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I have come across the problem of estimation of the limit $$\lim_{x\to+\infty}\frac{\int_0^x|\sin(s)|ds}{x}.$$ Since it is easy to check that the method of l'Hopital's Rule is incapable of it. I have tried the following: Since for every $x>\pi,$ there exist unique $n\in\mathbb{N}$ and $\theta\in[0, \pi),$ such that $$x=n\pi+\theta,$$ and $$x\to+\infty \Longleftrightarrow n\to\infty.$$ On account of the periodicity of the mapping $s\to |\sin(s)|,$ \begin{gather*} \begin{aligned} &\lim_{x\to+\infty}\frac{\int_0^x|\sin(s)|ds}{x}=\lim_{n\to\infty}\frac{\int_0^{n\pi+\theta}|\sin(s)|ds}{n\pi+\theta}=\lim_{n\to\infty}\frac{\int_0^{n\pi}|\sin(s)|ds+\int_{n\pi}^{n\pi+\theta}|\sin(s)|ds}{n\pi+\theta}\\ =&\lim_{n\to\infty}\frac{2n+\int_0^{\theta}|\sin(s)|ds}{n\pi+\theta}=\lim_{n\to\infty}\frac{2+\frac{1}{n}\cdot\int_0^{\theta}|\sin(s)|ds}{\pi+\frac{1}{n}\cdot\theta}\\ =&\frac{2}{\pi}. \end{aligned} \end{gather*}

I am not very sure that my trial is sound. Can anyone help me to check my method, or find another way to estimate this limit?

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Very nicely done, and excellently presented. – Stephen Montgomery-Smith Jan 3 '14 at 1:00
Thanks, Stephen Montgomery-Smith! Your comment gives me confidence. – nuage Jan 3 '14 at 1:11
Having a dangling free $\theta$ in your limit doesn't make sense. One way to fix it is define $n$ and $\theta$ as a short hand of $\lfloor\frac{x}{\pi}\rfloor$ and $x - n\pi$ and then take the limit in $x$ instead of $n$. – achille hui Jan 3 '14 at 1:30
Since $\theta$ is always in $[0,\pi)$, and $\frac{1}{n}\to 0, $ as $n\to\infty,$ we have $\frac{1}{n}\cdot \theta\to 0,$ as $n\to\infty.$ – nuage Jan 3 '14 at 1:55
It seems like it's only a notational problem and achille hui commented on how to fix that. The idea and the answer is still right. – Pratyush Sarkar Jan 3 '14 at 2:47
up vote 2 down vote accepted

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{x \to \infty}\bracks{{1 \over x}\int_{0}^{x}\verts{\sin\pars{s}} \,\dd s}:\ {\large ?}}$. $$ \int_{0}^{x}\verts{\sin\pars{s}}\,\dd s = x\verts{\sin\pars{x}} - \int_{0}^{x}s\sgn\pars{\sin\pars{s}}\cos\pars{s}\,\dd s =x\verts{\sin\pars{x}} - \int_{0}^{x}\sgn\pars{\sin\pars{s}}\varphi'\pars{s}\,\dd s $$ where $\ds{\varphi\pars{s} = \int_{0}^{s}t\cos\pars{t}\,\dd t}$. \begin{align} \int_{0}^{x}\verts{\sin\pars{s}}\,\dd s&= x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x} + \int_{0}^{x}\varphi\pars{s}\bracks{2\delta\pars{\sin\pars{s}}\cos\pars{s}}\,\dd s \\[3mm]&= x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x} + 2\int_{0}^{x}\varphi\pars{s}\cos\pars{s} \sum_{n = -\infty}^{\infty}\delta\pars{s - n\pi}\,\dd s \\[3mm]&= x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x} + 2\sum_{n = 1}^{\infty}\pars{-1}^{n}\varphi\pars{n\pi}\Theta\pars{x - n\pi} \end{align} However, $$ \varphi\pars{s} = s\sin\pars{s} - \int_{0}^{s}\sin\pars{t}\,\dd t = s\sin\pars{s} + \cos\pars{s} - 1\quad\imp\quad\varphi\pars{n\pi} = \pars{-1}^{n} - 1 $$ \begin{align}\color{#0000ff}{\large% \int_{0}^{x}\verts{\sin\pars{s}}\,\dd s = 2\sin^{2}\pars{x \over 2}\sgn\pars{\sin\pars{x}} + 4\sum_{n = 0}^{\infty}\Theta\pars{{x \over \pi} - 2n - 1}} \end{align}

\begin{align}\color{#0000ff}{\large% \lim_{x \to \infty}\bracks{{1 \over x}\int_{0}^{x}\verts{\sin\pars{s}}\,\dd s}} &= 4\lim_{x \to \infty}\bracks{{1 \over x}\sum_{n = 0}^{\infty} \Theta\pars{{x - \pi \over 2\pi} - n}} \\[3mm]&= 4\lim_{x \to \infty}\bracks{{1 \over 2\pi x + \pi}\sum_{n = 0}^{\infty} \Theta\pars{x - n}} = \color{#0000ff}{\large{2 \over \pi}} \end{align}
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