Complex integral

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  • #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.
 

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  • #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.
 
  • #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.
 
  • #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. :)
 
  • #5
quasar987
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I love ilumination. Let me try just that. :tongue2:
 
  • #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?
 
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  • #7
Tide
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quasar,

Way to go!
 

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