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The integral $$\int^{2\pi}_{0}\cos^k(x)dx$$comes up in one of Rudin's problems. He asked me to do a detailed approximation of it. I am wondering how to do it. This feels really elementary, but I need a lower bound that works for any $k$. Integrating by parts would not really help though the function is explicitly integrable.

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I am having trouble with your notation. Does $[x]$ denote the floor function? Is the power of $k$ being applied to $[x]$ or to the cosine? –  Qiaochu Yuan Dec 28 '12 at 21:48
    
Isn't this standard? Should I change this to $\cos^{k}(x)$? The power is on the cosine. –  Bombyx mori Dec 28 '12 at 21:49
    
Try Euler's identity: $\cos x = (e^{ix}+e^{-ix})/2$ and then binomial formula. –  user53153 Dec 28 '12 at 21:50
    
$\cos^k(x)$ is more standard. What you've written looks like it could be parsed as the cosine of the $k^{th}$ power of the floor of $x$, which didn't seem like your intent. –  Qiaochu Yuan Dec 28 '12 at 21:51
    
This is a good approach but I doubt if it will give me a good estimate. Let me try. –  Bombyx mori Dec 28 '12 at 21:51

1 Answer 1

up vote 6 down vote accepted

If $k$ is odd, then the integral is zero. If $k$ is even, note that $$\int_0^{2\pi} \cos^{2m}(x) dx = 4 \int_0^{\pi/2} \cos^{2m}(x) dx$$ Let $I_m = \displaystyle \int_0^{\pi/2} \cos^{2m}(x) dx$ Then \begin{align} I & = \int_0^{\pi/2} \cos^{2m-1}(x) d(\sin(x))\\ & = \left. \cos^{2m-1}(x) \sin(x)\right \vert_0^{\pi/2} + \int_0^{\pi/2}\sin(x) (2m-1) \cos^{2m-2}(x) \sin(x) dx\\ & = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) \sin^2(x) dx\\ & = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) (1-\cos^2(x)) dx\\ & = (2m-1) \int_0^{\pi/2} \cos^{2m-2}(x) dx - (2m-1) \int_0^{\pi/2} \cos^{2m}(x) dx\\ 2mI_{2m} & = (2m-1) I_{2m-2} \end{align} Hence, we get that $$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 I_0$$ Now $I_0 = \int_0^{\pi/2} 1 dx = \dfrac{\pi}2$. Hence, we get that $$I_{2m} = \dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \dfrac{\pi}2$$ Hence, $$\int_0^{2\pi} \cos^{2m}(x) dx = 2 \pi \left(\dfrac{2m-1}{2m} \cdot \dfrac{2m-3}{2m-2} \cdots \dfrac34 \cdot \dfrac12 \right) = 2\pi \dfrac{(2m)!}{4^m \cdot m! \cdot m!} = \dfrac{2 \pi}{4^m} \dbinom{2m}m$$ By Stirling, we have $$\dbinom{2m}m \sim \dfrac{2^{2m+1}}{\sqrt{4 \pi m}} = \dfrac{4^m}{\sqrt{\pi m}}$$ Hence, we have that $$\int_0^{2\pi} \cos^{2m}(x) dx \sim \dfrac{2 \pi}{\sqrt{\pi m}} = \dfrac{2 \sqrt{\pi}}{\sqrt{m}}$$

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Thanks! I guess I have to apply Stirling's formula then. –  Bombyx mori Dec 28 '12 at 21:57

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