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Integration of Sin[x^2]?

  1. Mar 29, 2008 #1
    I've attached my work, but basically, I'm trying to compute this:
    [tex]\int_{0}^{\infty}2\sin({x^2})dx[/tex].
    (The 2 is only there because when you expand it into an exponential function, it makes like easier.)
    You can look at my work to see what I did. My question now is, is it possible to express that limit in exact form? If so, what is it? Why does the limit have this value?
    [​IMG]
    I also tried doing this with double integrals and polar coordinates to calculate the square of the integral but I ran into a problem deep into it of evaluating [itex]\sin{\infty}[/itex] and [itex]\cos{\infty}[/itex].
    If you're curious why I care about this, it's because this integral is important in the diffraction of light. While only the approximate answer is important in physics, I'm just curious about the mathematics behind it.
    Thanks in advanced.
     
    Last edited: Mar 29, 2008
  2. jcsd
  3. Mar 29, 2008 #2
    It has no elementary anti derivative so it can't be evaluated in a closed form, but you can only get a pretty good approximation, as you have done
     
  4. Mar 29, 2008 #3

    tiny-tim

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    Hi gamesguru! :smile:

    (1) 1.2533 = √(π/2).

    (2) Do you know the trick for integrating [tex]\int_{0}^{\infty}e^{-x^2}dx[/tex] ?

    Try putting [tex]\int_{0}^{\infty}\sin{x^2}dx[/tex] = imaginary part of [tex]\int_{0}^{\infty}e^{i x^2}dx[/tex]. :smile:

    (or perhaps [tex]\lim{\epsilon\to0}\int_{0}^{\infty}e^{(i\,-\,\epsilon) x^2}dx[/tex], for positive real epsilon)
     
  5. Mar 29, 2008 #4
    @tiny-tim
    I tried that method, read my post. I got to a point where I was evaluating cos[∞] or sin[∞]. Try it for yourself, it won't work.
     
  6. Mar 29, 2008 #5

    tiny-tim

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    ah … I must have stopped reading the post when I started looking at the .jpg, and then forgotten where to resume reading. :redface:
    Yup … that's why I suggested the epsilon method.

    (I haven't actually checked it myself … but the result is right, so I expect it works … over to you! :smile:)
     
  7. Mar 29, 2008 #6
    I don't see how the epsilon method makes this more easy to integrate. I'd appreciate it if you should me how to do it, or at least get started.
     
  8. Mar 29, 2008 #7

    tiny-tim

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    Because you end up evaluating [tex]e^{-\epsilon x^2} e^{i x^2}[/tex] from 0 to ∞ (which converges), instead of just [tex]e^{i x^2}[/tex] (which doesn't). :smile:
     
  9. Mar 29, 2008 #8
    In, [tex]e^{-\epsilon x^2} e^{i x^2}[/tex] the first term goes to 1 (e^0=1), not 0. I still don't see how this helps...
     
  10. Mar 29, 2008 #9

    tiny-tim

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    Yes, but e^-∞ = 0, which is what matters! :smile:
     
  11. Mar 29, 2008 #10
    [tex]
    e^{-\epsilon x^2} e^{i x^2}=e^{-0x^2}e^{ix^2}=e^{ix^2}
    [/tex]
    ...what am i missing? I don't see a [itex]e^{-\infty}[/itex].
     
  12. Mar 29, 2008 #11

    tiny-tim

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    ah … in my post #7, I was talking about x going from 0 to ∞.

    I thought you had already found that that was the problem … that e^(ix^2), or sin(x^2), doesn't converge as x -> ∞.

    But multiply it by e^(-(epsilon)x^2), where epsilon is fixed, and it converges to 0. :smile:
     
  13. Mar 29, 2008 #12
    I'm not sure I follow you. I'd really like it if you just do it so I could see how it's done and possibly learn something. We're making no progress right now.
     
  14. Mar 29, 2008 #13
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