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Question: How to prove that $$\int_{0}^{\infty}\frac {\sin^{2n+1}x}{x} dx=\frac {π(2n)!} {2^{2n+1}(n!)^2},\qquad n\ge 0$$

This integral looks tricky, i tried many methods but all fails due to $(2n+1)$ in power which makes difficult to proceed further.

I'm very interested to see the proof.

C.F.G
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Paras
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    You could write $\int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} \ dx$ as the double integral $\int_{0}^{\infty} \int_{0}^{\infty} \sin^{2n+1}(x) e^{-xy} \ dy \ dx,$ and then reverse the order of integration (you will need to justify this) and use integration by parts (or use the complex definition of sine). – Vivek Kaushik May 09 '20 at 20:30
  • Also: https://math.stackexchange.com/q/388713/42969. – Martin R May 09 '20 at 20:39
  • A remark: when reversing the order of integration from my previous comment, you will end up having to calculate a Laplace Transform of a power of sine. See here on how it is done. https://math.stackexchange.com/questions/1756093/laplace-transforms-of-powers-of-cosine-solved or here: https://math.stackexchange.com/questions/1814913/prove-int-0-inftye-nx-sin2kxdx-2k-over-n-prod-j-1kn24j/2297177#2297177 – Vivek Kaushik May 09 '20 at 20:59

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