- #1

ryanwilk

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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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