Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus & Beyond Homework (http://www.physicsforums.com/forumdisplay.php?f=156)
-   -   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:

http://www.physicsforums.com/showthread.php?t=368440

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.


All times are GMT -5. The time now is 06:43 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums