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.


1 Answer 1


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

  • $\begingroup$ 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 $\mathbb R$? $\endgroup$
    – Romeo
    Commented Dec 1, 2012 at 16:20
  • $\begingroup$ $\sqrt{i}$ 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. $\endgroup$
    – Josh
    Commented Dec 1, 2012 at 16:58
  • $\begingroup$ 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. $\endgroup$
    – Romeo
    Commented Dec 1, 2012 at 17:28
  • $\begingroup$ Eventually, I've come to a solution: we have to use Fresnel integrals. With them we conclude immediately. Thanks for your help. $\endgroup$
    – Romeo
    Commented Dec 3, 2012 at 21:15

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