Solving Second Pole Residue ∫(dθ)/(a+bcosθ)^2

[davavsh]

∫(dθ)/(a+bcosθ)^2

Homework 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??).

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,

 

[clamtrox]

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.