Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 13F / Solution

Integrate the entire function $e^{i{z}^2}$ around the positively oriented sector boundary: the real segment $0\to R$, the arc $R{e}^{i\theta }$ for $0\le \theta \le \pi /4$, and the radial segment back to zero. Cauchy integral theorem givesThe last term follows from $z=r{e}^{i\pi /4}$ and $i{z}^2=-r^2$; its minus sign is the return orientation.

On the arc, $|e^{i{z}^2}|=e^{-R^2\sin (2\theta )}$. The supplied sine bound gives $\sin (2\theta )\ge 4\theta /\pi$ throughout this sector, soThe Gaussian integral then yields ${\int }_0^{\infty }{e}^{i{x}^2}dx=e^{i\pi /4}\sqrt{\pi }/2$. Taking real and imaginary parts proves both Fresnel integrals:The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.