There is a well-known theorem that states that for $f$ continuous and $f,\hat f$ integrable, $$f(0)=\frac{1}{\pi}\lim_{T\to\infty}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx.$$ So it should be that $$\frac{1}{\pi}\int_{-\infty}^\infty f(x)\frac{\sin(Tx)}{x}dx=f(0)+\text{(Error term)}.$$ Are there known explicit formulas for estimating this error term? Answers with added assumptions (e.g, smooth, even, etc.) are welcome.
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3$\begingroup$ 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. $\endgroup$– PigCommented Oct 23, 2015 at 3:20
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4$\begingroup$ 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 $f$. $\endgroup$– Christian RemlingCommented Oct 23, 2015 at 3:21
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2$\begingroup$ Meanwhile note that there's a factor $1/\pi$ missing (the integral over $\bf R$ of $\sin(Tx)/x$ is $\pi$, not $1$). $\endgroup$– Noam D. ElkiesCommented Oct 23, 2015 at 3:22
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$\begingroup$ If anyone would like to repost as an answer I'd be happy to accept! $\endgroup$– Tian AnCommented Oct 25, 2015 at 15:00
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