Timeline for Estimation of the integral $\int_a^b e^{2\pi i f(x)} dx $
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| Jul 21, 2018 at 4:34 | comment | added | Gerry Myerson | @alpoge, plus formatting: $$\int_a^be\bigl(f(x)\bigr)\,dx={e\bigl(f(b)\bigr)\over2\pi if'(b)}-{e\bigl(f(a)\bigr)\over2\pi if'(a)}+O\left(|b-a|\sup_x{|f''(x)|\over|f'(x)|^2}\right)$$ | |
| Jul 21, 2018 at 3:00 | comment | added | alpoge | Don’t you get that \int_a^b e(f(x)) dx = e(f(b))/[2\pi i f’(b)] - e(f(a))/[2\pi i f’(a)] + O(|b-a| \sup_x |f’’(x)| / |f’(x)|^2)? I’ve just written the integrand as [2\pi i f’(x) e(f(x))] / [2\pi i f’(x)] and integrated by parts (forgive me if this is nonsense! It’s late and I’m writing from my phone.). Dunno if this is useful to you, but at least it shows you the boundary term (basically) preventing you from having the desired bound. | |
| Jul 21, 2018 at 2:52 | comment | added | MichaelGaudreau | You may be interested in Chapter 8 of Stein's Harmonic Analysis or these notes by Tao math.ucla.edu/~tao/247b.1.07w/notes8.pdf | |
| Jul 21, 2018 at 2:15 | review | Close votes | |||
| Jul 31, 2018 at 8:18 | |||||
| Jul 21, 2018 at 2:07 | history | edited | GH from MO |
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| Jul 21, 2018 at 2:06 | comment | added | GH from MO | An obvious upper bound is $|a-b|$. As Christian Remling pointed out, this cannot be improved asymptotically for $b-a$ very small. | |
| Jul 21, 2018 at 1:55 | comment | added | Christian Remling | This can't work because the LHS is $\simeq b-a$ when you send $b-a\to 0$, no matter how large $f'$ is. | |
| Jul 21, 2018 at 1:53 | history | edited | user119197 | CC BY-SA 4.0 |
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| Jul 21, 2018 at 1:51 | history | edited | Christian Remling | CC BY-SA 4.0 |
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| Jul 21, 2018 at 1:49 | history | asked | user119197 | CC BY-SA 4.0 |