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I want to evaluate $$L=\lim_{x\to\infty}\frac1x\int_0^x\max\{\sin t,\sin(t\sqrt2)\}dt$$ My attempt $$L=\lim_{x\to\infty}\frac1{2x}\int_0^x\Big(\sin t+\sin(t\sqrt2)+\big|\sin t-\sin(t\sqrt2)\big|\Big)dt\\ =\lim_{x\to\infty}\frac1{2x}\int_0^x\big|\sin t-\sin(t\sqrt2)\big|dt\\ =\lim_{x\to\infty}\frac1x\int_0^x\bigg|\cos\frac{\sqrt2+1}2t\cdot\sin\frac{\sqrt2-1}2t\bigg|dt$$ Denote $s_n$ the $n$th zero point of $\cos\frac{\sqrt2+1}2t\cdot\sin\frac{\sqrt2-1}2t\ (t\ge0)$. Since $1$, $\sqrt2$ and $\pi$ are linear independent in $\mathbb Q$, the order of the zero points should be $1$. According to the squeeze theorem, we have $$L=\lim_{n\to\infty}\frac1{s_{n+1}}\sum_{k=0}^n(-1)^k\int_{s_k}^{s_{k+1}}\big(\sin t-\sin(t\sqrt2)\big)dt\\ =\lim_{n\to\infty}\frac1{s_{n+1}}\sum_{k=0}^n(-1)^k\bigg(\cos s_k-\cos s_{k+1}+\frac{\cos\sqrt2s_k-\cos\sqrt2s_{k+1}}{\sqrt2}\bigg)dt$$ I can't go further. I think the zero points of that function is the key point.

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  • $\begingroup$ I'm a bit puzzled by the first row in your attempt. Why is this true? And I suppose in the second row you split into three integrals and dumped zeros? $\endgroup$ – dEmigOd Jul 22 '18 at 10:15
  • $\begingroup$ @dEmigOd I guess the integral of first two terms are bounded and would vanish in the limit. $\endgroup$ – Szeto Jul 22 '18 at 10:18
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    $\begingroup$ @dEmigOd I used the identity $\max\{a,b\}=(a+b+|a-b|)/2$. The first and the second integral I splitted divided by $x$ is $0$ as $x\to\infty$. $\endgroup$ – Kemono Chen Jul 22 '18 at 10:19
  • $\begingroup$ Does the function $|\sin t-\sin(t\sqrt2)|$ have a period? $\endgroup$ – Szeto Jul 22 '18 at 10:35
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    $\begingroup$ @Szeto: of course not, $\sqrt2$ is irrational. $\endgroup$ – Yves Daoust Jul 22 '18 at 10:36
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As pointed out by OP, we can use $\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$ to discover that

$$ L = \lim_{x\to\infty} \frac{1}{x} \int_{0}^{x} \left|\cos\left(\frac{\sqrt{2}+1}{2}t\right)\sin\left(\frac{\sqrt{2}-1}{2}t\right)\right| \, dt. $$

Applying the substitution $\frac{\sqrt{2}+1}{2}t = \pi u$ and writing $\alpha=(\sqrt{2}-1)^2$ followed by the substitution $y = \frac{\sqrt{2}+1}{2\pi}x$, we see that

\begin{align*} L &= \lim_{y\to\infty} \frac{1}{y} \int_{0}^{y} \left| \cos(\pi u)\sin(\pi \alpha u) \right| \, du \\ &= \lim_{N\to\infty} \frac{1}{N} \int_{0}^{N} \left| \cos(\pi u)\sin(\pi \alpha u) \right| \, du \\ &= \lim_{N\to\infty} \int_{0}^{1} \left| \cos(\pi u) \right| \left( \frac{1}{N} \sum_{n=0}^{N-1} \left| \sin(\pi \alpha k + \pi \alpha u) \right| \right) \, du \end{align*}

Since $\alpha$ is irrational, the equidistribution theorem applied to $v \mapsto \left| \sin(\pi v + \pi \alpha u) \right|$ for each fixed $u$ tells that

$$ \forall u\in\mathbb{R} \ : \ \lim_{N\to\infty} \frac{1}{N} \sum_{n=0}^{N-1} \left| \sin(\pi \alpha k + \pi \alpha u) \right| = \int_{0}^{1} \left| \sin(\pi v) \right| \, dv. $$

Therefore by the dominated convergence theorem,

$$ L = \left( \int_{0}^{1} \left| \cos(\pi u) \right| \, du \right)\left( \int_{0}^{1} \left| \sin(\pi v) \right| \, dv \right) = \frac{4}{\pi^2}. $$

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  • $\begingroup$ I think you have a typo: $\max\{a,b\}=\frac{a+b}2+|\frac{a-b}2|$. $\endgroup$ – Kemono Chen Jul 23 '18 at 5:43
  • $\begingroup$ @LittleCuteKemono, Thank you for pointing out the typo. It is now fixed :) $\endgroup$ – Sangchul Lee Jul 23 '18 at 10:48
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I strongly conjecture that the limit is ${4\over\pi^2}=0.405285$. Integrating numerically over the interval $[0,1000]$ Mathematica obtained $0.406966$, but warned that the error might be larger than Mathematica's standard.

