Help me solving this complex integral

[sabbagh80]

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]

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]

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]

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]

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?