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Fresnel Integral

  1. Aug 14, 2006 #1
    Somebody could please tell me how to evaluate the integral:

    integral(sin(x^2)) from o to infinity
  2. jcsd
  3. Aug 14, 2006 #2


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    You can find a strategy here.
  4. Aug 14, 2006 #3
    well, I had already checked the link, but I don't undestand the strategy, could somebody please explain it to me?, or if you have a text where I can find some explanation about that kind of process, I would really apreciate if you help me.
  5. Aug 14, 2006 #4
    Since [itex]e^{iz^2}[/itex] is an analytic function on [itex]\mathbb{C}[/itex] we have:

    [tex]\int_{\Gamma_R} e^{iz^2} d z = 0[/tex] where [itex]\Gamma_R[/itex] is the "pizza-slice" countour given in the link.
    [tex]\int_{\Gamma_R} e^{iz^2} d z = \int_0^R e^{i t^2} d t + \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i \theta} d \theta + \int_R^0 e^{- t^2} e^{i \frac{\pi}{4}} d t [/tex].

    By taking the limit [itex]R \rightarrow \infty[/itex] and observing that

    [tex]\lim_{R\rightarrow \infty} \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i\theta} d \theta = 0 [/tex]

    we obtain

    [tex]\int_0^\infty e^{i t^2} d t = e^{i \frac{\pi}{4}} \int_0^\infty e^{- t^2} d t [/tex].

    By Euler's formula and gaussian integral we get:

    [tex]\int_0^\infty \sin(t^2) dt = \int_0^\infty \cos(t^2) dt = \sqrt{\frac{\pi}{8}}[/tex].
    Last edited: Aug 14, 2006
  6. Aug 15, 2006 #5


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    Because i'm lazy to write all that much, see pages 98 & 99 of Max Planck's "Theory of Light", Macmillan, 1932 (a.k.a. the IV-the volume of the "Introduction to Theoretical Physics" set).

  7. Mar 10, 2011 #6
    Reviving this tread, but why does

    [tex]\lim_{R\rightarrow \infty} \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i\theta} d \theta = 0 [/tex]

  8. Mar 23, 2011 #7
    This is why:

    \left\arrowvert\int_0^{\pi/4}e^{iR^2e^{i2\theta}}iRe^{i\theta}d\theta \right\arrowvert = \left\arrowvert\int_0^{\pi/4}exp\{iR^2(\cos(2\theta)+i \sin(2\theta))\}iRe^{i\theta}d\theta \right\arrowvert

    Evaluating the absolute value, this equals

    \int_0^{\pi/4}exp\{-R^2 \sin(2\theta)\}R d\theta

    Now find the maximum of [tex] g(\theta) = exp\{-R^2 \sin(2\theta)\} [/tex] by differentiation (check it's a max by second derivative test).
    This occurs at [tex] \theta = \pi/4 [/tex].
    Then the above is less than or equal to

    R \cdot exp\{-R^2\} \frac{\pi}{4}

    Take the limit as [tex] R\to \infty [/tex] to get 0.
    Absolute value to zero, original to zero.
  9. Jun 6, 2011 #8
    Buddy you are wrong the maximum of exp(-R^2sin(2thta)) is not at theta=pi/4 its maximum is at theta=0 ! So The maximum of g(theta) = exp{-R^2 *sin(2*theta)} is 1, not exp(-R^2), and this makes you proof wrong.!
    Any other proof that is right???
    this is my homework pls
    Last edited: Jun 6, 2011
  10. Jun 6, 2011 #9
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