Help verifying and simplifying $ \int_0^\infty \cos x^2 dx $ (the Fresnel integral) via complex contour integration

Question

Find $ \int_0^\infty \cos x^2 dx $.

Note: This is the Fresnel integral, whose derivation is available on this site and elsewhere. I'd like verification of my proof, as well any recommended improvements to the exposition, which to me is simpler than much of what's published.

Solution:

We use the technique of contour integrals in the complex plane, creating three contours: $\alpha$, from $0$ to $R$ along the real axis; $\beta$, from $R$ to $Re^{i \pi/4}$ in a circular arc, and $\gamma$, from $Re^{i \pi/4}$ to $0$ in a straight line.

We first show that $\lim_{R \to \infty} \int_\beta e^{-z^2}dz = 0$. $\int_\beta e^{-z^2}dz = \int_0^{\pi/4}e^{-R^2e^{2i\theta}}\cdot iRe^{i\theta}d\theta$. Since $|e^z| = |e^{\Re(z)}|$ and $|\int f(x) dx| \leq \int |f(x)| dx$, we have $|\int_\beta e^{-z^2}dz| \leq \int_0^{\pi/4} |Re^{-R^2\cos {2\theta}}|d\theta$. Since $0 \leq \theta \leq \pi/4$, $0 \leq \cos {2\theta} \leq 1$, and so this goes to $0$ as $R \to \infty$. From this, we conclude $\int_\gamma e^{-z^2}dz = - \int_\alpha e^{-z^2}dz$, since $e^{-z^2}$ is entire.

Since the Gaussian integral $\int^\infty_{-\infty} e^{-x^2}dx = \sqrt \pi$, and $e^{-x^2}$ is even, $\lim_{R \to \infty} \int_\gamma e^{-z^2}dz = - \int_0^\infty e^{-x^2}dx = - \sqrt \pi / 2$.

Also note that $\int_\gamma e^{-z^2} dz = \int_R^0 e^{-[re^{i\pi/4}]^2}e^{i\pi/4}dr = \frac{-\sqrt 2}{2}(1+i)\int_0^Re^{-ir^2}dr$. Thus, $\lim_{R \to \infty} \int_0^Re^{-ir^2}dr = \sqrt{2\pi}/4 + ki$ for some real $k$, and its complex conjugate $\int_0^Re^{ir^2}dr = \sqrt{2\pi}/4 - ki$.

Finally, we conclude $\int_0^\infty \cos x^2 dx = \int_0^\infty \frac{e^{ix^2} + e^{-ix^2}}{2} dx = \frac{\sqrt{2\pi}}{4}$, QED.

1. Is my proof correct?
2. I tried to skip mechanical steps while not omitting any conceptual leaps. My goal is for the exposition to be simple, clear, and direct. Did I succeed?
3. Could the writing and exposition be improved? How?

Answer 1

While it is true that $\lim_{R\to \infty}Re^{-R^2\cos(2\theta)}=0$ for $\theta\in [0,\pi/4)$, it is not true for $\theta =\pi/4$. Moreover, we need to justify interchanging the order of the limit and the integral.

One way to proceed, is to exploit the fact that for $\theta \in [0,\pi/2]$, $ \sin(\theta)\ge 2\theta/\pi$. Then, we have the following estimates

$$\begin{align} \int_0^{\pi/4}Re^{-R^2\cos(2\theta)}\,d\theta&=\frac R2\int_0^{\pi/2}e^{-R^2\sin(\theta)}\,d\theta\\\\ &\le \frac R2\int_0^{\pi/2}e^{-R^2 (2\theta/\pi)}\,d\theta\\\\ &=\frac \pi4 \left(\frac{1-e^{-R^2}}{R}\right) \end{align}$$

from which we find the limit as $R\to \infty$ is indeed equal to $0$.

As for the rest of the proof, it seems fine. But, since you found that

$$e^{i\pi/4}\lim_{R\to\infty}\int_0^R e^{-it^2}\,dt=\frac{\sqrt\pi}{2}$$

one you could instantly conclude that

$$\int_0^\infty e^{-it^2}\,dt=\frac{\sqrt\pi}{2}e^{-i\pi/4}$$

Now, simply take the real part of both sides to arrive at the coveted result.