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Here is a fun integral I am trying to evaluate:

$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$

I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.

But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.

$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$

or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.

Thanks everyone.

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    Have you tried it for small $n$, like $n=0$ and $n=1$?2012-07-17
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    @Arjang: Please try to space your edits. The front page looks like a lot of the same question.2013-03-01
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    @AsafKaragila : Thank you, by space do you mean time wise? I was just thinking I should wait between edits, or something else?2013-03-01

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