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Gaussian integral

  • Thread starter Pacopag
  • Start date
  • #1
197
4

Homework Statement


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?


Homework Equations





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].
 

Answers and Replies

  • #2
benorin
Homework Helper
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It holds for Re(a)>0.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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It holds for Re(a)>0.
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:
 
  • #4
197
4
Great. Thank you for your replies.
 
  • #5
197
4
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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