<s>[INST] Show that $\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$ for $n\ge 5.$ Question is to show that 
$$\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$$
when $n\geq5$, $x \in \mathbb R $ and $i$ is an imaginary unit.  
My effort:
$$|\int_{-n}^{n}e^{iy^2}dy|\leq \int_{-n}^{n}|e^{iy^2}|dy=\int_{-n}^{n}|\cos(y^2)+i\sin(y^2)|dy$$
$$ \leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|i||\sin(y^2)|dy$$
$$\leq \int_{-n}^{n}|\cos(y^2)|dy+\int_{-n}^{n}|\sin(y^2)|dy$$  
It is also known that $|\cos(x)|,|\sin(x)|\leq1$
but its leading nowhere since integral then evalutes to $0$..  
Any tips?
 [/INST] $$\left|\int_{-n}^{n}e^{iy^2}dy\right|^2 =\left|\int_{-n}^{n}\cos{y^2}dy +i\int_{-n}^{n}\sin{y^2}dy\right|^2 \\= 4\left(\int_{0}^{n}\cos{y^2}dy \right)^2 + 4\left(\int_{0}^{n}\sin{y^2}dy \right)^2\\=  4[I^2_n + J^2_n]$$
This is somehow related to this: Prove only by transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $.  Employing the change of variables $2u =x^2$ We get $$I_n=\int_0^n\cos(x^2) dx =\frac{1}{\sqrt{2}}\int^{n^2/2}_0\frac{\cos(2x)}{\sqrt{x}}\,dx$$ $$ J_n=\int_0^n \sin(x^2) dx=\frac{1}{\sqrt{2}}\int^{n^2/2}_0\frac{\sin(2x)}{\sqrt{x}}\,dx $$ Using the same  change of variables like here one readily get 
$$J_n = \frac{\sin^2 \frac{n^2}{2}}{n}+\frac{1}{2\sqrt{2}} \int^{\frac{n^2}{2}}_0\frac{\sin^2 x}{x^{3/2}}\,dx <  \frac{1}{n}+\frac{1}{2\sqrt{2}} \int^{\infty}_0\frac{\sin^2 x}{x^{3/2}}\,dx  $$
and
$$I_n =\frac{1}{2} \frac{\sin 2 n^2}{n} +\frac{1}{4 }\frac{\sin^2 \frac{n^2}{2}}{n} +\frac{3}{8\sqrt{2}} \int^{\frac{n^2}{2}}_0\frac{\sin^2 x}{x^{5/2}}\,dx \\< \frac{3}{4n} +\frac{3}{8\sqrt{2}} \int^{\infty}_0\frac{\sin^2 x}{x^{5/2}}\,dx  $$
Since $0\le \sin^2 x\le 1$. From this we have, 
$$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx = \frac{4\sqrt \pi}{3}$$
similarly, we have $$\int^{\infty}_0\frac{\sin^2 x}{x^{3/2}}\,dx =\sqrt\pi$$
Whence,
$$J_n < \frac{1}{n}+\frac{\sqrt\pi}{2\sqrt{2}}= \frac{1}{n}+ \sqrt{\frac{\pi}{8}} $$
and
$$I_n < \frac{3}{4n}+\frac{\sqrt\pi}{2\sqrt{2}} = \frac{3}{4n}+ \sqrt{\frac{\pi}{8}} $$ 
Since, 
$$\color{red}{\lim_{n\to\infty}4 \left[  \left(\frac{3}{4n}+ \sqrt{\frac{\pi}{8}} \right)^2+ \left(\frac{1}{n}+ \sqrt{\frac{\pi}{8}} \right)^2\right] = \pi <4}$$
That is for very large $n$ we get, 
$$\color{red}{\left|\int_{-n}^{n}e^{iy^2}dy\right| =  2[I^2_n + J^2_n]^{1/2} \le 2\left[  \left(\frac{3}{4n}+ \sqrt{\frac{\pi}{8}} \right)^2+ \left(\frac{1}{n}+ \sqrt{\frac{\pi}{8}} \right)^2\right]^{1/2}<2}$$
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