How do I solve complex contour integrals in complex analysis?

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Discussion Overview

The discussion revolves around solving complex contour integrals in complex analysis, specifically focusing on a particular integral involving poles and the residue theorem. Participants explore different methods for approaching the problem, including parametrization and the application of the residue theorem, while also addressing the importance of defining the contour of integration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about the approach to take for solving the integral and requests a step-by-step guide.
  • Another participant questions the definition of the contour and suggests that the roots should be enclosed by the contour for the residue theorem to apply.
  • A participant explains the residue theorem and provides the formula for calculating residues at second-order poles, indicating the need to find residues for the given roots.
  • One participant emphasizes the importance of knowing the contour before applying the residue theorem and suggests starting with basic definitions before using more advanced methods.
  • Another participant clarifies that when applying the residue theorem, the winding direction around each root is crucial, even if the contour is not well-defined.
  • A participant shares their method of obtaining the integral by changing variables and suggests that the contour is over the unit circle.
  • Another participant introduces a related problem involving a Whittaker contour integral and seeks assistance in deriving the function from the integral.

Areas of Agreement / Disagreement

Participants generally agree on the importance of defining the contour and the application of the residue theorem, but there are multiple approaches discussed, and no consensus is reached on the specific method to solve the integral or the correctness of the proposed solutions.

Contextual Notes

There are limitations regarding the assumptions about the contour and the roots, as well as the dependence on the definitions of the variables involved. The discussion includes unresolved mathematical steps and varying interpretations of the contour's properties.

matt_crouch
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I am trying to teach myself complex analysis . There seems to be multiple ways of achieving the same thing and I am unsure on which approach to take, I am also struggling to visualise the problem...Would someone show me step by step how to solve for example

$$\int_{\Gamma}\frac{2\lambda}{[(\lambda-\lambda_{+})(\lambda-\lambda_{-})]^{2}}\frac{d\lambda}{i}$$

where ##\lambda_{+}, \ \ \lambda_{-}## are the roots given by
##\lambda_{\pm}=\frac{-i\zeta\pm ir}{q}##

So i believe the answer should be ##\frac{x}{r^{3}}## or something like that but I can't get there. Could someone show me how to approach a problem like this... Like visualise the contour etc
 
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Do you know how the contour ##\Gamma## is defined? Should we assume that the roots, ##\gamma_{-}## and ##\gamma_{+}## are enclosed by the contour?
If they are then use the residue theorem to solve. It is very difficult to help you visualize a contour that is not defined.
Cauchy's Residue Theorem tells you that
##\int_\Gamma f(\gamma) \, d\gamma = 2\pi i \sum_{a \in \text{poles}}\mathrm{Res}(f,a)##
You already have the roots.
You should see that the poles are second order in this function, so use the higher order residue formula:
(From Wikipedia's page on Residue_(complex_analysis))
More generally, if c is a pole of order n, then the residue of f around z = c can be found by the formula:

fc16ae96cdbc7aea1023462e07d19753.png

For a 2nd order pole, this is simply

##\mathrm{Res}(f,a)=\lim_{\gamma \to a} \frac{d}{d\gamma} (\gamma - a)^2 f(\gamma) ##

Next, find the residues for ##\gamma_+## and ##\gamma_-##:
##\mathrm{Res}(f,\gamma_+)=\lim_{\gamma \to \gamma_+} \frac{d}{d\gamma}\left[ (\gamma-\gamma_+)^2\frac{2\gamma}{ i (\gamma-\gamma_+)^2 (\gamma-\gamma_-)^2}\right]##
Simplifying and taking the derivative, you get:
##\mathrm{Res}(f,\gamma_+)=\lim_{\gamma \to \gamma_+}\frac{2(\gamma_- +\gamma) }{i (\gamma_- - \gamma)^3 }=\frac{2(\gamma_- +\gamma_+) }{i (\gamma_- - \gamma_+)^3 }##
Due to the symmetry, you will get something very similar for ##\mathrm{Res}(f,\gamma_-)##.
Then, you plug in the expression you have for ##\gamma_+ , \gamma_-## to simplify even more.
Finally, you use the residue theorem to give you that the integral is equal to ##2\pi i ( \mathrm{Res}(f,\gamma_+)+\mathrm{Res}(f,\gamma_-))##.
And you are done.
 
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RUber's post should be very helpful. The most important point is that you can't even begin to think about a complex contour integral without knowing what contour you are integrating around.

If you are just beginning to learn about contour integrals, it's probably a good idea to do some "by hand" — i.e., just going back to the definitions — before graduating to the extremely useful method of residues.

For instance, suppose you want to integrate f(z) = 1/z about the contour given by the unit circle C about the origin in ℂ. So: parametrize the contour C, by say

z(t) = eit, 0 ≤ t ≤ 2π,​

which turns the original integral

C dz/z​

into

∫ (1/eit) i eit dt = ∫ i dt​

(where the integral is from t = 0 to t = 2π) , since dz = i eit dt.

Thus we have shown

C dz/z = 2πi,​

"by hand".
 
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To fill in a couple of important points regarding the comments above. Assuming that the contour is a closed curve. When applying the residue theorem, you do not need to know a lot about the contour. You do need to know how many times and in what direction it winds around each root, clockwise or counterclockwise.
 
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Ok thanks for the replies everyone...

So the way I obtained the integral in the first place was to go from

$$\vartheta \rightarrow \lambda,$$ by using $$\lambda = e^{i\vartheta}$$

So i guess the contour is an integration over the unit circle like Zinq said.

Ok thanks for your comments
 
So further from this, my original problem was to calculate the so called Whittaker contour integral

so we start from

$$f=\int_{0}^{2\pi} \frac{1}{(x+izcos(\vartheta)+iysin(\vartheta))^{2}}d\vartheta$$

This should give $$f=2\pi/r^(3)$$

could someone show me how? Essentially i changed from $$\vartheta \rightarrow \lambda$$ where $$\lambda=e^{i\vartheta}$$ and then i tried the contour integral but I still don't see how one can get the function f above?
 
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