How can we compute $\int^{\infty}_0\sin(x^2)dx$ using Fourier transform?

I had an idea in my mind. To use the $\text{sinc}$ function and take its inverse Fourier Transform or something like that. I also tried using the inverse Fourier Transform of $\frac1{x^2}$, but I cannot use the fact to this improper integral.

Can someone point me out how can we use the Fourier Transform here?

Please only use Fourier Transform, nothing else.

  • $\begingroup$ Oh, you're using Riemann integral then. But no matter what my comment was wrong... $\endgroup$
    – Tryss
    Jan 5, 2016 at 15:03
  • $\begingroup$ @Tryss, no problem. (y) But how can I solve it?? $\endgroup$ Jan 5, 2016 at 15:42

2 Answers 2


Only a partial answer, but maybe useful. Assuming $\int_0^\infty\sin(x^2)dx=\int_0^\infty\cos(x^2)dx$ (which is numerically verified and possibly easy to prove), and employing the Fourier transform $$ \hat{f}(\xi)=\int_{-\infty}^\infty dx \frac{1}{\sqrt{|x|}}e^{-2\pi\mathrm{i}x\xi}=2\mathrm{Re}\int_0^\infty dx \frac{1}{\sqrt{x}}e^{-2\pi\mathrm{i}x\xi}=\frac{1}{\sqrt{|\xi|}}\ , $$ we have easily $$ I=\int_0^\infty dx\cos(x^2)=\mathrm{Re}\int_0^\infty dx\ e^{\mathrm{i}x^2}=\frac{1}{2}\mathrm{Re}\int_0^\infty dy\ \frac{e^{\mathrm{i}y}}{\sqrt{y}}=\frac{1}{4}\hat{f}\left(-\frac{1}{2\pi}\right)=\sqrt{\frac{\pi}{8}}\ . $$

  • $\begingroup$ I didn't understand the last step. Can you explain it please? How you got $-\frac1{2\pi}$ there? $\endgroup$ Jan 6, 2016 at 4:03
  • $\begingroup$ From the first line, you read $$ \mathrm{Re}\int_0^\infty dx\frac{1}{\sqrt{x}}e^{-2\pi\mathrm{i}x\xi}=\frac{1}{2}\hat{f}(\xi)$$. So if you compute $(1/2)\hat{f}(\xi)$ at $\xi=-1/(2\pi)$, you precisely obtain $\mathrm{Re}\int_0^\infty dy \frac{e^{\mathrm{i}y}}{\sqrt{y}}$, which is the integral you need on the second line. $\endgroup$ Jan 6, 2016 at 6:54
  • $\begingroup$ Can we use the unnormalized sinc function also? Is there a way with that? $\text{sinc}(x^2)=\frac{\sin(x^2)}{x^2}$. $\endgroup$ Jan 6, 2016 at 8:45
  • $\begingroup$ The equality $\int_0^{\infty}\sin(x^2)dx=\int_0^{\infty}\cos(x^2)dx$ is indeed established. These integrals (with variable upper limit) are the Fresnel integrals, both of which have the limit $\sqrt{\pi/8}$ (see here). $\endgroup$
    – Matt L.
    Jan 6, 2016 at 9:19
  • 1
    $\begingroup$ @Matt. True. But if we assume we know about Fresnel integral (which of course I do), the whole question is pointless. My point is: suppose we don't know anything about Fresnel. Is it possible to show [using only algebraic manipulations] that $\int_0^\infty dx\sin(x^2)=\int_0^\infty dx\cos(x^2)$ [no matter what the actual value of the integral is]? If yes, then use my answer from that point onwards. $\endgroup$ Jan 6, 2016 at 9:27

I hope the Laplace transform is allowed too, since it is just a "rotated variant" of the Fourier transform. We have: $$ \mathcal{L}(\sin x) = \frac{1}{1+s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\sqrt{\pi s}}\tag{1}$$ hence:

$$ \begin{eqnarray*}\int_{0}^{+\infty}\sin(x^2)\,dx &=& \frac{1}{2}\int_{0}^{+\infty}\frac{\sin x}{\sqrt{x}}\,dx\\&=&\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(1+s^2)\sqrt{s}}\\&=&\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{1+s^4}\tag{2}\end{eqnarray*}$$ and the last integral can be computed in many ways, for instance through the residue theorem and/or through partial fraction decomposition ($1+4s^4 = (1-2s+2s^2)(1+2s+2s^2)$).

That leads to:

$$ \int_{0}^{+\infty}\sin(x^2)\,dx = \frac{1}{\sqrt{\pi}}\cdot\frac{\pi}{\sqrt{8}} = \color{red}{\sqrt{\frac{\pi}{8}}.}\tag{3}$$

  • $\begingroup$ Oh sorry, Fourier Transform was supposed to solve the problem in 3-4 steps. That is why I am after it. I can do nothing more than an upvote to this. Thanks for helping anyway. $\endgroup$ Jan 5, 2016 at 16:48

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