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-   -   Contour integral, exp(-z^2) (http://www.physicsforums.com/showthread.php?t=388318)

 NT123 Mar20-10 08:04 PM

Contour integral, exp(-z^2)

1. The problem statement, all variables and given/known data Integrate exp(-z^2) over the rectangle with vertices at 0, R, R + ia, and ia.

2. Relevant equations

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

3. 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.

 ideasrule Mar20-10 08:11 PM

Re: Contour integral, exp(-z^2)

Isn't the contour integral equal to 0 if there are no poles?

 NT123 Mar20-10 09:36 PM

Re: Contour integral, exp(-z^2)

Quote:
 Quote by ideasrule (Post 2633527) 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.

 Count Iblis Mar20-10 10:02 PM

Re: Contour integral, exp(-z^2)

See here:

for some explanations.

 ideasrule Mar20-10 10:29 PM

Re: Contour integral, exp(-z^2)

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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