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Complex integral

  1. Sep 25, 2005 #1

    quasar987

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    Now I have to evaluate

    [tex]\int_{-\infty}^{\infty} e^{-Bx^2} e^{-iAx} dx[/tex]

    Splitting it in two using Euler's identity show that the imaginary part is 0 (cuz integrand is odd). Remains the real part

    [tex]2 \int_0^{\infty} cos(-Ax) e^{-Bx^2} dx[/tex]

    for which integration by parts leads nowhere.
     
  2. jcsd
  3. Sep 26, 2005 #2

    Tide

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    I recommend completing the square in the exponential then suitably choosing a loop around which to integrate in the complex plane.
     
  4. Sep 26, 2005 #3

    quasar987

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    I should have mentionned that I have no knowledge of complex analysis whatesoever. I don't know anything about residues, integration in the complex plane, and stuff like that.
     
  5. Sep 26, 2005 #4

    Tide

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    Since a complex number appears in the exponential you are at least familiar with some of the basics. I still recommend completing the square in the exponential. It should provide some illumination. :)
     
  6. Sep 26, 2005 #5

    quasar987

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    I love ilumination. Let me try just that. :tongue2:
     
  7. Sep 26, 2005 #6

    quasar987

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    Is it just [itex]\sqrt{\pi / B}[/itex]?

    After completing the square, I'm left with

    [tex]\mbox{exp}(-A^2 / 4B) \int \mbox{exp}(-B(x+Ai/2B)^2) dx[/tex]

    And so with substitution y = x+Ai/2B, I get the integral

    [tex]\mbox{exp}(-A^2 / 4B) \int_{-\infty}^{\infty} \mbox{exp}(-By^2) dy[/tex]

    which is [itex]\sqrt{\pi / B}[/itex] as pointed out by another thread by Tom Mattson.

    Is this valid with complex too?
     
    Last edited: Sep 26, 2005
  8. Sep 26, 2005 #7

    Tide

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    quasar,

    Way to go!
     
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