Contour Integrals

[monomocoso]

By Cauchy's theorem show


\(\displaystyle \int^\infty_0 sin (x^2)\,dx = \int^\infty_0 cos (x^2)\,dx =\sqrt{ (2\pi/}4)\)

Apply Cauchy's theorem to \(\displaystyle f(z) = e^{-z^2}\) where the contour T is the path consisting
of the three smooth curves: (1) the line segment \(\displaystyle [ 0,R ]\), (2) the arc \(\displaystyle z = Re ^ {i\theta}, 0 \le\theta \le \pi/4\),
(3) the line segment from \(\displaystyle Re^{i\pi/4} \) to 0. Then let R go to infinity.

 

[galactus]

This is a classic CA integral. I remember having to do it at one time.

Consider \(\displaystyle \int_{\gamma} e^{iz^{2}}dz\), where \(\displaystyle \gamma\) is in the diagram.

It is holomorphic.

\(\displaystyle \displaystyle \int_{0}^{R}e^{ix^{2}}dx+\int_{0}^{\frac{\pi}{4}}e^{iR^{2}e^{2i\theta}}iRe^{i\theta}d\theta-\int_{0}^{R}e^{it^{2}e^{\frac{{\pi}i}{2}}}e^{\frac{{\pi}i}{4}}dt=I_{1}+I_{2}-I_{3}=0\)

But, \(\displaystyle \displaystyle |e^{iR^{2}e^{2i\theta}}iRe^{i\theta}|=|Re^{R^{2}(icos2\theta-sin2\theta)}|=Re^{-R^{2}sin2\theta}\leq Re^{-4R^{2}\theta/{\pi}}\)

Thus, \(\displaystyle \displaystyle I_{2}\leq R\int_{0}^{\frac{\pi}{4}}e^{-4R^{2}\theta/ {\pi}}d\theta=\frac{\pi}{4R}\left(1-e^{-R^{2}}\right)\)

which goes to 0 as \(\displaystyle R\to \infty\)

Which gives:

\(\displaystyle \displaystyle\int_{0}^{\infty}e^{ix^{2}}dx=\lim_{R\to \infty}I_{3}=\frac{1+i}{\sqrt{2}}\int_{0}^{\infty}e^{-t^{2}}dt=\frac{(1+i)\sqrt{\pi}}{2\sqrt{2}}\)

Take real and imaginary parts:

\(\displaystyle \displaystyle\int_{0}^{\infty}cos(x^{2})dx=\int_{0}^{\infty}sin(x^{2})dx=\sqrt{\frac{\pi}{8}}\)

We can also do this without CA if you're interested.