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Contour integral, exp(-z^2)

  • Thread starter NT123
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  • #1
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Homework Statement

Integrate exp(-z^2) over the rectangle with vertices at 0, R, R + ia, and ia.



Homework Equations



int(0, inf)(exp(-x^2)) = sqrt(pi/2)

The Attempt at a Solution

I really don't have much of an idea here - the function is analytic so has no residues... The part from 0 to R is just the real integral, but for the other 3 sides I'm not too sure on how to proceed.
 

Answers and Replies

  • #2
ideasrule
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Isn't the contour integral equal to 0 if there are no poles?
 
  • #3
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Isn't the contour integral equal to 0 if there are no poles?
This is what I would have thought, but I'm supposed to be using the integral of e^(-z^2) to evaluate the real integral int(0,inf)((e^(-x^2))*cos(2ax)), which is apparently equal to
sqrt(pi)*exp(-a^2)/2.
 
  • #5
ideasrule
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Ah, that makes much more sense.

If we want to integrate from R+ia to ia, just integrate e^(-z^2)dz=e^-(x+ia)^2 dx from R to 0. Do the same for the other 3 sides. You won't get an analytic answer, but that's OK; just write out the entire contour integral first and you'll see where this is going.
 

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