TITLE: Fourier transform of $\cos x^2$
QUESTION [5 upvotes]: Consider the real valued function $f(x):=\cos{(x^2)}$. How can we calculate its Fourier transform? 
In other words, I have to calculate 
$$
\hat{f}(\omega):=\frac{1}{2\pi}\int_{\mathbb R}\cos{(x^2)}e^{-i\omega x}dx.
$$
Any ideas? I'm sincerely stuck... I tried to calculate 
$$
\int_{\mathbb R}e^{ix^2-ikx}dx
$$
in order to get the Fourier transforms of both $\cos x^2$ and $\sin x^2$ but I do not know how to begin. Mathematica says that the Fourier transform of $f$ has this simple expression: 
$$
\frac{1}{2} \left[\cos\left(\frac{\omega^2}{4}\right)+\sin\left(\frac{\omega^2}{4}\right)\right]
$$
Thanks in advance.

REPLY [3 votes]: You almost finished. You just need to complete the square in the exponential term, and use a Gaussian integral
$$
\int_{-\infty}^{\infty}e^{ix^2-ikx}dx=e^{\frac{(-ik)^2}{4i}}\sqrt{\frac{\pi}{-i}}=e^{\frac{-ik^2}{4}}\sqrt{i\pi}
$$