Complex integral

  • #1

Main Question or Discussion Point

I read in some text or book that the integral

[tex] \int_{-\infty}^{\infty} \cos(x^2) + \sin(x^2) \, \mathrm{d}x = \sqrt{2\pi}[/tex]

I was wondering how this is possible. I read on this site that one such possible way was to start by integrating

[tex] e^{-i x^2} = \cos(x^2) - i \cos(x^2)[/tex]

My knowledge about complex analysis is rather limited. Could anyone expain to me how the integral at the top is evaluated? (I know one could start of by noticing the symmetry about the y-axis. )

https://www.physicsforums.com/showthread.php?t=139465

Agomez, shows one way to do it. But it is not exactly the same as the integral above. Sigh, I feel stupid for not seeing this one...
 

Answers and Replies

  • #2
morphism
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Because cos(x^2)+sin(x^2) is an even function, we have
[tex]\int_{-\infty}^\infty \cos(x^2)+\sin(x^2)\, dx = 2 \int_0^\infty \cos(x^2)+\sin(x^2) \,dx.[/tex]
Now use the results from agomez's calculations.
 
  • #3
My main problem is that

$$ e^{-ix^2} = \cos(x^2) - i \sin(x^2) $$

Whilst this integral is [itex] \cos(x^2) + \sin(x^2) [/itex]. Does the minus sign change anything?
 
  • #4
morphism
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It changes nothing. All I'm saying is that you can use agomez's evaluation of [itex]\int_0^\infty \cos(x^2)\, dx[/itex] and [itex]\int_0^\infty \sin(x^2)\, dx[/itex].
 

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