The limit of the integral of $\text{sinc}(x)^n$

Question

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 $?$.

Answer 1

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.

Answer 2

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 !

Answer 3

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 $n\to\infty$.

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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, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\int_0^\infty\sinc^n(x)\,\mathrm{d}x=\sqrt{\frac{3\pi}2}}\tag7 $$

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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 $$ \begin{align} \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\int_0^\infty\sinc^n(x)\,\mathrm{d}x} &=\lim_{n\to\infty}\int_0^\infty\sinc^n\left(x/\sqrt{n}\right)\,\mathrm{d}x\tag{15}\\ &=\int_0^\infty e^{-x^2/6}\,\mathrm{d}x\tag{16}\\ &=\bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{3\pi}2}}\tag{17} \end{align} $$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{% n^{y}\int_{0}^{\infty}\on{sinc}\pars{x}^{n}\,\dd x}}: \ {\Large ?}}$. 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,

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\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}

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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}