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Complex Analysis Integral

  • Thread starter ryanwilk
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  • #1
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Homework Statement



Use the residue theorem to compute [tex]\int_0^{2\pi} sin^{2n}\theta\ d\theta[/tex]

Homework Equations



[tex]\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right)[/tex]

The Attempt at a Solution



I started with the substitution [tex]z = e^{i\theta}[/tex]

so that [tex]sin\theta= \frac{1}{2i} (z - \frac{1}{z})[/tex]

and [tex]d\theta = \frac{dz}{iz}.[/tex]

Therefore, the integral becomes: [tex]\oint_C \frac{1}{iz} \frac{1}{{(2i)}^{2n}} (z - \frac{1}{z})^{2n}\ dz = \frac{1}{i(2i)^{2n}} \oint_C \frac{(z - \frac{1}{z})^{2n}}{z}\ dz[/tex]

so there's a pole at z=0 of order 2n+1.

The residue is [tex]\frac{1}{i(2i)^{2n}(2n)!} \lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} z^{2n} (z - \frac{1}{z})^{2n} = \frac{1}{i(2i)^{2n}(2n)!} \lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} (z^2-1)^{2n}.[/tex]

But I have no idea how to evaluate this...
 
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Answers and Replies

  • #2
tiny-tim
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hi ryanwilk! :smile:

in that final differential, you only need the coefficient of z2n in z(z2 - 1)2n … but isn't that zero? :confused:
 
  • #3
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hi ryanwilk! :smile:

in that final differential, you only need the coefficient of z2n in z(z2 - 1)2n … but isn't that zero? :confused:
Ah I see! It should be just the coefficient of z2n in (z2 - 1)2n though, which is non-zero but looks complicated.
 
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  • #4
Dick
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tiny-tim is saying that the only term that contributes to the limit as z->0 of the 2n'th derivative of (z^2-1)^(2n) is the term containing z^(2n). In other words, use the binomial theorem on (z^2-1)^(2n). I think he got the extra z from an earlier version of your post.
 
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  • #5
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tiny-tim is saying that the only term that contributes to the limit as z->0 of the 2n'th derivative of (z^2-1)^(2n) is the term containing z^(2n). In other words, use the binomial theorem on (z^2-1)^(2n). I think he got the extra z from an earlier version of your post.
Oh right, so [tex]\lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} (z^2-1)^{2n}[/tex] seems to be [tex](-1)^n(2n)!{2n \choose n}.[/tex]

The residue is [tex]\frac{(-1)^n{2n \choose n}}{i(2i)^{2n}}[/tex] so the integral is just [tex]\frac{2\pi(-1)^n{2n \choose n}}{(2i)^{2n}}[/tex] ?
 
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  • #6
tiny-tim
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Yup! :biggrin:

EDIT: oh, you altered it :redface:

my "yup" was for the first line, i haven't checked the actual residue
 
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  • #7
Dick
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Oh right, so [tex]\lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} (z^2-1)^{2n}[/tex] seems to be [tex](-1)^n(2n)!{2n \choose n}.[/tex]

The residue is [tex]\frac{(-1)^n{2n \choose n}}{i(2i)^{2n}}[/tex] so the integral is just [tex]\frac{2\pi(-1)^n{2n \choose n}}{(2i)^{2n}}[/tex] ?
I checked the whole thing and that's correct. Of course, you can also cancel the (-1)^n in the numerator with the (i)^(2n). You know the result should be positive.
 
  • #8
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Yup! :biggrin:

EDIT: oh, you altered it :redface:
Yup, sorry about that!

I checked the whole thing and that's correct. Of course, you can also cancel the (-1)^n in the numerator with the (i)^(2n). You know the result should be positive
Aha, so the final result is: [tex]\int_0^{2\pi} sin^{2n}\theta\ d\theta = \frac{\pi*{2n \choose n}}{(2)^{2n-1}} [/tex].

Thanks a lot, tiny-tim and Dick! :smile:
 
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