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I am required to prove that $\displaystyle\int_{0}^{2\pi} (\cos\theta)^{2n}d\theta= 2\pi.\frac{1 \times 3 \times \dots \times(2n-1)}{2 \times 4 \times \dots \times(2n)}$.

Can someone guide me along for this question? My intuition tells me that i am able to convert this real integral to a complex function and (possibly) the limits to a simple closed curve with domain $D$ on $\mathbb{C}$. Thereafter, we may probably use Cauchy-Goursat or Cauchy Integral Formula to make our computations easy.

Any help or hints will be highly appreciated!

Cleytus
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  • This has been asked hundreds of times and it is not difficult, actually. Just write $\cos\theta$ as $\frac{1}{2}(e^{i\theta}+e^{-i\theta})$, then exploit the binomial theorem and the fact that $\int_{0}^{2\pi} e^{2mi\theta},d\theta = 2\pi\delta(m)$. – Jack D'Aurizio Oct 20 '17 at 12:24
  • Duplicate of https://math.stackexchange.com/questions/1366304/integration-of-int-02-pi-cos2ntdt – Jack D'Aurizio Oct 20 '17 at 12:25
  • Hi Jack, thank you for the link. I appreciate your feedback. – Cleytus Oct 20 '17 at 12:27

2 Answers2

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Notice that after the substitution $z=e^{i \theta}$ one needs to evaluate $$\frac{1}{i}\int_{C(0;1)}(\frac{z+z^{-1}}{2})^{2n}\frac{1}{z}dz$$ Now notice that $$\frac{1}{z}\left(\frac{z^2+1}{2z}\right)^{2n} = \frac{1}{2^{2n}z^{2n+1}}(z^2+1)^{2n} = \frac{1}{2^{2n}}\sum_{k=0}^{2n}{2n\choose k}z^{2k-2n-1}$$

Now one has that $2k-2n-1 = -1$ and therefore that $k = n$

From which one deduces that $$\mathrm{Res}\left(\frac{1}{z}\left(\frac{z^2+1}{2z}\right)^{2n},0\right) = \frac{1}{2^{2n}}{2n\choose n}$$

  • Hi Kevin, thanks for your help! May i ask: (1) What does your notation $C(0;1)$ mean and (2) how did you establish the equality: $2k-2n-1=-1$? – Cleytus Oct 20 '17 at 12:22
  • @Cleytus You're welcome. Firstly $C(0,1)$ is a circle of radius $1$ centred on the origin oriented in the positive manner. Secondly, I have expanded as a Laurent Series, with $|z| >1$. –  Oct 20 '17 at 13:04
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Hint:

This is a Wallis' integral. The standard proof uses integration by parts:

Set $I_{2n}=\displaystyle\int_0^{2\pi}\cos^{2n}\theta\,\mathrm d \theta$, $\;u=\cos^{2n-1}\theta$, $\;\mathrm d v=\sin\theta\,\mathrm d \theta$ and establish the recurrence relation: $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}.$$

Bernard
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