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Prove that for each $n\in\mathbb{N}$

$$\frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$

My attempt:

I know by the cauchy form:

$f^{(n)}(z_0)=\frac{1}{2\pi i}\displaystyle\int_0^{2\pi}\frac{f(z)}{(z-z_0)^{n+1}}dz$

Here I'm stuck. Can someone help me?

Shaun
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rcoder
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  • Also: https://math.stackexchange.com/q/1366304/42969, https://math.stackexchange.com/q/2481356/42969, https://math.stackexchange.com/q/3011547/42969 – all found with Approach0 – Martin R Mar 06 '20 at 13:21
  • If you want to do it by complex analysis note that the integrand is $(z+\frac{1}{z})^{2n}\frac{dz}{z}$ up to some $i$ and the residue at $0$ is precisely the middle coefficient of the binomial (in other words the constant term because of the $\frac{dz}{z}$ which unsurprisingly is the answer! – Conrad Mar 06 '20 at 13:34

1 Answers1

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$$\int_{0}^{2a} f(x) dx=2\int_{0}^{a} f(x) dx,~ if ~ f(2a-x)=f(x).$$ $$I=\frac{2^{2n}}{2\pi} \int_{0}^{2\pi} \cos^{2n}t dt=4\frac{2^{2n}}{2\pi} \int_{0}^{\pi/2} \cos^{2n}t dt=\frac{2^{2n+1}}{\pi} \int_{0}^{\pi/2} \cos^{2n} t dt=\frac{2^{2n}}{\pi} \frac{\Gamma(1/2) \Gamma(n+1/2)}{n!}.$$ $$\implies I=\frac{2^{2n}}{\sqrt{\pi}} \frac{\Gamma(n+1/2)}{n!}=\frac{(2n)!}{(n!)^2}.$$ Here we have used the beta integral: $$\int_{0}^{\pi/2} \sin^p x \cos^q x dx=\frac{1}{2} \frac{\Gamma((p+1)/2)\Gamma((q+1)/2)}{\Gamma((p+q+2)/2)}$$

Z Ahmed
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