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: I'll write $I = \int_{-\infty}^{\infty} \left(\frac{\sin z}{z} \right)^n dz$
First, to simplify matters let's take $n$ odd and $\geq 3$. Let $C_{\epsilon}^+$ be the contour along the real line that takes a semicircular detour into the upper half plane about the origin, and let $C_{\epsilon}^-$ be the same for the lower half plane. We use continuity of the integrand to argue that
$$
I = \lim_{\epsilon \rightarrow 0} \int_{C_{\epsilon}^{\pm}} = \frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( \int_{C_{\epsilon}^+} + \int_{C_{\epsilon}^-} \right)
$$
Now think about $(\sin x)^n$: it's a sum of exponential terms of the form $e^{i l x}$ for $-n \leq l \leq n$ with some coefficients. You should convince yourself that any $l < 0$ term is killed by $\int_{C_{\epsilon}^-}$ and any $l > 0$ term is killed by $\int_{C_{\epsilon}^+}$. Moreover by completing these contours with large semicircles, you can derive ($l > 0$):
$$
\int_{C_{\epsilon}^{\mp}} \frac{e^{\pm i l x}}{x^n} dx = \mp 2 \pi i \frac{(\pm i l)^{n-1}}{(n-1)!}
$$
Summing everything up and noticing that there is no $\epsilon$ dependence, and keeping track of signs (which I failed to do on a first pass) we've shown that,
$$
I = \frac{\pi }{2^{n-1} (n-1)!} \sum_{l = 0}^{(n-1)/2} (-1)^{n-1-l}\left(\begin{array}{c}n \\ l \end{array} \right) (n-2l)^{n-1}
$$
I hope that wasn't too much (or too little).
A: I have a generalized elementary method for this problem，If f (x) is an even function, and the period is $\pi$,we have:
$$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx     \qquad (1)$$
Where the $g_n(x)$ in (1) is as follows 
$$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\csc x\right),& \text{for n is odd $n\in\Bbb N$ and}\\[2ex]
\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\cot x\right),& \text{ for n is even .}
\end{cases}$$
——————————————————————————————————————————————————
Proof:
\begin{align}
\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x}\right)^ndx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin x}{x}\right)^ndx\\
&=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx\\
&=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin x}{x-k\pi}\right)^ndx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^n}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^n}+\frac{1}{(x-k\pi)^n}\right]\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_n(x)dx
\end{align}
We know by the Fourier series
\begin{align}
\csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\
\end{align}
and
\begin{align}
\cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)
\end{align}
Take the n-1 order derivative,thus we obtain $g_n(x)$.
——————————————————————————————————————————————————
Example:
\begin{align}
(1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\
&=\frac{\pi}{4}\\
\end{align}
\begin{align}
(2.)
\int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\
\end{align}
\begin{align}
(3.)
\int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\
(4.)
\int_{0}^{\infty}\frac{1}{3+\cos2x}\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{3+\cos2x}dx
=\frac{\pi}{4\sqrt{2}}\\
\end{align}
A: 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. 
A: Just to verify oen's post (since there is a post with a different answer), I will post the answer I got.
$|\sin(z)|\le e^{|\mathrm{Im}(z)|}$; therefore, on the strip $|\mathrm{Im}(z)|\le1$, we have $|\sin(z)|\le e$. Thus, $\left(\frac{\sin(z)}{z}\right)^n$ vanishes as $|z|\to\infty$ in that strip and therefore,
$$
\int_{-\infty}^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z
=\int_{-\infty-i}^{\infty-i}\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z\tag{1}
$$
Next define two contours $\gamma^+$ and $\gamma^-$. $\gamma^+$ goes from $-R-i$ to $R-i$ then circles back through the upper half plane along $|z+i|=R$. $\gamma^-$ goes from $-R-i$ to $R-i$ then circles back through the lower half plane along $|z+i|=R$.
Using the binomial theorem, we get
$$
\left(\frac{\sin(z)}{z}\right)^n=\frac1{(2iz)^n}\sum_{k=0}^n(-1)^k\binom{n}{k}e^{(n-2k)iz}\tag{2}
$$
Integrate the terms where $n-2k\ge0$ along $\gamma^+$ and the others along $\gamma^-$. Since $\gamma^-$ doesn't enclose any singularities, we can ignore that integral. Therefore,
$$
\begin{align}
\int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z
&=\frac12\int_{\gamma^+}\frac1{(2iz)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}e^{(n-2k)iz}\mathrm{d}z\\
&=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\mathrm{Res}\left(\frac{e^{(n-2k)iz}}{z^n},0\right)\\
&=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\frac{(n-2k)^{n-1}i^{n-1}}{(n-1)!}\\
&=\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1}\tag{3}
\end{align}
$$
A: There is an excellent result related with this integral, enjoy !
$$I=\int_{0}^{\infty }x^{p}\ \left ( \frac{\sin(x)}{x} \right )^ndx\quad, n=1,2,3...... , \quad 0\geq p\geq -1\\
\\
\\
I=\frac{\pi }{2(2i)^{n}\Gamma (n-p)}\sum_{m=0}^{n }(-1)^{n-m}\frac{n!}{m!(n-m)!}\left | n-2m \right |^{n-p-1}\left ( \frac{1}{\sin(\frac{n-p+1}{2})\pi }-\frac{\text{sgn}(n-2m)}{\sin(\frac{n-p}{2})\pi}i \right )$$
A: 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 $n\to+\infty$.
