The limit of the integral of $\text{sinc}(x)^n$ I saw this exercise, and I'm wondering what methods could be used to solve it.
$$
\lim_{n\to\infty}\left[%
n^{y}\int_{0}^{\infty}
\operatorname{sinc}^n\left(x\right)\,
\mathrm{d}x\right]
$$
Any naïve approach I've tried has failed pretty fast, such as splitting $\operatorname{sinc}^{n}\left(x\right)$ to $\sin^{n}\left(x\right)\cdot\frac{1}{x^{n}}$ and integrating by parts, so what method would you use to tackle this problem $?$.
 A: The idea is $\newcommand{\sinc}{\operatorname{sinc}}$ that $\sqrt{n}\int_0^\infty\sinc^n x\,dx=\int_0^\infty\sinc^n(t/\sqrt{n})\,dt$ tends to $\int_0^\infty e^{-t^2/6}\,dt=\color{blue}{\sqrt{3\pi/2}}$ as $n\to\infty$, according to DCT. This gives the answer for $y=1/2$, and obviously for other values.
Here is a more detailed explanation. Clearly, $$\sqrt{n}\left|\int_\pi^\infty\sinc^n x\,dx\right|\leqslant\sqrt{n}\int_\pi^\infty\frac{dx}{x^n}=\frac{\sqrt{n}}{(n-1)\pi^{n-1}}\underset{n\to\infty}{\longrightarrow}0.$$ Hence, if the (first) limit exists, $$\lim_{n\to\infty}\sqrt{n}\int_0^\infty\sinc^n x\,dx=\lim_{n\to\infty}\sqrt{n}\int_0^\pi\sinc^n x\,dx=\lim_{n\to\infty}\int_0^{\pi\sqrt{n}}\sinc^n\frac{t}{\sqrt{n}}\,dt.$$ Since $\sinc x=1-x^2/6+o(x^2)$ as $x\to 0$, we have $\lim\limits_{n\to\infty}\sinc^n(t/\sqrt{n})=e^{-t^2/6}$ for fixed $t$.
It remains to exhibit a dominating function for DCT to apply. But $$0\leqslant\sinc x\leqslant 1-x^2/\pi^2\leqslant e^{-x^2/\pi^2}$$ for $0\leqslant x\leqslant\pi$ (the middle inequality follows immediately from the infinite product for $\sin x$, and perhaps may be shown an easier way). Thus, $e^{-t^2/\pi^2}$ is a suitable dominating function.
A: Too long for comments.
$$I_n=\frac 1{\pi}\int_0^\infty\big[\text{sinc}(x)\big]^n dx$$ generate the sequence
$$\left\{\frac{1}{2},\frac{1}{2},\frac{3}{8},\frac{1}{3},\frac{115}{384},\frac{11}{40
   },\frac{5887}{23040},\frac{151}{630},\frac{259723}{1146880},\frac{15619}{72576},
   \frac{381773117}{1857945600},\cdots\right\}$$ The numerators are sequence $A049330 $ and the denominators are sequence $A049331$ in $OEIS$.
According to Vladimir Reshetnikov
$$I_n=\frac 1 {2^n (n-1)!}\sum _{k=0}^{\frac{n}{2}} (-1)^k  \binom{n}{k} (n-2 k)^{n-1}$$
It seems that quite decent approximations could be
$$I_n=\sum_{p=1}^q a_p \,n^{-\frac p2}$$ So, for large values of $n$
$$I_n \sim \frac C {\sqrt n} \qquad \text{with} \qquad C \sim 0.6910$$
Making $n=10^m$, some results
$$ \left(
\begin{array}{cc}
 m & \sqrt{10^m}\,I_{10^m} \\
 1 & 0.680550247659969 \\
 2 & 0.689951020377500 \\
 3 & 0.690884642683269 \\
 4 & 0.690977934037989 \\
 5 & 0.690987262459421
\end{array}
\right)$$
Edit
As @metamorphy and @robjohn commented,as we say in French, I reinvented the warm water !
A: In this answer I computed that
$$
\int_0^\infty\left(\frac{\sin(x)}x\right)^n\,\mathrm{d}x=\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1}\tag0
$$
However, in this form, it is difficult to discern the asymptotic behavior as $n\to\infty$.

The Fourier Transform and Central Limit Theorem
$\newcommand{\sinc}{\operatorname{sinc}}\newcommand{\Res}{\operatorname*{Res}}$The Fourier Transform of $\sinc(x)$ is
$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{ix}-e^{-ix}}{2ix}e^{-2\pi ix\xi}\,\mathrm{d}x
&=\int_\gamma\frac{e^{iz}-e^{-iz}}{2iz}e^{-2\pi iz\xi}\,\mathrm{d}z\tag1\\
&=\pi\Res_{z=0}\left(\frac{e^{iz(1-2\pi\xi)}}{z}\right)-\pi\Res_{z=0}\left(\frac{e^{-iz(1+2\pi\xi)}}{z}\right)\tag2\\[3pt]
&=\pi\left[\zeta\le\frac1{2\pi}\right]-\pi\left[\zeta\le-\frac1{2\pi}\right]\tag3\\[6pt]
&=\pi\left[-\frac1{2\pi}\le\zeta\le\frac1{2\pi}\right]\tag4
\end{align}
$$
Since $\widehat{\!fg}=\widehat{\!f}{\ast}\widehat{\vphantom{f}g}$, the Fourier Transform of $\sinc^n(x)$ is the convolution of $n$ copies of $(4)$.
$(4)$ is the PDF for a probability distribution with mean $0$ and variance $\frac1{12\pi^2}$. The convolution of $n$ copies of this distribution with itself approaches a normal distribution with mean $0$ and variance $\frac{n}{12\pi^2}$, which has the PDF
$$
f_n(\xi)=\sqrt{\frac{6\pi}n}\,e^{-6\pi^2\xi^2/n}\tag5
$$
Because $\int_{-\infty}^\infty f(x)\,\mathrm{d}x=\widehat{\!f}(0)$,
$$
\int_{-\infty}^\infty\sinc^n(x)\,\mathrm{d}x\sim f_n(0)=\sqrt{\frac{6\pi}n}\tag6
$$
and therefore, since $\sinc(x)$ is even,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\int_0^\infty\sinc^n(x)\,\mathrm{d}x=\sqrt{\frac{3\pi}2}}\tag7
$$

