Using the Residue Theorem for Complex Analysis Integrals

[ryanwilk]

Homework Statement

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

Homework Equations

\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)

The Attempt at a Solution

I started with the substitution z = e^{i\theta}

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

and d\theta = \frac{dz}{iz}.

Therefore, the integral becomes: \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

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

The residue is \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}.

But I have no idea how to evaluate this...

 

[tiny-tim]

hi ryanwilk!

in that final differential, you only need the coefficient of z²ⁿ in z(z² - 1)²ⁿ … but isn't that zero?

 

[ryanwilk]

hi ryanwilk!

in that final differential, you only need the coefficient of z²ⁿ in z(z² - 1)²ⁿ … but isn't that zero?

Ah I see! It should be just the coefficient of z²ⁿ in (z² - 1)²ⁿ though, which is non-zero but looks complicated.

 

[Dick]

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.

 

[ryanwilk]

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 \lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} (z^2-1)^{2n} seems to be (-1)^n(2n)!{2n \choose n}.

The residue is \frac{(-1)^n{2n \choose n}}{i(2i)^{2n}} so the integral is just \frac{2\pi(-1)^n{2n \choose n}}{(2i)^{2n}} ?

 

[tiny-tim]

Yup!

EDIT: oh, you altered it …

my "yup" was for the first line, i haven't checked the actual residue

 

[Dick]

Oh right, so \lim_{z \to 0} \frac{d^{2n}}{dz^{2n}} (z^2-1)^{2n} seems to be (-1)^n(2n)!{2n \choose n}.

The residue is \frac{(-1)^n{2n \choose n}}{i(2i)^{2n}} so the integral is just \frac{2\pi(-1)^n{2n \choose n}}{(2i)^{2n}} ?

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.

 

[ryanwilk]

Yup!

EDIT: oh, you altered it …

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: \int_0^{2\pi} sin^{2n}\theta\ d\theta = \frac{\pi*{2n \choose n}}{(2)^{2n-1}}.

Thanks a lot, tiny-tim and Dick!