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Integration using residue theory

  1. Dec 22, 2009 #1
    I'm trying to evaluate the integral:
    [tex]\int_0^{2\pi} \frac{d\theta}{24-6sin\theta}
    [/tex]
    using calculus of residues.

    I've tried this so far:
    Let [tex]z=e^{i\theta}[/tex] so

    [tex]d\theta=\frac{dz}{iz}[/tex].
    Also, using the exponential definition of sine,

    [tex]sin\theta=\frac{z^2-1}{2iz}[/tex]

    This gives messy expressions for the singularities of the function, and therefore the residue I got is messy and not useful(it gives the wrong answer). The same kind of calculation is done in my book with cosine instead of sine, and it works out nicely, so I think I must be missing something in this question.

    Anything in specific I should try?
     
  2. jcsd
  3. Dec 22, 2009 #2
    Yes there is the standard substitution for these kinds of integrals. And the substitution is tan (@/2)= t. Differentiating that u get d@=2/(1+t^2)dt. Because differential of tan(@/2) is 1/2*1/cos(@/2)^2 and this is equal to1*2( 1+Tan(@/2)^2).This is the usual identity just refresh you knowledge. Then we have to express sin@ in terms of tan(@/2). U do this by following
    sin(@)=2sin(@/2)cos(@/2)/1
    This 1 can be written as 1= sin^2(@/2)+ cos^2(@/2).Substitute this into your fraction and then divide top and bottom by cos^2(@/2) and you get sin in terms of tan.
    sin(@/2)=2t/(t^2+1).
    Now u have everything and substitute these into your main integral and u should be in business.
    Hope this is what u asked for, I have no idea what residue theory is:)
     
  4. Dec 22, 2009 #3
    It is a simple matter of algebra. Solve the quadratic by setting the denominator to zero (i.e. after you've multiplied the 1/(i z) factor with the denominator that originally contained the sin). Then determine which of the two zeroes is inside the unit circle. You can then write the integrand in the form:

    A/[(z-z1)(z-z2)]

    where z1 and z2 are the two poles. The residue at z1 is then obviously A/(z1-z2).

    So, you already know that the value of the integral is proportional to the reciprocal of the difference of the two zeroes. Now, if you consider the formula for the roots of the general quadratic equation and you take the difference of the two solutions, you get a simple square root. Also, you can see that taking the wrong root for the pole inside the unit circle can only yield an erroneous minus sign. You can also see this by considering the contour integral, if you take the integral to be a circle of radius R and consider the limit of R to infinity, then the integral tends to zero. But the integral is 2 pi i times the sum of both residues ion that case, implying that the two residues differ in sign.
     
  5. Dec 23, 2009 #4
    One question I have about this is why use a unit circle as a contour? If you use a circle of radius 10 say, both roots are inside, so obviously this will give a different answer if I don't change the way I do the question. How should using a larger radius contour change how the question is done, since there can only be one answer to the original question?
     
  6. Dec 23, 2009 #5

    Mute

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    Homework Helper

    You choose a circular contour of radius 1 when you set [itex]z = \exp[i\theta][/itex], since |z| = 1. You could chose [itex]z = r\exp[i\theta][/itex], with r any nonzero positive number you wanted, which would correspond to choosing a circular contour of radius r, but this would change the form of the integral you do and hence where the poles are; however, all the r dependence will cancel out in the end.
     
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