# Complex contour integrals

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1. Feb 25, 2016

### matt_crouch

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

Last edited by a moderator: Feb 25, 2016
2. Feb 25, 2016

### RUber

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.

3. Feb 25, 2016

### 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".

4. Feb 25, 2016

### FactChecker

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.

5. Feb 26, 2016

### matt_crouch

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.

6. Feb 26, 2016

### matt_crouch

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?

Last edited: Feb 26, 2016