How do I solve complex contour integrals in complex analysis?

In summary, the person is trying to teach themselves complex analysis and is struggling to visualize the problem. They ask for help and are provided with a summary of how to solve the equation.
  • #1
matt_crouch
161
1
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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  • #2
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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  • #3
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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  • #4
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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  • #5
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
 
  • #6
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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1. What is a complex contour integral?

A complex contour integral is a type of integral that is computed along a specified path, called a contour, in the complex plane. It involves the integration of a complex-valued function along this path, and is used to evaluate complex functions in various mathematical and scientific fields.

2. How is a complex contour integral different from a regular integral?

A complex contour integral is different from a regular integral in that it involves integration along a specific path in the complex plane, rather than over a continuous interval on the real number line. This allows for more flexibility in the types of functions that can be integrated and the resulting values that can be obtained.

3. What are some applications of complex contour integrals?

Complex contour integrals have numerous applications in mathematics, physics, and engineering. They are commonly used in the evaluation of complex-valued functions, in the solution of differential equations, and in the analysis of physical systems such as electric circuits and fluid flow.

4. How is a complex contour integral computed?

The computation of a complex contour integral involves breaking down the contour into smaller, simpler segments and then using various techniques such as the Cauchy integral formula or the residue theorem to evaluate the integral along these segments. These values are then combined to obtain the final result.

5. Can complex contour integrals be evaluated numerically?

Yes, complex contour integrals can be evaluated numerically using methods such as the trapezoidal rule or Simpson's rule. However, these methods may not always provide accurate results and are typically used as approximations in cases where the integral cannot be evaluated analytically.

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