Help me solving this complex integral

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Discussion Overview

The discussion revolves around solving a complex integral involving exponential functions and poles, specifically the integral \(\oint \frac{e^{-(a+b)+az+\frac{b}{z}}}{z(z-1)}dz\) over the unit circle, with \(a\) and \(b\) being positive constants. The conversation includes technical details about contour integration and residue calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant identifies two poles in the integral, one at the center and another on the boundary, suggesting that contour deformation may be necessary.
  • Another participant expresses concern about the pole at \(z=0\), questioning whether it is of infinite order.
  • A participant proposes that the integral can be simplified to \(e^{-(a+b)}\oint_{\gamma}\frac{e^{az+\frac{b}{z}}}{z(z-1)}dz\) and mentions the need to compute Laurent series for the involved functions.
  • One participant provides a proposed answer involving residues at the poles, detailing the contributions from both \(z=1\) and \(z=0\), and presents a final expression for the integral.

Areas of Agreement / Disagreement

Participants express various viewpoints regarding the handling of the poles and the necessary calculations, but there is no consensus on the correctness of the proposed solution or the interpretation of the poles.

Contextual Notes

The discussion includes assumptions about the behavior of the integral near the poles and the convergence of the series involved, which remain unresolved.

sabbagh80
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Hi,

could you please help me solving this integral:

[itex]\oint \frac{e^{-(a+b)+az+\frac{b}{z}}}{z(z-1)}dz[/itex]

over the unit circle, where [itex]a, b[/itex] are two positive constants (it is not a homework)
thanks a lot in advance
 
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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.
 
hunt_mat said:
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 [itex]z=0[/itex]. it is of order infinity. Am I right?
 
So the integral basically becomes:
[tex] e^{-(a+b)}\oint_{\gamma}\frac{e^{az+\frac{b}{z}}}{z(z-1)}dz[/tex]
According to the sources I have read, you have to compute the Laurent series for [itex]e^{z}[/itex] and te Laurent series of [itex]e^{\frac{1}{z}}[/itex] along with all the other functions involved and just pick out the coefficient of the [itex]1/z[/itex] term. Sorry, but it is going to take a lot of algebra on this one.
 
The answer is as follows:

Residue at pole z=1 is [itex]2\pi \frac{1}{2}[/itex]
and residue at pole z=0 is [itex]-2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}[/itex]

So, we conclude the result as:

[itex]\pi - 2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}[/itex]

Is everything Ok?
 

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