Real Approach
Using $(7)$ from this answer, it follows that
$$
\begin{align}
\frac1x-\cot(x)
&=\sum_{k=1}^\infty\frac{2x}{k^2\pi^2-x^2}\tag8\\
&=\sum_{k=1}^\infty\frac{2x}{k^2\pi^2}\sum_{j=0}^\infty\left(\frac{x^2}{k^2\pi^2}\right)^j\tag9\\
&=\sum_{j=0}^\infty\frac{2\zeta(2j+2)}{\pi^{2j+2}}x^{2j+1}\tag{10}\\
-\log(\sinc(x))
&=\sum_{j=0}^\infty\frac{\zeta(2j+2)}{\pi^{2j+2}}\frac{x^{2j+2}}{j+1}&&\text{where }|x|\lt\pi\tag{11}\\
\sinc^n\left(x/\sqrt{n}\right)
&=\prod_{j=1}^\infty e^{-\frac{\zeta(2j)}{\pi^{2j}}\frac{x^{2j}}{jn^{j-1}}}&&\text{where }|x|\lt\pi\sqrt{n}\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: apply $\cot(x)=\sum\limits_{k\in\mathbb{Z}}\frac1{k\pi+x}=\frac1x-\sum\limits_{k=1}^\infty\frac{2x}{k^2\pi^2-x^2}$
$\phantom{1}(9)$: apply the sum of a geometric series
$(10)$: change the order of summation
$(11)$: integrate
$(12)$: exponentiate
Thus, $(12)$ shows that $\sinc^n\left(x/\sqrt{n}\right)\left[|x|\lt\pi\sqrt{n}\right]$ increases to $e^{-x^2/6}$.
Furthermore,
$$
\begin{align}
\left|\,\int_{\pi\sqrt{n}}^\infty\sinc^n\left(x/\sqrt{n}\right)\,\mathrm{d}x\,\right|
&\le\int_{\pi\sqrt{n}}^\infty\left(\frac{\sqrt{n}}x\right)^n\,\mathrm{d}x\tag{13}\\
&=\frac{\sqrt{n}}{(n-1)\pi^{n-1}}\tag{14}
\end{align}
$$
The Monotone Convergence Theorem, $(12)$, and $(14)$ show that
$$
\begin{align}
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\int_0^\infty\sinc^n(x)\,\mathrm{d}x}
&=\lim_{n\to\infty}\int_0^\infty\sinc^n\left(x/\sqrt{n}\right)\,\mathrm{d}x\tag{15}\\
&=\int_0^\infty e^{-x^2/6}\,\mathrm{d}x\tag{16}\\
&=\bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{3\pi}2}}\tag{17}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{%
n^{y}\int_{0}^{\infty}\on{sinc}\pars{x}^{n}\,\dd x}}:
\ {\Large ?}}$. Note that the integrand main contribution comes from values of $\ds{x \gtrsim 0}$ because $\ds{\verts{\on{sinc}\pars{x}} \leq 1}$ and
$\ds{\on{sinc}\pars{0} = 1}$. That suggests the use of Laplace Method. Namely,

\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{%
n^{y}\int_{0}^{\infty}\on{sinc}\pars{x}^{n}\,\dd x}}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{y}\int_{0}^{\infty}
\exp\pars{n\ln\pars{\on{sinc}\pars{x}}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{y}\int_{0}^{\infty}
\exp\pars{n\ln\pars{1 - {x^{2} \over 6}}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{y}\int_{0}^{\infty}
\exp\pars{-{nx^{2} \over 6}}\,\dd x}
\\[5mm] = &\
\root{3\pi \over 2}
\lim_{n \to \infty}n^{y - 1/2}\,\,
=
\bbx{\left\{\begin{array}{lclcl}
\ds{0} & \mbox{if} & \ds{y} & \ds{<} & \ds{1 \over 2}
\\
\ds{\root{3\pi \over 2}} & \mbox{if} & \ds{y} & \ds{=} & \ds{1 \over 2}
\\
\ds{\infty} & \mbox{if} & \ds{y} & \ds{>} &
\ds{1 \over 2}
\end{array}\right.} \\ &
\end{align}

It's interesting to note that
\begin{align}
&\int_{0}^{\infty}
\exp\pars{n\ln\pars{\on{sinc}\pars{x}}}\,\dd x
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
\root{3\pi \over 2}\pars{{1 \over n^{1/2}} -
{3 \over 20}\,{1 \over n^{3/2}}}
\end{align}
