The question is
Prove that for any $n\in\mathbb N$, $\int\limits _0^\infty\frac {\sin^{2n+1}x}{x}dx=\frac 1 {4^n}\binom{2n}{n}\int_0^\infty \frac {\sin{x}}{x}dx$
I don't have any ideas how to solve it and I'll be glad to get a hint.
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Did you try integrating by parts recursively? By induction, the step $n=0$ is trivial. What if you had a reduction formula? – Patrick Da Silva Jul 06 '13 at 06:54
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$\frac{\sin{x}}{x}$ is not trivial and can't be integrated without taylor (or something similier...). – Jul 06 '13 at 06:57
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I didn't say compute the integral, I said use the integration by parts formula to reduce the integral with $\sin^{2n+1}(x)/x$ to the integral $\sin(x)/x$. Did you try that? What happened? – Patrick Da Silva Jul 06 '13 at 06:59
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1@CoarguAliquis Actually, it can be integrated pretty directly - see Aryabhata's answer at http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2. This is not necessary for the question, though. – Cocopuffs Jul 06 '13 at 07:00
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I think it's a pain to go through integration by parts, but it was my only guess. Sorry if I'm not helping. Perhaps maybe residue theory could be helpful but I'm not into that these days. – Patrick Da Silva Jul 06 '13 at 07:01
1 Answers
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See Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction.
We have
$$
\frac{\pi\binom{2n}{n}}{2^{2n+1}}=\frac{1}{4^n}\binom{2n}{n}\int_0^\infty\frac{\mathrm{sin}x}{x}dx
$$
since
$$
\int_0^\infty\frac{\mathrm{sin}x}{x}dx=\frac{\pi}{2}.
$$
