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Problem

Compute $$\lim\limits_{x \to +\infty}\dfrac{\ln x}{\displaystyle \int_0^x \dfrac{|\sin t|}{t}{\rm d}t}.$$

Comment

Maybe, we can solve it by L'Hospital's rule, but there still exists a difficulty here. Though $x \to +\infty$ implies $\ln x \to +\infty$, we do not know the limit of the denominator. How to solve it?

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  • $\begingroup$ the limit of the denominator is $+\infty$. $\endgroup$ – mathworker21 Mar 24 at 9:53
  • $\begingroup$ @mathworker21 how you know that? $\endgroup$ – mengdie1982 Mar 24 at 9:56
  • $\begingroup$ The set of $t$ for which $|\sin t| \ge 1/2$ (say) is some positive proportion of all $t$, so the integral of $1/t$ over that set is infinite. $\endgroup$ – mathworker21 Mar 24 at 9:57
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    $\begingroup$ Hint: we can write the denominator as $\int_0^1 \frac{|\sin t|}{t}dt+\int_1^x \frac{|\sin t|-\frac{2}{\pi}}{t}dt +\frac{2\ln x}{\pi}.$ Show that $\int_1^x \frac{|\sin t|-\frac{2}{\pi}}{t}dt$ is bounded in $x$ using the fact that $\int_a^{a+\pi}|\sin t|-\frac{2}{\pi} dt =0$ for all $a$. $\endgroup$ – Song Mar 24 at 10:05
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    $\begingroup$ @mathworker21 The L'Hopital's rule says that if the limit for $\frac{f'}{g'}$ exists then it is equal to the original limit. It does not say anything about case when the limit does not exist. $\endgroup$ – lEm Mar 24 at 10:31
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It helps to visualise the functions involved. A cartoonish picture of the integrand is a sequence of bumps of smaller and smaller amplitude. Specifically, for $t$ between $n\pi$ and $(n+1)\pi$, you have $\frac{|\sin t|}{(n+1)\pi}\leq \frac{|\sin t|}{t} \leq \frac{|\sin t|}{n\pi}$. For the first bump, i.e. when $n=0$, we need to take some care: let us replace the upper bound by $\sin t$ in that case.

The integral of $|\sin t|$ over each such interval is $2$ (just integrate $\sin t$ from $0$ to $\pi$, and note that $|\sin t|$ is periodic with period $\pi$ – as always, pictures help), so if we write $f(x)$ for your denominator, then $\sum_{1\leq n\leq x-1}\frac{2}{n\pi}\leq f(x) \leq 2+\sum_{1\leq n\leq x}\frac{2}{n\pi}$. The harmonic sum converges to $\ln x$, so the whole limit is $\pi/2$.

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  • $\begingroup$ THX! But could you please elaborate the last few steps? $\endgroup$ – mengdie1982 Mar 24 at 13:40
  • $\begingroup$ What specifically would you like me to elaborate on? $\endgroup$ – Alex B. Mar 24 at 14:06
  • $\begingroup$ please check the solution I write. $\endgroup$ – mengdie1982 Mar 25 at 6:08
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Thanks to @Alex B.'s hint , I complete the solution. Please correct me if I'm wrong.

For any $x>0$, we can choose some $n \in \mathbb{N}$ such that $n \pi\leq x<(n+1)\pi$. Thus, we obtain $$\int_0^{n\pi}\frac{|\sin t|}{t}{\rm d}t \leq \int_0^x \frac{|\sin t|}{t}{\rm d}t<\int_0^{(n+1)\pi}\frac{|\sin t|}{t}{\rm d}t.$$

On one hand, notice that \begin{align*} \int_0^{n \pi} \frac{|\sin t|}{t}{\rm d}t&=\int_0^\pi \frac{|\sin t|}{t}{\rm d}t+\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\ &> \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\ & > \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{(k+1)\pi}{\rm d}t\\ &=\sum_{k=1}^{n-1}\frac{\int_{k\pi}^{(k+1)\pi}|\sin t|{\rm d}t}{(k+1)\pi}\\ &=\frac{2}{\pi}\sum_{k=2}^{n}\frac{1}{k}. \end{align*}

On the other hand, likewise, \begin{align*} \int_0^{(n+1) \pi} \frac{|\sin t|}{t}{\rm d}t&=\int_0^\pi \frac{|\sin t|}{t}{\rm d}t+\sum_{k=1}^{n}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\ &< \int_0^\pi {\rm d}t+\sum_{k=1}^{n}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{k\pi}{\rm d}t\\ &=\pi+\sum_{k=1}^{n}\frac{\int_{k\pi}^{(k+1)\pi}|\sin t|{\rm d}t}{k\pi}\\ &=\pi+\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}. \end{align*}

