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.

A complete asymptotic expansion for large $n$ may be derived as follows. We write
$$
\int_0^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t}  = \int_0^\pi  {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t}  + \int_\pi ^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} .
$$
Notice that
$$
\int_\pi ^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t}  = \frac{\pi }{{\pi ^n }}\int_1^{ + \infty } {\left( {\frac{{\sin (\pi s)}}{s}} \right)^n {\rm d}s}  = \mathcal{O}(\pi ^{ - n} )
$$
as $n\to+\infty$. By Laplace's method,
$$
\int_0^{ \pi } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t}  = \int_0^\pi  {\exp \left( { - n\log (t\csc t)} \right){\rm d}t}  \sim \sqrt {\frac{{3\pi }}{{2n}}} \sum\limits_{k = 0}^\infty  {\frac{{a_k }}{{n^k }}} 
$$
as $n\to+\infty$, where
$$
a_k  = \left( {\frac{3}{2}} \right)^k \frac{1}{{k!}}\left[ {\frac{{{\rm d}^{2k} }}{{{\rm d}t^{2k} }}\left( {\frac{{t^2 }}{{6\log (t\csc t)}}} \right)^{k + 1/2} } \right]_{t = 0} .
$$
Accordingly,
\begin{align*}
\int_0^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} &\sim \sqrt {\frac{{3\pi }}{{2n}}} \sum\limits_{k = 0}^\infty  {\frac{{a_k }}{{n^k }}} \\ & = \sqrt {\frac{{3\pi }}{{2n}}} \left( {1 - \frac{3}{{20n}} - \frac{{13}}{{1120n^2 }} + \frac{{27}}{{3200n^3 }} + \frac{{52791}}{{3942400n^4 }} +  \ldots } \right)
\end{align*}
as $n\to+\infty$.
