calculus - A sine integral $int_0^{infty} left(frac{sin x }{x }right)^n,mathrm{d}x$

The following question comes from Some integral with sine post
$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!

Sis.

Answer

Here's another approach.

We have
$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,
\left(\frac{\sin x}{x-i\epsilon}\right)^n \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,

\frac{1}{(x-i\epsilon)^n}
\left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty dx\,
\frac{1}{(x-i\epsilon)^n}
\sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^n (-1)^k {n \choose k}
\int_{-\infty}^\infty dx\, \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n}.

\end{eqnarray*}$$
If $n-2k \ge 0$ we close the contour in the upper half-plane and pick up the residue at $x=i\epsilon$.
Otherwise we close the contour in the lower half-plane and pick up no residues.
The upper limit of the sum is thus $\lfloor n/2\rfloor$.
Therefore, using the Cauchy differentiation formula, we find
$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n
&=& \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{eqnarray*}$$
The sum can be written in terms of the hypergeometric function but the result is not particularly enlightening.