Need help finding contour for this integral

  • Thread starter quasar_4
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Homework Statement



Hi everyone. I'm trying to integrate the function

[tex]\int_0^{\pi} \frac{1}{(1+\cos{\theta}) \sqrt{1+q^2 \cos^2{\theta}+2q\cos{\theta}-p^2 \sin^2{\theta}}} d\theta[/tex]

analytically, where p and q are constants satisfying [tex] p^2 + q^2 =1 [/tex].

Homework Equations



[tex] p^2 + q^2 =1 [/tex]
[tex] \cos{\theta} = \frac{1}{2}(e^{i\theta}+e^{-i\theta}), \sin{\theta} = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) [/tex]
Residue theorem, Cauchy's theorem, etc

The Attempt at a Solution



It seemed to me that the only way to do this was as a contour integral, so I made the substitution

[tex] z = e^{i\theta} \rightarrow dz = iz d\theta \rightarrow d\theta = -i/z dz [/tex].

Using the above trig substitutions, I simplified and got the integral

[tex] \int \frac{1}{(1+z)^2 \sqrt{4+2(q^2-p^2)+\frac{1}{z^2}+z^2+\frac{4q}{z}+4qz} }dz[/tex].

It's still rather ugly, but I can see that there is a pole of order 2 at z=-1. I think I can neglect the roots of the stuff in the square root because each root has a factor of (-1+p^2+q^2) = 0 in the denominator, making all roots infinite. The issue now (assuming my pole analysis is ok) is to choose a contour, but because the sqrt in the denominator is not single valued I need to make a branch cut. Also, I'm integrating from 0 to pi, so that's a half circle in the complex plane. I can't see how to make a contour that includes both the branch cut and goes around the pole at z=-1. Can anyone help, or at least tell me if what I've done so far is valid? :)
 
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Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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Homework Statement



Hi everyone. I'm trying to integrate the function

[tex]\int_0^{\pi} \frac{1}{(1+\cos{\theta}) \sqrt{1+q^2 \cos^2{\theta}+2q\cos{\theta}-p^2 \sin^2{\theta}}} d\theta[/tex]

analytically, where p and q are constants satisfying [tex] p^2 + q^2 =1 [/tex].

A little simplification of the radicand could make things much nicer for you:

[tex]\begin{aligned}1+q^2 \cos^2{\theta}+2q\cos{\theta}-p^2 \sin^2{\theta} & = 1+(1-p^2) \cos^2{\theta}+2q\cos{\theta}-p^2 \sin^2{\theta} \\ & = 1+\cos^2{\theta}+2q\cos{\theta}-p^2 (\sin^2\theta+\cos^2\theta) \\ & = 1+ \cos^2{\theta}+2q\cos{\theta}-p^2 \\ & = q^2 +\cos^2{\theta}+2q\cos{\theta}\end{aligned}[/tex]

...and that looks a lot like a perfect square to me :wink:
 
  • #3
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ooo, that makes me VERY happy!! :-) :rofl:

Now it's cute and simple and just has a pole of order 4 at z=-1. No icky branch cuts! Yay!

THANK YOU!
 
  • #4
gabbagabbahey
Homework Helper
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ooo, that makes me VERY happy!! :-) :rofl:

Now it's cute and simple and just has a pole of order 4 at z=-1. No icky branch cuts! Yay!

THANK YOU!

You might still need to be a little careful since

[tex]\sqrt{a^2}=|a|=\left\{\begin{array}{lr}a, & a \geq 0 \\ -a, & a<0\end{array}\right.[/tex]

Unless you know for certain that [itex]q\geq 1[/itex], your radicand will have two different branches.
 

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