An odd trigonometric integral

Let $n \in \mathbb{N}$ . Prove that

$\int_{0}^{\infty} \frac{\sin^{2n+1} y}{y}\, \mathrm{d}y = \frac{\pi}{2^{2n+1}} \binom{2n}{n}$

Solution

Since $\displaystyle \sin x =\frac{e^{ix}-e^{-ix}}{2i}$ we deduce that

$\sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^{n} (-1)^r \binom{2n+1}{r}\sin(2r+1)x$

Hence,

$\begin{aligned} \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right) \end{aligned}$

since $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x} \, \mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x} \, \mathrm{d}x=\frac{\pi}{2}$ and $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ . Finally, we note that the sum telescopes hence the result.

An odd trigonometric integral

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