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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?

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  • $\begingroup$ My first thought is that since $e^{iy^2}$ is even, it suffices to show $|\int_0^n e^{iy^2}| \le 1$. $\endgroup$ – Alex Vong Dec 5 '17 at 13:32
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    $\begingroup$ Maybe that helps ... $$\int_{-n}^{n}e^{iy^2}dy = \int_0^{n^2}\frac{e^{iu}}{\sqrt u} du$$ $\endgroup$ – Gabriel Romon Dec 5 '17 at 13:42
  • $\begingroup$ I just noticed that the title was changed. I worked so long trying to show this for all $n$, until I plotted it and saw that what I had was about as good as it gets. $\endgroup$ – robjohn Dec 6 '17 at 16:04
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$$\left|\int_{-n}^{n}e^{iy^2}dy\right|^2 =\left|\int_{-n}^{n}\cos{y^2}dy +i\int_{-n}^{n}\sin{y^2}dy\right|^2 \\= 4\left(\int_{0}^{n}\cos{y^2}dy \right)^2 + 4\left(\int_{0}^{n}\sin{y^2}dy \right)^2\\= 4[I^2_n + J^2_n]$$ 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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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 $$ \forall n\geq 2\pi,\qquad \left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\leq \sqrt{\pi}+\frac{1}{2\sqrt{n}}.\tag{H} $$

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    $\begingroup$ The bound that the OP wants is 2 not 3 $\endgroup$ – Guy Fsone Dec 5 '17 at 19:50
  • $\begingroup$ @GuyFsone: I guess the constant $2$ is not that important (also because it is not optimal) and the OP is mainly interested in techniques for controlling oscillating integrals, like integration by parts and the Laplace transform. By the way, OP's upper bound does not hold at $n=3$, for instance. $\endgroup$ – Jack D'Aurizio Dec 5 '17 at 19:59
  • $\begingroup$ How do you prove $\left|\int_{-n}^{n}e^{-iy^2}\,dy\right|\to \sqrt{\pi}$ ? $\endgroup$ – Gabriel Romon Dec 5 '17 at 20:04
  • $\begingroup$ @GabrielRomon: through the Laplace transform. That is a Fresnel integral. $\endgroup$ – Jack D'Aurizio Dec 5 '17 at 20:05
  • $\begingroup$ Yes I do hope too the bound is not important. But I may highlight that he put the constraint $n\ge 5. $ and this more hard. $\endgroup$ – Guy Fsone Dec 5 '17 at 20:05
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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. $$ \begin{align} \left|\,\int_{-n}^ne^{iy^2}\,\mathrm{d}y\,\right| &\le\left|\,\int_{-n(1+i)}^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right| +\left|\,\int_n^{n(1+i)}e^{iy^2}\,\mathrm{d}y\,\right| +\left|\,\int_{-n(1+i)}^{-n}e^{iy^2}\,\mathrm{d}y\,\right|\\ &\le\sqrt2\int_{-n}^ne^{-2y^2}\,\mathrm{d}y +2\int_0^1e^{-2n^2t}n\,\mathrm{d}t\\[3pt] &\le\sqrt\pi+\frac1n \end{align} $$


This is pretty good

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

enter image description here

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

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  • $\begingroup$ Shifting the integration line is a very slick idea, (+1). $\endgroup$ – Jack D'Aurizio Dec 6 '17 at 17:17

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