Fresnel Integrals: $\lim_{R\to\infty}\int_0^{\pi/4} \mathrm{e}^{-(Re^{i\theta})^{2}}{iRe^{i\theta}}\,\mathrm{d}\theta=?$

Question

I'm having trouble proving that the arc from $R$ to $Re^{i\pi/4}$ in the Fresnel contour goes to zero. Currently I have $$\int_0^{\pi/4} \mathrm{e}^{-(Re^{i\theta})^{2}}{iRe^{i\theta}}\,\mathrm{d}\theta$$ and I've tried a couple of things but I can't seem to find an inequality that works.

Answer 1

If $z = R e^{i\theta}$, $|e^{-z^2}| = \exp(-\text{Re}(z^2)) = \exp(-R^2 \cos(2\theta))$. If $\cos(2\theta) > 0$, this goes to $0$ as $R \to +\infty$...

Answer 2

Let's start be taking the absolute value and seeing where it leads us. \begin{align} \Biggl\lvert\int_0^{\pi/4}e^{-iR^2e^{2i\theta}}iRe^{i\theta}d\theta\Biggr\rvert & \leq\int_0^{\pi/4}\bigl\lvert e^{-iR^2e^{2i\theta}}\bigr\rvert\lvert iRe^{i\theta}\rvert d\theta\\ &=R\int_0^{\pi/4}\bigl\lvert e^{-iR^2e^{2i\theta}}\bigr\rvert d\theta \end{align} Now $e^{2i\theta} = \cos(2\theta) + i\sin(2\theta)$. Then $ie^{2i\theta}=i\cos(2\theta) - \sin(2\theta)$ so we have \begin{align} R\int_0^{\pi/4}\bigl\lvert e^{-iR^2e^{2i\theta}}\bigr\rvert d\theta &= R\int_0^{\pi/4}\bigl\lvert e^{-R^2\sin(2\theta)}\bigr\rvert\bigl\lvert e^{iR^2\cos(2\theta)}\bigr\rvert d\theta\\ &=R\int_0^{\pi/4}\bigl\lvert e^{-R^2\sin(2\theta)}\bigr\rvert d\theta \end{align} Let $\phi=2\theta$. Then we have $$ \frac{R}{2}\int_0^{\pi/2} e^{-R^2\sin(\phi)} d\phi\leq \frac{R}{2}\int_0^{\pi/2} e^{-2\phi R^2/\pi} d\phi = \frac{\pi}{4R}(1-e^{-R^2}) $$ The limit as $R\to\infty$ is zero.

Answer 3

This is a very nice question.

First obsersve that, $$|e^{-iR^2e^{2i\theta}}| = e^{-R^2 \sin{2\theta}} $$

And using the change of variable, $ u = 2\theta $ we get, $$ \left| \int_0^{\frac\pi{4}}e^{-iR^2e^{2i\theta}}iRe^{i\theta} d\theta\right| \le R\int_0^{\frac\pi{4}}|e^{-iR^2e^{2i\theta}}| d\theta = R\int_0^{\frac\pi{4}}e^{-R^2\sin2\theta}d\theta \\= \frac{R}{2}\int_0^{\frac{\pi}{2}} e^{-R^2\sin\theta}d\theta$$

afterward shows by studying the function $[0,\frac{\pi}{2}]\ni\theta \mapsto\frac{\sin\theta}{\theta}$ that

$$ \color{blue}{\sin\theta \geq \frac{2}{\pi}\theta ~~ \forall \theta\in [0,\frac{\pi}{2}] } $$ therefore we get that $$\lim_{R\to\infty}\left| \int_0^{\frac\pi{4}}e^{-iR^2e^{2i\theta}}iRe^{i\theta} d\theta\right| \leq \lim_{R\to\infty}\frac{R}{2}\int_0^{\frac{\pi}{2}} e^{-R^2\sin\theta}d\theta\\ \le \lim_{R\to\infty}\frac{R}{2}\int_0^{\frac{\pi}{2}} e^{-\frac{2R^2}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{4R}(1-e^{-R^2}) =0$$