# Complex contour integrals

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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RUber
Homework Helper
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 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:

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.

matt_crouch and zinq
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".

matt_crouch
FactChecker
Gold Member
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.

matt_crouch and zinq
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.

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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