludwig | I see, thanks. So, assuming that I could get rid of the $e^{n\epsilon}$ term, what I get is the upper bound $I(n)\le (k-\sqrt{k^2-1})^{2n}$. Is it possible to show that such a bound is tight? (btw, I'm not completely sure about the $\sqrt{n}$ factor: it's just a numerical guess) | |
| Feb 25, 2020 at 14:31 | comment | added | Willie Wong | bounded above by, up to a constant depending on $\epsilon$. (So I guess I could've written: for every $\epsilon > 0$ there exists a $C(\epsilon)$ (independent of $n$) such that $I(n) \leq C(\epsilon) e^{-2an + n\epsilon}$.) In general the constant $C(\epsilon)$ may blow-up as $\epsilon \searrow 0$; in your case since you have a very explicit integrand you may be able to show by doing the contour integral that there is a uniform bound on $C(\epsilon)$ in which case you can get rid of the $e^{n\epsilon}$ loss. I am not sure how to get the $\sqrt{n}$ factor though. | |
| Feb 25, 2020 at 14:29 | comment | added | Ludwig | Thanks! One quick question: what does $\lesssim_\varepsilon$ mean? | |
| Feb 25, 2020 at 14:21 | history | answered | Willie Wong | CC BY-SA 4.0 |