$$\int_{0}^{\infty }\sin(x^{2})dx$$

I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.

I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions (without complex analysis)?

  • $\begingroup$ I'm not sure, would $\dfrac{1}{x^2} \times (-cos(x^2))$ be correct? $\endgroup$ – Niklas R May 1 '13 at 8:38
  • 2
    $\begingroup$ This is called a Fresnel integral. The standard method to evaluate it uses complex analysis, but I am sure there are more elementary methods available if you Google for them. $\endgroup$ – Potato May 1 '13 at 8:38
  • $\begingroup$ @Aryabhata: I am not sure it is an exact duplicate. Unless I have missed something, the answer to the question being asked here cannot just be read off... $\endgroup$ – user1729 May 1 '13 at 9:32
  • $\begingroup$ @Aryabhata It can be read off the very first sentence so long as you know you need to use complex analysis. So there is a leap. But I suppose that is covered by Potato's comment. $\endgroup$ – user1729 May 1 '13 at 9:38
  • $\begingroup$ I edited my question. $\endgroup$ – please delete me May 1 '13 at 12:01

$\displaystyle \int_{0}^{\infty}\sin(x^{2})dx$

Let $\displaystyle u=x^{2}, \;\ x=\sqrt{u}, \;\ dx=\frac{1}{2}u^{-1/2}du$


Now, the Gamma function comes in handy:


But, since $a=1/2$, we have:



Let $t=\sqrt{z}$:


This is a rather famous integral and can be found here and there. I leave it's evaluation to the reader.

But, it evaluates to $\displaystyle \frac{1}{\sqrt{\pi}}\cdot \frac{\pi\sqrt{2}}{4}=\frac{\sqrt{2\pi}}{4}$, and so does the Fresnel in question.

  • $\begingroup$ I'm not completely following how you get equation 3 from equation 2. Can you help me out? $\endgroup$ – Bitrex May 3 '13 at 23:04
  • $\begingroup$ It comes from the Gamma function. You can use this as a parameter in the integral. $\frac{1}{u^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{\infty}z^{a-1}e^{-uz}dz$ $\endgroup$ – Cody May 4 '13 at 11:24
  • $\begingroup$ I follow now. Thanks! $\endgroup$ – Bitrex May 4 '13 at 15:25
  • $\begingroup$ An nice integration, if you're interested, is to derive $\int_{0}^{\infty}\cos(x^{a})dx, \;\ a>1$. You can do it with Laplace transforms instead of the classic residue method. $\endgroup$ – Cody May 5 '13 at 13:33

The integral does not have an elementary antiderivative. If you want to avoid complex analysis, try this method.


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