# Definite Integration using Residue Theorem

• Apr 30th 2013, 12:16 AM
CuriosityCabinet
Definite Integration using Residue Theorem
Evaluate $\int_{0}^{\infty} cos(x^2) dx$.

What complex integral and branch cut should I use??
• Jun 14th 2013, 12:30 PM
Phantasma
Re: Definite Integration using Residue Theorem
Quote:

Originally Posted by CuriosityCabinet
Evaluate $\int_{0}^{\infty} cos(x^2) dx$.

What complex integral and branch cut should I use??

I'd evaluate it using the fact that the residue of $\cos(z^2)=\frac{e^{iz^2}+e^{-iz^2}}{2}$ is 0.

Using that, we can set our integral's contour, C, over the real line from 0 to infinity and arc around back.

Thus, we'll get $\oint\limits_{C}\cos(z^2)\,dz = 0 = \int\limits_{[0,\infty]}cos(z^2) \, dz - \frac{in}{2} \lim_{n\to\infty^+}\int\limits_{[0, \pi]}\cos(\frac{n^2}{4}(1+e^{i\theta})^2)e^{i\theta}\, d\theta.$. Make sense?