Contour integration

  • Thread starter bwinter
  • Start date
  • #1
27
1

Homework Statement


Evaluate [tex]\int_{0}^{2\pi} \frac{d\theta}{1+\epsilon cos\theta}[/tex]

where [tex]\left|\epsilon\right|<1[/tex], by letting [tex]z=e^{i\theta}[/tex] and [tex]cos\theta = (z+z^{-1})/2[/tex] and choosing contour [tex]\left|z\right| = 1[/tex], a unit circle.


Homework Equations


I know this has something to do with contour integration, but there are no poles as far as I'm aware and I'm used to going from [tex]dz[/tex] to [tex]d\theta[/tex] that I'm a bit confused here as to what to do.


The Attempt at a Solution


I've gotten as far as subbing in the relevent replacement, but I'm lost otherwise. Residues don't apply here, right?
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
Can you show us what you got after you rewrote the integral in terms of z?
 
  • #3
27
1
I'm not sure how to do that either. I just replaced the cosine with what's written. Not sure how dtheta changes into dz.
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
You do it like you normally do. You start with [itex]z=e^{i\theta}[/itex] and differentiate to get [itex]dz=ie^{i\theta}\,d\theta=iz\,d\theta[/itex].
 
  • #5
27
1
Oh, ok I think I got this. Just needed a push in the right direction. If I do that I get a quadratic in the denominator which results in two poles, then I can choose one and integrate around it which results in a residue which gives the answer. Sounds right?
 
  • #6
Cyosis
Homework Helper
1,495
0
You can't just choose one and integrate around it. You're integrating along the unit circle [itex]|z|=1[/itex]. Therefore you must take the pole that lies within this circle.
 
  • #7
27
1
I've found that the poles exist at [tex]z=\frac{-1\pm\sqrt{1-\epsilon^2}}{\epsilon}[/tex] which don't fall inside the circle. Does this imply there are no poles I have to be concerned about? I'm super confused what I'm supposed to be doing here.
 
Last edited:
  • #8
Cyosis
Homework Helper
1,495
0
I haven't checked your entire solution, but it seems you missed the 'obvious' z=0 pole.
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
Your roots aren't correct. I think you made an algebra mistake somewhere. What did you get in the denominator? I got [itex]\epsilon z^2+2z+\epsilon[/itex].
 
  • #10
27
1
I haven't checked your entire solution, but it seems you missed the 'obvious' z=0 pole.
That's actually not a pole here, I don't think. Plugging in z=0 gives me [tex]\epsilon[/tex] in the denominator.

Your roots aren't correct. I think you made an algebra mistake somewhere. What did you get in the denominator? I got [itex]\epsilon z^2+2z+\epsilon[/itex].

Yep, that's what I got after I factored out the 2/i, in which case my roots are correct, as I double-checked them with Wolfram.
edit: Actually one of my roots does fall inside the circle after all. Do I find a residue for that pole?
 
Last edited:
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,164
1,763
OK, I see you fixed the expression for the roots in your earlier post. I wasn't sure if you got the wrong roots or just entered the wrong LaTeX. In any case, our roots agree now.

So, yeah, as you found, one pole is inside the contour. The contour integral will equal 2πi times the residue at that pole, so you want to find it.
 

Related Threads on Contour integration

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
10
Views
769
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
2
Views
821
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
1K
Top