Consider the torus $T:=\bigl({\mathbb R}/(2\pi{\mathbb Z})\bigr)^2$and on $T$ the function $$f(x,y):=\max\{\sin x,\sin y\}\ .$$ Draw a figure of the fundamental domain $[-\pi,\pi]^2$ in order to identify the parts of $T$ where $\sin x$, resp. $\sin y$, is larger. Then compute the required double integrals and obtain $$\int_T f(x,y)\>{\rm d}(x,y)=16\ .$$ This means that the "space average" $E$ of $f$ is given by $$E(f)={16\over {\rm area}(T)}={4\over\pi^2}\ .$$ Now the orbit $$t\mapsto \bigl(x(t),y(t)\bigr):=(t,\>\sqrt{2}\,t)$$ projects to a line with irrational slope on $T$. In such a situation an "ergodic principle" is at work. According to this principle the "time average" of $f$ coincides with the "space average" $E(f)$. Theoretical basis for this to happen is the fact that the multiples $k/\sqrt{2}$ $(k\in{\mathbb N})$ are uniformly distributed mod $1$.

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For simplicity I'll notate $\alpha = \frac{\sqrt{2}+1}{2}$ and $\beta = \frac{\sqrt{2}-1}{2}$. Now note that the integral basically only depends on the asymptotic behaviour of the integrand, meaning that for all $y$:

$$ \begin{align} L =\lim_{x\to\infty} \frac1x \int_0^x \bigg| \cos \alpha t \cdot \sin \beta t\bigg| \,\mathrm{d}t &= \lim_{x\to\infty} \frac1x \int_y^x \bigg| \cos \alpha t \cdot \sin \beta t\bigg| \,\mathrm{d}t\\ &= \lim_{x\to\infty} \frac1{x+y} \int_0^{x} \bigg| \cos \alpha (t+y) \cdot \sin \beta (t+y)\bigg| \,\mathrm{d}t\\ &= \lim_{x\to\infty} \frac1{x} \int_0^x \bigg| \cos \alpha (t+y) \cdot \sin \beta (t+y)\bigg| \,\mathrm{d}t \end{align} $$

Letting $y = \frac{2\pi n}{\alpha}$ for $n \in \mathbb{N}$ this implies:

$$ L = \lim_{x\to\infty} \frac1{x} \int_0^x \bigg| \cos \alpha t \bigg| \cdot\bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg| \,\mathrm{d}t $$

now since $\alpha$ and $\beta$ are irrational we get (using the ergodic theorem):

$$ \lim_{N\to\infty} \frac1N \sum_{n=0}^N \bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg| = \frac1{2\pi}\int_0^{2\pi} \bigg| \sin(s) \bigg| \,\mathrm{d}s = \frac{2}{\pi} $$

for almost all $t$. With this we can now solve the original integral as follows:

$$ \begin{align} L =\lim_{x\to\infty} \frac1x \int_0^x \bigg| \cos \alpha t \cdot \sin \beta t\bigg| \,\mathrm{d}t &= \lim_{N\to\infty} \frac1N \sum_{n=0}^N \lim_{x\to\infty} \frac1{x} \int_0^x \bigg| \cos \alpha t \bigg| \cdot\bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg| \,\mathrm{d}t\\ &= \lim_{N\to\infty} \lim_{x\to\infty} \frac1{x} \frac1N \sum_{n=0}^N\int_0^x \bigg| \cos \alpha t \bigg| \cdot\bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg| \,\mathrm{d}t\\ &= \lim_{N\to\infty} \lim_{x\to\infty} \frac1{x} \int_0^x \bigg| \cos \alpha t \bigg| \cdot \frac1N \sum_{n=0}^N \bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg| \,\mathrm{d}t\\ &= \lim_{x\to\infty} \frac1{x} \int_0^x \bigg| \cos \alpha t \bigg| \cdot \frac2{\pi} \,\mathrm{d}t\\ &= \frac{4}{\pi^2} \end{align} $$

this second to last step still requires some justification however. This can be done by noting that ergodic theorem guarantees that $\frac1N \sum_{n=0}^N \bigg| \sin \beta \Bigl(t+\frac{2\pi n}{\alpha}\Bigr)\bigg|$ converges in the $L^2$ sense on $[0, 2\pi / \beta]$ to the constant function $\frac{2}{\pi}$. This is enough since, if we have $g$ bounded and $f_n$ periodic with period $2\pi / \beta$ and $f_n \to f$ in the $L^2$ sense over that period then:

$$ \begin{align} \lim_{N\to\infty} \left| \lim_{x\to\infty} \frac1{x} \int_0^x g \cdot f_n \,\mathrm{d}t - \lim_{x\to\infty} \frac1{x} \int_0^x g \cdot f \,\mathrm{d}t \right| &\le \lim_{N\to\infty} \lim_{x\to\infty} \frac1{x} \int_0^x | g \cdot f_n - g \cdot f |^2 \,\mathrm{d}t\\ &\le \lim_{N\to\infty} \lim_{x\to\infty} \frac1{x} \int_0^x C | f_n - f |^2 \,\mathrm{d}t\\ &\le \lim_{N\to\infty} \lim_{x\to\infty} \frac{C}{x} \left\lceil\frac{x}{2\pi / \beta}\right\rceil \| f_n - f \|_2^2\\ &\le \lim_{N\to\infty} \frac{C}{2\pi / \beta} \| f_n - f \|_2^2 \to 0 \end{align} $$

It's possible there's a less fiddly proof to show convergence, but combining asymptotic density and ergodic theory does make things a bit tricky.

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