# Complex Analysis Integral

## 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...

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tiny-tim
Homework Helper
hi ryanwilk!

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

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.

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Dick
Homework Helper
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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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 $$\frac{2\pi(-1)^n{2n \choose n}}{(2i)^{2n}}$$ ?

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tiny-tim
Homework Helper
Yup!

EDIT: oh, you altered it

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

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Dick
Homework Helper
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.

Yup!

EDIT: oh, you altered it
Aha, so the final result is: $$\int_0^{2\pi} sin^{2n}\theta\ d\theta = \frac{\pi*{2n \choose n}}{(2)^{2n-1}}$$.