# Is there a simple way to bound this contour integral?

One method of evaluating $\int_0^\infty \cos x^2 \, dx$ and $\int_0^\infty \sin x^2 \, dx$ is to consider the complex function $f(z) = e^{iz^2}$ and evaluate it along a contour shaped like a $45^\circ$ sector of a circle, centered at the origin and with radius $R$. (More precisely, one that starts at $0$, moves to $R$, then to $Re^{i\pi/4}$, then back to the origin.) Applying Cauchy's theorem and letting $R \to \infty$ gives the desired results.

One problem I have is with bounding the circular part of the contour integral. We have to show that to show that $\int_0^{\pi/4} f(Re^{i\theta}) Ri e^{i\theta} \, d\theta \to 0$ as $R \to \infty$. I found a way to do this, but it is pretty tedious.

First we have \begin{align*} \int_0^{\pi/4} f(Re^{i\theta}) Ri e^{i\theta} \, d\theta = \int_0^{\pi/4} e^{-R^2e^{i2\theta}} Ri e^{i\theta} \, d\theta = \int_0^{\pi/4} e^{-R^2(\cos2\theta + i\sin2\theta)} Ri e^{i\theta} \, d\theta \end{align*} So: \begin{align*} \left| \int_0^{\pi/4} f(Re^{i\theta}) Ri e^{i\theta} \, d\theta \right| \leq \int_0^{\pi/4} e^{-R^2\cos 2\theta} R \, d\theta \end{align*} In the interval $[0, \frac{\pi}{4}]$, we have $\cos 2\theta \geq 1 - \frac{4}{\pi}\theta$, so: \begin{align*} \left| \int_0^{\pi/4} f(Re^{i\theta}) Ri e^{i\theta} \, d\theta \right| \leq \int_0^{\pi/4} e^{-R^2(1-\frac{4}{\pi}\theta)} R \, d\theta = \frac{\pi R}{4} \int_{0}^1 e^{-R^2\phi} \, d\phi = \frac{\pi R}{4} \frac{1-e^{-R^2}}{R^2} \end{align*}

At this point, we can easily see that the right hand side goes to $0$ as $R \to \infty$. However, is there a nicer argument? The fact that I have to use crude inequalities like $\cos 2\theta \geq 1 - \frac{4}{\pi}\theta$ makes me feel like there would be. (Or maybe I am not used to using these kinds of inequalities in analysis yet!)

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It would be nice if you clarify what you mean by nicer argument. Do you mean a more elementary argument? Or a more natural argument which may perhaps require less elementary facts for the justification? Or maybe a sharper estimation? Actually, this estimation is both simple and sharp (in magnitude), in that actually we have $$\int_{0}^{\pi/4}e^{-R^2 \cos2\theta} R \; d\theta = \frac{1}{2R} + o\left(\frac{1}{R}\right).$$ – Sangchul Lee Sep 1 '12 at 8:21
Thanks for your comment. I realized that I was having a hard time phrasing my question. Now that I think about it, I wanted to be able to get a sharper estimation - I wanted to be able to get a feel for how $\int e^{-R^2 \cos2\theta} R \, d\theta$ behaves for large values of $R$ (even though to evaluate the original integral, we just needed to show this was $o(1)$). But like you said, this estimation isn't too bad. How did you get that estimate of yours? – Alan C Sep 1 '12 at 20:26