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Deduce that $\int_{0}^{2\pi}\cos^{2n}\theta\ d\theta = {{2n}\choose{n}}\frac{\pi}{2^{2n -1}}$

So first we make the following substitution $z = e^{i\theta}$ and hence $\cos = \frac{z + z^{-1}}{2}$. Then we have the following;

$\frac{1}{i}\int_{C(0;1)}(\frac{z+z^{-1}}{2})^{2n}\frac{1}{z}dz$;

multiply "top and bottom on the outside" by $2\pi$, we then arrived at

$2\pi$Res($(\frac{z+z^{-1}}{2})^{2n}\frac{1}{z}$, $0$).

And now I'm totally stuck, I have no idea how to even "magically" make the combinatoric function appear.

Is there some standard result I need to use here or is there a trick I'm not seeing here.

Any help or insight is deeply appreciated (I really mean it :))

EDIT: I realize there is a mistake in my working, hence it's not making sense to some. I edited it.

some1fromhell
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  • We can use https://math.stackexchange.com/questions/193435/prove-int-cosn-x-dx-frac1n-cosn-1x-sin-x-fracn-1n-int-cos – lab bhattacharjee May 03 '17 at 07:03
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    Just multiply it out using the binomial theorem. Only one of the terms in the resulting sum will have a nonzero residual – Bananach May 03 '17 at 07:20

2 Answers2

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$$\frac{1}{z}\left(\frac{z^2+1}{2z}\right)^{2n} = \frac{1}{2^{2n}z^{2n+1}}(z^2+1)^{2n} = \frac{1}{2^{2n}}\sum_{k=0}^{2n}{2n\choose k}z^{2k-2n-1}$$

Now we must have $2k-2n-1 = -1$ which implies $k = n$

Hence $$\mathrm{Res}\left[\frac{1}{z}\left(\frac{z^2+1}{2z}\right)^{2n},0\right] = \frac{1}{2^{2n}}{2n\choose n}$$

Zaid Alyafeai
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In terms of special functions, this is $2 B(1/2, n+1/2)$, where $B(x, y)$ is the Beta function.

Using that $$ B(x, y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, $$ where $\Gamma(x)$ is the Gamma function, one has $$ \begin{align} 2B(1/2, n+1/2) &= \dfrac{2\Gamma(1/2)\Gamma(n+1/2)}{\Gamma(n+1)} \\ &= \dfrac{2\sqrt{\pi}}{n!}\dfrac{(2n)!\sqrt{\pi}}{4^n n!} \\ &= \dfrac{\pi}{2^{2n-1}}\binom{2n}{n}. \end{align} $$

Marcelo Forets
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