View Full Version : help me solving this complex integral
sabbagh80
Jul9-11, 10:07 AM
Hi,
could you please help me solving this integral:
\oint \frac{e^{-(a+b)+az+\frac{b}{z}}}{z(z-1)}dz
over the unit circle, where a, b are two positive constants (it is not a homework)
thanks a lot in advance
hunt_mat
Jul9-11, 01:09 PM
There are two poles, one in the centre (no problem) and on on the boundary (not that much of a problem, you just have to deform your contour slightly. I think you main problem is going to be exp(b/z) term.
sabbagh80
Jul10-11, 01:49 AM
There are two poles, one in the centre (no problem) and on on the boundary (not that much of a problem, you just have to deform your contour slightly. I think you main problem is going to be exp(b/z) term.
I think the problem is exactly related to the pole which is placed at z=0 . it is of order infinity. Am I right?
hunt_mat
Jul10-11, 04:45 AM
So the integral basically becomes:
e^{-(a+b)}\oint_{\gamma}\frac{e^{az+\frac{b}{z}}}{z(z-1)}dz
According to the sources I have read, you have to compute the Laurent series for e^{z} and te Laurent series of e^{\frac{1}{z}} along with all the other functions involved and just pick out the coefficient of the 1/z term. Sorry, but it is going to take a lot of algebra on this one.
sabbagh80
Jul10-11, 11:56 AM
The answer is as follows:
Residue at pole z=1 is 2\pi \frac{1}{2}
and residue at pole z=0 is -2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}
So, we conclude the result as:
\pi - 2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}
Is everything Ok?
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