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Homework Help: Real integral using Residue Theorem

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Convert to suitable contour integral and use Cauchy's Residue Theorem to evaluate it

    ##\int_0^{2\pi} \frac{\sin \theta d\theta}{5-4\sin \theta}##

    2. Relevant equations

    Res ##f(z)_{z=z_0} = Res_{z=z_0} \frac{p(z)}{q(z)}=\frac{p(z_0)}{q'(z_0)}##

    3. The attempt at a solution

    ## \displaystyle \int_0^{2\pi} \frac{\sin \theta d\theta}{5-4\sin \theta}=\oint_C \frac{(1/2i)(z-1/z)dz/iz}{5-4(z-1/z)}=\oint_C \frac{(z-1/z)dz}{8z^2-10z-8}##

    The denominator has simple poles at ##z_1=5/8+\sqrt{89}/8## and ##z_2=5/8-\sqrt{89}/8## with z_2 being inside the unit circle hence

    Res ##f(z)_{z=z_0} = Res_{z=z_0} \frac{z-1/z}{8z^2-10z-8}=\frac{z-1/z}{16z-10}|_{z=5/8-\sqrt{89}/8}##....?
  2. jcsd
  3. May 7, 2012 #2
    You got to move that 1/z downstairs. So write it as
    [tex] I = \frac{-1}{2} \oint \frac{(z^2 -1) dz}{5z^2-2/i (z^3-z)} [/tex]
  4. May 9, 2012 #3
    Not sure what you are suggesting. I have checked mine with Mathematica and it seems right.
    Perhaps I am not applying the Theorem correctly?


  5. May 9, 2012 #4
    The point of residue calculus is that you need to find all the poles of the integrand. An easy find to way the poles is to write your integrand in the form of polynomial divided by polynomial. Now you are missing the pole at z=0, because you don't have a polynomial in the numerator.
  6. May 9, 2012 #5
    But I thought we find all the poles in the denominator and in this case it will be 2 poles because of quadratic...?
  7. May 9, 2012 #6
    You got that entirely wrong and you're not missing it. It's still there just in the numerator. The objective is to find all the singular point of the function however it is written, numerator, denominator, polynomial, trig, special function, whatever, wereever they are and then by the Residue Theorem, an integral of the function over a closed contour where the function is single-valued and analytic except for issolated singular points like poles is 2pi i times the sum of the residues. Now it does make it look nicer if you put that 1/z factor in the denominator but just don't put it there right? Is that insulting? I mean by valid algebraic means. It's then easier to compute the residue by standard techniques.
  8. May 10, 2012 #7
    SO my original integrand in post #1 was correct as above link, the simple pole z=0 was in the numerator and the other two in the denominator


  9. May 10, 2012 #8
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