Timeline for Error term for a Fourier integral
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2015 at 15:00 | comment | added | Tian An | If anyone would like to repost as an answer I'd be happy to accept! | |
| Oct 23, 2015 at 3:56 | history | edited | Tian An | CC BY-SA 3.0 |
added 13 characters in body
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| Oct 23, 2015 at 3:47 | history | edited | Tian An | CC BY-SA 3.0 |
added 13 characters in body
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| Oct 23, 2015 at 3:22 | comment | added | Noam D. Elkies | Meanwhile note that there's a factor $1/\pi$ missing (the integral over $\bf R$ of $\sin(Tx)/x$ is $\pi$, not $1$). | |
| Oct 23, 2015 at 3:21 | comment | added | Christian Remling | There's the obvious estimate $\int_{|t|\ge T} |\widehat{f}(t)|\, dt$ on the error, and that seems to be about all you can say without extra smoothness assumptions on $f$. | |
| Oct 23, 2015 at 3:20 | comment | added | Pig | Try to use Plancherel, so that your integral equals $\int_{\infty}^{\infty} \widehat{f}(x) \widehat{\frac{\sin Tx}{x}} dx$. The fourier transform of $\frac{\sin Tx}{x}$ is given explicitly as some indicator function on an interval of the scale $[-T,T]$. Approximate this indicator function with identity on real line, and you get your main term $f(0)$ by Fourier inversion. The error term would then take the shape of $\int_{[T,\infty] \cup [-\infty, T]} \widehat{f}(x)$ and depending on what you know about $\widehat{f}$, you should be able to give estimates on that. | |
| Oct 23, 2015 at 2:24 | history | asked | Tian An | CC BY-SA 3.0 |