Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
Great, thanks a lot, it seems a good idea. Just one question: would you please explain a little more how do you reduce to a Gaussian integral? What is $\sqrt{i}$? Isn't it a bit "dangerous" to perform a complex substitution while we are calculating an integral over $\mathbb R$? –  Romeo Dec 1 '12 at 16:20
$\sqrt{i}$ can be written as $\sqrt{e^{j\pi/2}}$ which equals to $e^{j\pi/4}$. You right. The change of variables is little problematic, but, you can use simple complex analysis to argue that. A better option is to use the identity $\cos^2(x) = (1+cos(2x))/2$, which readily solves the problem. –  Josh Dec 1 '12 at 16:58
Well, thanks again; the problem now is that I do not know - even using complex analysis - how to justify rigorously the calculations I perform. I'm sorry but I disagree with the last part of your comment. I'm not looking for the Fourier transform of $\cos^2(x)=\cos(x)\cos(x)$, but for the Fourier transform of $\cos(x^2)=\cos(x \cdot x)$: these two things are very different. –  Romeo Dec 1 '12 at 17:28
Eventually, I've come to a solution: we have to use Fresnel integrals. With them we conclude immediately. Thanks for your help. –  Romeo Dec 3 '12 at 21:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.