<s>[INST] Evaluating $\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x$ I was generalizing the following integral:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x \hspace{40pt} n\geq 0$$
We can start by noting that $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$ and thus $\displaystyle \sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\sin(2r+1)x$.
This implies $$\begin{aligned}\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right)\end{aligned}$$
Where I have used the well known result $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x}\mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ and the property of binomial coefficients that $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$
Since the above sum telescopes, we have $$\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x=\frac{(-1)^n\pi}{2^{2n+1}}(-1)^n\binom{2n}{n}=\frac{\pi}{2^{2n+1}}\binom{2n}{n} ~\forall ~ n\in \mathbb{Z^{+}}$$
I would like to know other methods for evaluating this integral.
 [/INST] One can use the recurrence relation of Gradshteyn and Ryzhik to give, for some $p, q, \in \mathbb{N}$
\begin{align}
\displaystyle \int_0^\infty \frac{\sin^p x}{x^q}\,\mathrm{d}x &= \frac{p}{q-1}\displaystyle \int_0^\infty \frac{\sin^{p-1} x}{x^{q-1}}\cos x\,\mathrm{d}x \qquad (p > q-1>0)\\
&= \frac{p(p-1)}{(q-1)(q-2)}\displaystyle \int_0^\infty \frac{\sin^{p-2} x}{x^{q-2}}\,\mathrm{d}x - \frac{p^2}{(q-1)(q-2)}\int_0^\infty \frac{\sin^{p} x}{x^{q-2}}\,\mathrm{d}x\qquad (p > q-1>1)
\end{align}
Now, so if $q$ and $p$ are both even or both odd, with $p \geq q$ to assure convergence of the integral, we arrive at integrals of the form
$$\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}$$
$$\int_0^\infty \frac{\sin^{2n}x}{x^2}\,dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
Which I think satisifes your requirement of another method.
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