Using residues to evaluate ∫_{-π}^{π} dθ/(1+sin²θ)

  • Thread starter Thread starter gtfitzpatrick
  • Start date Start date
  • Tags Tags
    Method
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
14 replies · 2K views
gtfitzpatrick
Messages
372
Reaction score
0

Homework Statement



evaluate using the method of residues [itex]\int^{\pi}_{-\pi}[/itex] [itex]\frac{d\theta}{1+sin^2\theta}[/itex] (=[itex]\pi\sqrt{2})[/itex]

Homework Equations


The Attempt at a Solution

[itex]sin^2 \theta = \frac{1}{2i}(z^2 - \frac{1}{z^2})[/itex] so our integral becomes [itex]\int \frac{2iz^2}{z^4+2iz^2-1}\frac{dz}{iz}[/itex]

but before i continue on from here I am not sure what the story with the limits. Normally I've just seen limits of 0 to 2[itex]\pi[/itex] so should iI put a [itex]\frac{1}{2}[/itex] in from of the integral?
 

Discussion

Physics news on Phys.org
sorry got that wrong [itex]sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2[/itex] should be [itex]\int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}[/itex] but still unsure about my limits...
 
As theta varies from -pi to pi, how much of the unit circle does it sweep out?
 
I think its Half a sweep, that's why i think i put a half in front of the integral
 
a whole sweep is 2[itex]\pi[/itex], half a sweep is o to [itex]\pi[/itex] so is [itex]\pi[/itex] to -[itex]\pi[/itex] half a sweep...backwards? so i put -[itex]\frac{1}{2}[/itex] in front of it?
 
gtfitzpatrick said:
sorry got that wrong [itex]sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2[/itex] should be [itex]\int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}[/itex] but still unsure about my limits...
^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.
 
awkward said:
What is pi - (-pi)?

2[itex]\pi[/itex] thanks :)
 
uart said:
^ Also, double check your denominator there, you've made a mistake in the algebra.

When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.

[itex]\int \frac{-4z^2}{z^4-2z^2+2}\frac{dz}{iz}[/itex]

thanks a million
 
**** that's not right either
 
Now I'm getting [itex]\int \frac{-4z^2}{z^4-6z^2+2}\frac{dz}{iz}[/itex] which doesn't work out that easy. So I am guessing I got it wrong again. I am going to use formula and see what it throws up anywho.thanks guys. Do either of you have a clue what my other post (jordans lemma) is about? i haven't a clue...
 
I think you need to double-check your algebra, one more time.
 
yes your right, it should be [itex]\int \frac{-4z^2}{z^4-6z^2+1}\frac{dz}{iz}[/itex]