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!:)



Is there a uniform upper bound for this oscillatory integral?
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](http://ams.org/mathscinet-getitem?mr=2163589) 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.


