Timeline for Asymptotic behavior of a double oscillatory integral
Current License: CC BY-SA 4.0
18 events
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| Aug 1, 2022 at 20:47 | vote | accept | Medo | ||
| Aug 1, 2022 at 20:45 | comment | added | Iosif Pinelis | @Medo : This additional question is for another post, possibly on another forum. At this point, I can say two things: (i) for (12) to hold, the integral $\int_{\mathbb R}dx\,\frac{\psi(x)}{|x|}$, defining $c_\psi$, must obviously exist in some sense and (ii) again, given your condition "$\psi$ is a smooth real-valued function with compact support", $\psi(x)/x$ will be always integrable if $\psi(0)=0$. | |
| Aug 1, 2022 at 20:35 | comment | added | Medo | I genuinely appreciate your patience. The question is simple now; The integrability of $x\mapsto \psi(x)/x$ is sufficient for the asymptotic $(12)$. Is it necessary ? (hence the numerical experiments...) | |
| Aug 1, 2022 at 20:21 | comment | added | Iosif Pinelis | Previous comment continued: (ii) One can hardly rely here on numerical experiments. Indeed, $\ln t$ is varying very slowly (and even $\sqrt t$ is varying slowly), whereas the integral for large $t$ is highly oscillatory. Anyhow, I do not see the point of the "Numerical experiments" sentence in your latter comment. | |
| Aug 1, 2022 at 20:20 | comment | added | Iosif Pinelis | @Medo : (i) "The assumption that $\psi(x)/x$ is integrable is strong. [...] I think we should only need $\psi(x)/x$ to be locally integrable away from zero, which is the case of course." -- In fact, $\psi(x)/x$ will always be integrable if $\psi(0)=0$, given your condition that "$\psi$ is a smooth real-valued function with compact support". (Moreover, given these conditions, $\psi(x)/x$ will also be a smooth real-valued function with compact support.) I have now added a corresponding comment to the answer. | |
| Aug 1, 2022 at 20:05 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2022 at 19:39 | comment | added | Medo | Thank you so much. The assumption that $\psi(x)/x$ is integrable is strong. Numerical experiments suggest that $I(t)\sim \log{t}/\sqrt{t}$ when tested with $\psi$ a positive bump centered at $x=0$ or a Gaussian. I think we should only need $\psi(x)/x$ to be locally integrable away from zero, which is the case of course. | |
| Aug 1, 2022 at 12:59 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2022 at 12:45 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2022 at 12:41 | comment | added | Iosif Pinelis | @Medo : (i) The condition $A\to\infty$ was used to conclude that $\sqrt{1-2 i y^2}\sim y\sqrt{-2 i}$ for $y>A$. (ii) I have now added the case $\psi(0)=0$. | |
| Aug 1, 2022 at 12:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2022 at 8:55 | comment | added | Medo | Now, it remains to remove the assumption that $\psi(0)\ne 0$. I appreciate your effort and I really want to accept your answer but I fear the question will no longer be of interest to others. So (+1) for now. | |
| Aug 1, 2022 at 8:07 | comment | added | Medo | The statement ''So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$'' is not really necessary I think. One can take $A=1$ and get a precise estimate for $\int_0^1=o(1/\sqrt{t})$. | |
| Aug 1, 2022 at 8:05 | vote | accept | Medo | ||
| Aug 1, 2022 at 8:45 | |||||
| Aug 1, 2022 at 2:41 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Aug 1, 2022 at 0:37 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2022 at 23:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 31, 2022 at 22:21 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |