Timeline for Oscillatory integrals

11 events

when toggle format	what		by	license	comment
Apr 19, 2019 at 23:22	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 87 characters in body
Apr 19, 2019 at 22:08	comment	added	Ali		yes this is what I meant as well.
Apr 19, 2019 at 21:33	comment	added	Iosif Pinelis		@Ali : I have fixed the mistake. In fact, all values of $n\in(0,1)$ are like $n=1/2$.
Apr 19, 2019 at 21:30	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 1149 characters in body
Apr 19, 2019 at 15:40	comment	added	Ali		I don't think your solution is correct for the case when $n=\frac{1}{2}$, but it is correct for all the other cases. For the case $n=\frac{1}{2}$ there always is a strong singularity at one of the end points depending on the branch cut that will cancel out the $\epsilon^{\frac{1}{2}}$.
Apr 13, 2019 at 8:18	vote	accept	Ali
Apr 12, 2019 at 22:57	comment	added	Iosif Pinelis		@Ali : Oops! Previously, by mistake, I put $\|\sin u\|^n$ into the denominator, rather than the numerator. Now it's only significantly simpler, and I do get your $2\pi$ for $n=1$.
Apr 12, 2019 at 22:55	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 455 characters in body
Apr 12, 2019 at 19:12	comment	added	Ali		This does not work out when $n=1$ as you should not get the $\epsilon^{-1}$ asymptotic.
Apr 12, 2019 at 18:33	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 37 characters in body
Apr 12, 2019 at 18:25	history	answered	Iosif Pinelis	CC BY-SA 4.0