1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second Pole Residue

  1. Mar 3, 2013 #1
    ∫(dθ)/(a+bcosθ)^2


    2. Relevant equations
    I'm trying to find the above integral (from 0-2pi) using Cauchy's Residue theorem. After closing the contour and re-writing the integrant, I know that I have singularity at (-a/b)+(√(a/b)^2-1)- (double pole or is it??).

    3. The attempt at a solution
    I have tried both single pole and double pole residues using the limit approach but the calculation gets very cumbersome and i don't arrive at the write answer [its (2a*pi)/(a^2-b^2)^(3/2)]. According the Mathematica Res[(-a/b)+(√(a/b)^2-1)]=b*(b^2-a^2)*((a/b)^2-1)^(1/2).

    Any tips/suggestions would be appreciate on how to arrive at the result given by Mathematica (which agrees with answer given in the book).

    Thanks,
     
  2. jcsd
  3. Mar 4, 2013 #2
    Well, what you are calculating with Mathematica makes no sense. You should be calculating the residue of [tex] \frac{z}{(az + b(z^2+1)/2)^2}. [/tex]

    Your work seems correct as far as you write it... So the integrand has a double pole at [itex] z = -\frac{a}{b} + \sqrt{\frac{a^2}{b^2}-1} [/itex] assuming a/b>0.

    You can write the integral as [tex] I = \frac{4i}{b^2} \int \frac{z dz }{ (z^2 + \frac{2a}{b} z + 1)^2} = \frac{4i}{b^2} \int \frac{z dz }{(z-c_{+})^2(z-c_{-})^2} [/tex] where [itex] z_{\pm} = -\frac{a}{b} \pm \sqrt{\frac{a^2}{b^2} -1} [/itex]. Finding the residue from here should not be too difficult.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Second Pole Residue
  1. Poles and residues (Replies: 12)

Loading...