# Evaluate $\int_{0}^{\infty} \cos(x^2)dx$ [duplicate]

Prove that the above integral is equal to $\frac{\sqrt{2\pi}}{2}$

I have already tried expanding using $\cos$ identity and also taking Laplace for it. I am getting nowhere with this.

## marked as duplicate by Tomás, Jack D'Aurizio, apnorton, mrf, user147263 Feb 11 '15 at 0:32

• It should be $\frac{\sqrt{2\pi}}{4}$, no? Have you seen, or know the/some computation of $\int_{0}^{\infty}e^{-x^2}dx$? From this one you can compute yours, using that $\cos(a)=\frac{e^{ai}+e^{-ai}}{2}$. – Carol Feb 10 '15 at 20:21
Let $f(z) = \exp( iz^2)$ be an entire function of a complex variable with $\Gamma$ the contour consisting of the circle wedge of angle $\pi/4$. In other words, it consists of three pieces. First the real segment $[0,R]$, followed by the circular arc $Re^{i\theta}$ where $0\leq \theta \leq \frac{\pi}{4}$ and then returns back, in a straight line, to the origin. So it looks like a pizza.
Thus, $$\oint_{\Gamma} f(z) ~ dz = 0$$ Now break up this integral into three pieces. The circular arc goes to zero as $R\to \infty$ by a standard estimation argument. The two segments of the integral can be computed.