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I am wondering whether the following uniform upper bound holds:

$|\int_a^{2a}\frac1t\sin(N b^2t)\exp(iNbt^2)dt|\le Cab^2,$

where $0<a<b<1$, $N>N_0(a,b)\gg1$, and $C$ is a constant independent of $N,a,b$.

Remark: (1) A trivial upper bound is 1; (2) If we write $\frac1t\sin(Nb^2t)=b^2\int_0^N \cos(ub^2t)du$, then we get an easy upper bound $Nab^2$; (3) This uniform upper bound can be verified by Matlab but I have no idea how to prove it; (4) Equivalently, $\limsup_{N\to\infty}\frac1{ab^2}|\int_a^{2a}\frac1t\sin(N b^2t)\exp(iNbt^2)dt|\le C.$

Any comments are welcome!:)

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    $\begingroup$ What do you mean by $N\gg 1$? If this means larger than an absolute constant (i.e. independent of $a,b$), then take $b=\epsilon, a = \epsilon^2, N = \epsilon^{-3}$ and $\epsilon\ll 1$. The integral, upon the change of variable $t\mapsto \epsilon t$, is $\int_{\epsilon}^{2\epsilon} dt \frac{\sin{t}}{t} \exp(i t^2) = \epsilon + O(\epsilon^3)$, whereas the bound you ask for is of order $\epsilon^4$ (caveat: if I haven't made a mistake!). If we're allowed to take $N$ sufficiently large depending on $a,b$, then I think repeated integration by parts (the stationary phase method) will give something. $\endgroup$ – alpoge Mar 30 '17 at 3:56
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    $\begingroup$ I agree with alpoge that the standard methods for oscillatory integrals should give you something, but it's probably not totally straightforward because (after writing the sine in terms of exponentials) you potentially have the critical point $t=b/2$ of the phase inside your region of integration. $\endgroup$ – Christian Remling Mar 30 '17 at 19:21
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    $\begingroup$ Agreed! Even so you save a $\sqrt{N}$ from the resulting Fresnel integral about the critical point (on the "tails" you save arbitrarily many powers of $N$) and if you're allowed $N\gg_{a,b} 1$ that should be enough. (I got a notification for a comment that said something like we're allowed to take $N\gg_{a,b} 1$, but it doesn't seem to exist! --- pardon me if I'm mistaken, I should know better by now than to try to comment while on my phone!) $\endgroup$ – alpoge Mar 30 '17 at 20:19
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    $\begingroup$ @alpoge: Yes, that was my feeling too, that the limsup (= lim) should just be zero, but I didn't think it through at that level of detail. $\endgroup$ – Christian Remling Mar 30 '17 at 22:00
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Yes, it holds with $C=1$ for all $N\geqslant 16 a^{-4}b^{-5}$. Here is a proof along the lines of alpoge's and Christian Remling's comments. Let $f(t)$ be one of $$ f_\pm(t)=N(bt^2\pm b^2t). $$ Then $f''(t)=2Nb$, so van der Corput's lemma gives $ \smash{\left|\int_a^z e^{if(t)}dt\right|\leqslant 4(Nb)^{-1/2}}$ for all $a, z\in\mathbf R$. Integrating by parts with $v(t)=\int_a^te^{if(s)}ds$, we obtain $$ \left|\int_a^z \frac{e^{if(t)}dt}t\right| = \left|\frac{v(z)}z+\int_a^z\frac{v(t)\,dt}{t^2}\right| \leqslant \frac4{\sqrt{Nb}}\left\{\frac1z+\int_a^z\frac{dt}{t^2}\right\} =\frac4{a\sqrt{Nb}} $$ and therefore $$ \left|\int_a^z\frac{\sin(N b^2t)e^{iNbt^2}}t\,dt\,\right| =\left|\int_a^z \frac{e^{if_+(t)}-e^{if_-(t)}}{2it}\,dt\,\right| \leqslant \frac4{a\sqrt{Nb}} $$ whenever $0<a<z$. So your lim sup is zero, and $\sqrt{Nb}\geqslant 4a^{-2}b^{-2}$ gives the claimed bound. Note that your assumptions $z=2a$ and $a<b<1$ are not needed.

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