Is the integral $\int e^{2 \pi i z^2} dz$ uniformly bounded for any interval of $\mathbb{R}$? I was wondering if there exists a constant $C$ such that 
$| \int_I e^{2 \pi i z^2} dz | \leq C $ for any interval $I$ of $\mathbb{R}$?
Here I want $C$ to be independent of the choice of the interval $I$. 
Also $I$ is not necessarily a finite interval. 
I would any appreciate hint! Thank you!
 A: So we should prove
$$
\left|\int_0^x e^{2\pi i z^2}\,dz\right| <  \frac{1}{2}
$$
for all $x$.  Then we get ${} < 1$ for all intervals.
A: We define $$f(x)=\int_0^x \cos(2\pi z^2) dz,$$ then $f(x)$ is continuous and  $f(0)=0$. Furtheremore 
\begin{align}
\lim_{x\to \infty} f(x)&=\int_0^\infty \cos(2\pi z^2) dz =\frac{1}{\sqrt{2\pi}}\int_0^\infty \cos t^2 dt\\
&=\frac{1}{4},
\end{align}
where$$\int_0^\infty \cos t^2 dt=\int_0^\infty \sin t^2 dt=\sqrt{\frac{\pi}{8}}$$ are well known as  Fresnel integrals.
Thus $f(x)$ is a bounded function, so  there exists a positive $M$ such that $|f(x)|\le M$ for all $x\ge 0$. Similarly $|f(x)|\le M$ for all $x<0$. We have for arbitrary $a, b$ that 
\begin{align}
\left|\int_a^b \cos(2\pi z^2) dz\right|&=\left|\int_0^b \cos(2\pi z^2) dz-\int_0^a\cos(2\pi z^2) dz\right|\\
&=\left|\int_0^b \cos(2\pi z^2) dz\right|+\left|\int_0^a\cos(2\pi z^2) dz\right|\\
&\le 2M.
\end{align}
The same argument applies to $$g(x)=\int_0^x \sin(2\pi z^2) dz$$ and we see that there exists $M^\prime $ such that $|g(x)|\le M^\prime $ holds for all $x$.  So we have that $$\left|\int_a^b \sin(2\pi z^2) dz\right|\le 2M^\prime$$
for arbitrary  $a$ and $b$.
Now we see that
\begin{align}
\left|\int_a^b e^{2\pi iz^2} dz \right|&=\left|\int_a^b \cos(2\pi z^2) dz+ i\int_a^b \sin(2\pi z^2) dz\right|\\
&\le\left|\int_a^b \cos(2\pi z^2) dz\right|+\left|\int_a^b \sin(2\pi z^2) dz\right|\\
&\le 2M+2M^\prime=C,
\end{align}
which completes our arguments.
A: Since $\left|\sin(x)\right|\le\min\!\left(|x|,1\right)$, integration by parts yields
$$
\begin{align}
\left|\int_0^xe^{2\pi iz^2}\mathrm{d}z\right|
&=\left|\int_0^x\frac1{4\pi iz}\,\mathrm{d}\!\left(e^{2\pi iz^2}-1\right)\right|\\
&=\left|\frac{e^{2\pi ix^2}-1}{4\pi ix}+\int_0^x\frac{e^{2\pi iz^2}-1}{4\pi iz^2}\,\mathrm{d}z\right|\\
&\le\underbrace{\left|\frac{\sin\left(\pi x^2\right)}{2\pi x}e^{\pi ix^2}\right|}_{\le\min\left(\frac{|x|}2,\frac1{2\pi|x|}\right)}
+\int_0^x\underbrace{\left|\frac{\sin\left(\pi z^2\right)}{2\pi z^2}e^{\pi iz^2}\right|}_{\le\min\left(\frac12,\frac1{2\pi z^2}\right)}\mathrm{d}z\\
&\le\frac1{2\sqrt\pi}+\left(\frac1{2\sqrt\pi}+\frac1{2\sqrt\pi}\right)\\[3pt]
&=\frac3{2\sqrt\pi}
\end{align}
$$
Therefore,
$$
\left|\int_Ie^{2\pi iz^2}\mathrm{d}z\right|\le\frac3{\sqrt{\pi}}
$$
