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Gaussian integral

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data
    We know that
    [tex]\int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\pi \over a}[/tex].

    Does this hold even if [tex]a[/tex] is complex?


    2. Relevant equations



    3. The attempt at a solution
    In the derivation of the above equation, I don't see any reason why we must assume that [tex]a[/tex] be real. So I think it does hold for complex [tex]a[/tex].
     
  2. jcsd
  3. Jun 3, 2008 #2

    benorin

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    It holds for Re(a)>0.
     
  4. Jun 3, 2008 #3

    tiny-tim

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    Yes … it relies on [tex]e^{-ax^2}\arrowvert_{\infty}\,=\,0[/tex] .

    If a = b + ic, then [tex]e^{-ax^2}=\,e^{-bx^2}e^{-icx^2}[/tex] , which is 0 if b > 0, and really wobbly if b ≤ 0. :smile:
     
  5. Jun 3, 2008 #4
    Great. Thank you for your replies.
     
  6. Jun 3, 2008 #5
    Hi again. I'd just like to make a remark that is bothering me. I created this thread because I was trying to find a Green function and arrived at
    [tex]G(x,x',t) = {1\over{2\pi}}e^{im(x-x')\over{2t}}\int_{-\infty}^\infty e^{-{it\over{2m}}\left(p-{m(x-x')\over t}\right)^2}[/tex].
    Please ignore everything except the integral.
    When I naiively use the gaussian integral formula in my original post, I get the correct answer. But according to what you said, I should not be able to do this since (i.e. Re(a)<0) in this case. Can you see any reason for this?
     
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