Remind me how to do this complex integral

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Discussion Overview

The discussion revolves around the evaluation of complex integrals, specifically focusing on the integral of the function e^z/(1-cosz) over a circular contour and the related properties of integrals involving singularities. Participants explore methods for calculating residues and the implications of singularities on the integral's value.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about integrating e^z/(1-cosz) over a circle of radius 2 and mentions a singularity at z=0, suggesting that the solution involves 2*pi*i times the residue at that point.
  • Another participant questions whether there is a theorem stating that the integral of 1/(1-sinz) over a circle of radius 2 would be zero, noting the presence of two singularities at equal distances.
  • A third participant cites Wolfram Alpha, indicating that the residue for e^z/(1-cosz) is 2, and seeks clarification on this result.
  • One participant reflects on the theorem regarding the sum of residues, suggesting that the integral would equal zero only if the residues at the singularities are negatives of each other.
  • Another participant corrects their earlier statement, asserting that only the singularity at z=0 is inside the contour, implying that other singularities do not affect the integral.
  • One participant expresses difficulty in finding a solution through various methods, including series expansion, without success.
  • Another participant notes that the residue can be identified as the coefficient of 1/z in the series expansion, as all other terms would integrate to zero.
  • One participant discusses their approach of expanding e^z and 1-cosz into series but struggles to isolate the 1/z term.
  • A later reply clarifies that they were not integrating but rather stating that term-by-term integration would yield only the 1/z term as nonzero.

Areas of Agreement / Disagreement

Participants express various methods and approaches to the problem, with some uncertainty regarding the application of theorems and the identification of residues. There is no consensus on the best method or final solution, and multiple viewpoints on the nature of the singularities and their impact on the integral remain present.

Contextual Notes

Participants mention different singularities and their locations, as well as the potential for residues to affect the integral's value. There are unresolved aspects regarding the series expansions and the identification of the relevant terms for integration.

catsarebad
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integrate e^z/(1-cosz) dz over circle of radius, say 2

i can't seem to recall how it is done.

singularity at z=0

2*pi*i * res (at z=0) would be the solution

any shortcut to find this residue?
 
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btw, is there any kind of theorem saying integral of 1/(1-sinz) dz over circle of radius 2 would be zero? this one has 2 singularities at equal distance apart. i feel like there's a theorem somewhere. anymore care to refresh my memory?
 
catsarebad said:
btw, is there any kind of theorem saying integral of 1/(1-sinz) dz over circle of radius 2 would be zero? this one has 2 singularities at equal distance apart. i feel like there's a theorem somewhere. anymore care to refresh my memory?

The only one I know is the one where you use the sum of the residues at each of the singularities, so it would integrate to 0 only if the residue at one singularity is the negative
of the other one.
 
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nvm what i said before. the only singularity inside our contour is z = 0. rest are outside and don't even matter.

and yes

integral = 2*pi*i * [sum of residues at each singularities]
 
i've tried several methods (that i can think of) for this problem and none of them seem to work.

also tried expanding as series and didn't get anywhere.
 
If you have done the expansion, the residue is the coefficient of 1/z , since all other terms integrate to 0.
 
why are you integrating?

this is what i was doing

e^z = 1+z+z^2/2 + O(z^3)
1-cosz = z^2/2-z^4/4! + O(z^6)

i couldn't really single out 1/z part
 
No I'm not integrating, just stating that if you did term-by-term integration of the series, the only term that would be nonzero is the term containing 1/z. Let me think it thru a bit more.
 

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