Show that $\left|\int_{-n}^{n}e^{iy^2}dy\right|\le 2$ for $n\ge 5.$

Question

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?

Answer 1

$$\left|\int_{-n}^{n}e^{iy^2}dy\right|^2 =\left|\int_{-n}^{n}\cos{y^2}dy +i\int_{-n}^{n}\sin{y^2}dy\right|^2 \\= 4\left(\int_{0}^{n}\cos{y^2}dy \right)^2 + 4\left(\int_{0}^{n}\sin{y^2}dy \right)^2\\= 4[I^2_n + J^2_n]$$ 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, $$\color{red}{\left|\int_{-n}^{n}e^{iy^2}dy\right| = 2[I^2_n + J^2_n]^{1/2} \le 2\left[ \left(\frac{3}{4n}+ \sqrt{\frac{\pi}{8}} \right)^2+ \left(\frac{1}{n}+ \sqrt{\frac{\pi}{8}} \right)^2\right]^{1/2}<2}$$

Answer 2

By symmetry and the change of variable $y^2=x$ we have $$ \int_{-n}^{n}e^{-iy^2}\,dy = \int_{0}^{n^2}e^{-ix}\frac{dx}{\sqrt{x}}.\tag{A}$$ By integration by parts the RHS of $(A)$ turns into $$ \left[\frac{-i+ie^{-ix}}{\sqrt{x}}\right]_0^{n^2}+\int_{0}^{n^2}\frac{-i+ie^{-ix}}{2x\sqrt{x}}\,dx\tag{B} $$ and the function $\left|\frac{-i+ie^{-ix}}{2x\sqrt{x}}\right|=\frac{\left|\sin\frac{x}{2}\right|}{x\sqrt{x}}$ is bounded by $\frac{1}{x\sqrt{x}}$ for any $x\geq \pi$ and by $\frac{1}{2\sqrt{x}}$ over the interval $(0,\pi)$. By the triangle inequality it follows that

$$ \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\leq \frac{2}{n^2}+\int_{0}^{\pi}\frac{dx}{2\sqrt{x}}+\int_{\pi}^{+\infty}\frac{dx}{x\sqrt{x}}\leq 3\tag{C} $$ for any $n\geq 5$. The last inequality can be further refined by applying an extra step of integration by parts after $(B)$. Actually $$ \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\to \sqrt{\pi} \tag{D}$$ as $n\to +\infty$, hence for any $\varepsilon>0$ the inequality $$ \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\leq \sqrt{\pi}+\varepsilon \tag{E}$$ holds for any $n$ large enough. We also have $$ \int_{n^2}^{+\infty}\frac{e^{-iz}}{\sqrt{z}}\,dz \stackrel{\mathcal{L}}{=} e^{-in^2}\int_{0}^{+\infty}\frac{e^{-n^2 s}}{(i+s)\sqrt{\pi s}}=\frac{2 e^{-in^2}}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{e^{-n^2 t^2}}{t^2+i}\,dt \tag{F}$$ so by the triangle inequality and the Cauchy-Schwarz inequality it follows that $$ \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\leq \sqrt{\pi}+\frac{\pi^{1/4}}{\sqrt{2n}}.\tag{G} $$ By mixing these two approaches (integration by parts, Laplace transform) we may improve $(G)$ up to $$ \forall n\geq 2\pi,\qquad \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\leq \sqrt{\pi}+\frac{1}{2\sqrt{n}}.\tag{H} $$

Answer 3

A Contour Integral Estimate

This estimate is valid for all $n\gt0$, and shows that the bound is $2$ for $n\ge4.5$ (since $\sqrt\pi\doteq1.77245385$). The contours in the complex plane are straight lines, and are parametrized linearly. $$ \begin{align} \left|\,\int_{-n}^ne^{iy^2}\,\mathrm{d}y\,\right| &\le\left|\,\int_{-n(1+i)}^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right| +\left|\,\int_n^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right| +\left|\,\int_{-n(1+i)}^{-n}e^{iy^2}\,\mathrm{d}y\,\right|\\ &\le\sqrt2\int_{-n}^ne^{-2y^2}\,\mathrm{d}y +2\int_0^1e^{-2n^2t}n\,\mathrm{d}t\\[3pt] &\le\sqrt\pi+\frac1n \end{align} $$

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This is pretty good

Here is the plot of the actual value of the integral vs the estimate above vs $2$:

Note that the first maximum after $n=4.5$ is the first one that is less than $2$.