The boundedness of an integral Is there a constant $C$ which is independent of real numbers $a,b,N$, such that 

$$\left| {\int_{-N}^N \dfrac{e^{i(ax^2+bx)}-1}{x}dx} \right| \le C?$$

 A: These integrals are indeed uniformly bounded. As the only effect of changing the signs of $a,b,N$ on the integral is a sign change and/or complex conjugation, we can restrict to $a,b,N\gt0$. Setting $M=Nb$ and $\alpha=a/b^2$, we have
$$
\int_{-N}^N\left(e^{i(ax^2+bx)}-1\right)\frac{dx}x=\int_0^M\left(e^{i(\alpha x^2+x)}-e^{i(\alpha x^2-x)}\right)\frac{dx}x.
$$
However, the integrand can be written as $2ie^{i\alpha x^2}\frac{\sin x}x$, so is bounded by $2$ in absolute value. Therefore, fixing any constant $K\gt0$, the integral is bounded by
$$
\begin{align}
2K+\int_K^{K\vee M}e^{i(\alpha x^2+x)}\frac{dx}x-\int_K^{K\vee M}e^{i(\alpha x^2-x)}\frac{dx}x.&&{\rm(1)}
\end{align}
$$
I'll prove that this is bounded with the help of a lemma.

Lemma: (van der Corput lemma) If $f\colon[A,B]\to\mathbb{R}$ is convex with $\lvert f^\prime(x)\rvert\ge\lambda\gt0$ and $g\colon[A,B]\to\mathbb{C}$ is differentiable then,
  $$\left\lvert\int_A^Be^{if(x)}g(x)dx\right\rvert\le\frac2\lambda\left(\lvert g(B)\rvert+\int_A^B\lvert g^\prime(x)\rvert dx\right)$$

As I will only be interested in intervals $[A,B]\subset[K,\infty)$ with $g(x)=1/x$, the bound in the lemma can be written as
$$
\begin{align}
\left\lvert\int_A^Be^{if(x)}\frac{dx}x\right\rvert\le\frac2{\lambda K}&&{\rm(2)}
\end{align}
$$
Taking $f(x)=\alpha x^2+x$, this shows that the first integral in (1) is bounded by $2/K$. For the second integral, take $f(x)=\alpha x^2-x$, fix any $0\lt\epsilon\lt1/2$, and first look at value of the integral with integration range restricted to $[0,\epsilon/\alpha]$. In this range, we have $f^\prime(x)\le f^\prime(\epsilon/\alpha)=-(1-2\epsilon)$. So, by inequality (2), this part of the integral is bounded by $2K^{-1}(1-2\epsilon)^{-1}$.
Next, for any fixed $\gamma\gt1/2$, the value of the last integral in (1), restricted to the range $[\epsilon/\alpha,\gamma/\alpha]$ is bounded by
$$
\int_{\epsilon/\alpha}^{\gamma/\alpha}\frac{dx}{x}=\log(\gamma/\epsilon).
$$
Finally, look at the last integral in (1) restricted to the range $[\gamma/\alpha,\infty)$. As $f^\prime(x)\ge f^\prime(\gamma/\alpha)=(2\gamma-1)$ in this range, inequality (2) shows that this part of the integral is bounded by $2K^{-1}(2\gamma-1)^{-1}$.
Putting these together shows that the set of integrals in the question is bounded above by a constant. The upper bound obtained here is
$$
2K+\frac2K+\frac2{K(1-2\epsilon)}+\frac2{K(2\gamma-1)}+\log(\gamma/\epsilon)
$$
for arbitrary positive constants $K,\epsilon,\gamma$ with $\epsilon\lt1/2\lt\gamma$.
A: Note that $$ {\int_{-N}^N ({e^{i(ax^2+bx)}-1)\frac{1}{x}dx} }=2i\int_{0}^N e^{iax^2}\frac{\sin bx}{x}dx $$ As I suppose you've already proved. 

Maybe, we can approach in the following way
$$I=2i\int_{0}^N e^{iax^2}\frac{\sin bx}{x}dx=-2\int_{0}^{N}\sin ax^2 \frac{\sin bx}{x}dx+2i\int_{0}^{N}\cos ax^2\frac{\sin bx}{x}dx$$
Hence $$\left|I\right|\le 2\sqrt{I_1^2+I_2^2}$$ where $$I_1=\int_{0}^{N}\sin ax^2 \frac{\sin bx}{x}dx,\ I_2=\int_{0}^{N}\cos ax^2 \frac{\sin bx}{x}dx$$ 

Now, if we can show that both $I_1,\ I_2$ are bounded by some constant independent of $a,b,N$ then $I $ is also bounded.
A: we can use theorem
for $a \le b$
$ |\int _a^b f(x)dx|\le \int_a^b|f(x)|dx$
$\int _{-N}^N|(e^{i(ax^2+bx)}-1)\frac1x|dx\le \int _{-N}^N|(e^{i(ax^2)}-1)\frac1x|dx\ $
you can play around with the integral to make it more intergreable
