A closed form for a generalized sin integral

Let $n \in \mathbb{N}$. Deduce a closed form for the integral:

$$\mathcal{J}_n=\int_0^\infty \left( \frac{ \sin x}{x} \right)^n \, {\rm d}x$$

Solution

We are proving that:

$$\mathcal{J}_n = \frac{\pi}{2^n (n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$

We have successively:

\begin{align*}
\int_{0}^{\infty} \frac{\sin^n x}{x^n}\, {\rm d}x &=\lim_{\epsilon\to 0^+} \frac{1}{2} \int_{-\infty}^\infty \left(\frac{\sin x}{x-i\epsilon}\right)^n \, {\rm d}x \\
&=\lim_{\epsilon\to 0^+}\frac{1}{2} \int_{-\infty}^\infty \frac{1}{(x-i\epsilon)^n}\left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n\, {\rm d}x \\
&= \lim_{\epsilon\to 0^+}\frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty
\frac{1}{(x-i\epsilon)^n} \sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \, {\rm d}x\\
&= \lim_{\epsilon\to 0^+}\frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^n (-1)^k {n \choose k} \int_{-\infty}^\infty \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n} \, {\rm d}x
\end{align*}

If $n-2k \geq 0$ we close the contour in the upper half-plane and pick up the residue at $x=i \epsilon$. Otherwise we close the contour in the lower half-plane and pick up no residues. The upper limit of the sum is thus $\left \lfloor \frac{n}{2} \right \rfloor$. Thus, using the Cauchy differentiation formula, we find:

\begin{align*}
\int_0^\infty \left(\frac{\sin x}{x}\right)^n \, {\rm d}x &= \frac{1}{2} \frac{1}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k}\frac{2\pi i}{(n-1)!} \left.\frac{d^{n-1}}{d x^{n-1}} e^{i x(n-2k)}\right|_{x=0}\\
&=\frac{1}{2} \frac{1}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} \frac{2\pi i}{(n-1)!} (i(n-2k))^{n-1} \\
&=\frac{\pi}{2^n (n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}
\end{align*}

and the calculations are complete. 

This answer was taken from MSE (math.stackexchange.com)