Therefore $$\frac{2}{\pi}\sum_{k=2}^{n}\frac{1}{k} <\int_0^x \frac{|\sin t|}{t}{\rm d}t<2+\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}.$$

Since $$\ln n\pi\leq \ln x<\ln(n+1)\pi,$$ we have $$\dfrac{\ln n\pi}{\pi+\dfrac{2}{\pi}\sum\limits_{k=1}^{n}\dfrac{1}{k}}<\dfrac{\ln x}{\int_0^x \dfrac{|\sin t|}{t}{\rm d}t}<\dfrac{\ln(n+1)\pi}{\dfrac{2}{\pi}\sum\limits_{k=2}^{n}\dfrac{1}{k}}.$$

Applying the subsitution as follows $$\sum_{k=1}^n \frac{1}{k}=\ln n+\gamma+\varepsilon_n,$$ (in fact, we only need to recall that $\sum\limits_{k=1}^n \dfrac{1}{k}$ and $\ln n$ are equivalent infinities), we can readily infer that the limits of the both sides in the last expression are both equal to $\dfrac{\pi}{2}$ under the process $n \to \infty$(i.e. $x \to +\infty$). Hence, according to the squeeze theorem, we can conclude that $$\frac{\ln x}{\int_0^x \frac{|\sin t|}{t}{\rm d}t} \to \frac{\pi}{2}(x \to +\infty),$$which is what we want to evaluate.

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Note that L'Hopital is not helpful here:

We have $|\sin t| \le \frac12$ if and only if $\left(k+\frac16\right)\pi \le t \le \left(k+\frac56\right)\pi$ for some $k \in \mathbb{N}$. Therefore $$\int_0^\infty \frac{|\sin t|}{t}\,dt \ge \frac12 \sum_{k=1}^\infty \int_{\left(k+\frac16\right)\pi}^{\left(k+\frac56\right)\pi} \frac{dt}{t} = \frac12\sum_{k=1}^\infty \ln\left(\frac{k+\frac56}{k+\frac16}\right) = +\infty $$ since $\ln\left(\frac{k+\frac56}{k+\frac16}\right) \xrightarrow{k\to\infty} 1$.

Hence we can apply L'Hopital to obtain $$\lim_{x\to\infty} \frac{\frac{1}{x}}{\frac{|\sin x|}{x}} = \lim_{x\to\infty} \frac1{|\sin x|}$$ but this limit doesn't exist, e.g. consider $x_n = \frac{n\pi}2$ and $x_n = n\pi$. This doesn't tell us anything about whether the original limit exists.

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The denominator can be compared to the harmonic sum $1+1/2+1/3+\dots$ (How?), which is divergent. Surely the nominator also diverges. A moment's reflection shows that it suffices to consider $(M_1, M_2)$ where $M_1=m_1\pi,M_2=m_2 \pi$ for any $0 < m_1 <m_2, m_1, m_2 \in \mathbb{N}$. Set $k=m_2-m_1$. $$ I :=\lim_{x \to \infty} \frac {\ln x} {\int^x_0 |\sin t| /t} =\lim_{x \to \infty} \frac {\int^x_{1} (1/t) dt} {\int^x_{1} (|\sin t| /t) dt} =\lim_{x \to \infty} \frac {\int^x_{M_1} (1/t) dt} {\int^x_{M_1} (|\sin t| /t) dt} $$ Set $$ N(M) =\int_{M}^{M+\pi} \frac{1}{t} dt \\ D(M) =\int_{M}^{M+\pi} \frac{|\sin t|}{t} dt $$ We see that $$ I =\lim_{m_2 \to \infty} \frac{\sum_{m'=m_1}^{m_2} N(\pi m')}{\sum_{m'=m_1}^{m_2} D(\pi m')} $$ Since $$ \left| M N(M) -M \frac{\pi}{M} \right| \leq \frac{\pi}{M+1} \\ \left| M D(M) -M \frac{2}{M} \right| \leq \frac{2}{M+1} $$ Cesaro summation tells us $$ I =\lim_{k \in \mathbb{N}:\; k \to \infty} \frac{(1/k)\sum_{M'=M_1}^{M_2} M' N(M')}{(1/k)\sum_{M'=M_1}^{M_2} M' D(M')} =\frac{\pi}{2} $$

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As for the denominator, note that $\int_0^x dt\,\frac{|\sin(t)|}{t}=\int_0^{x/\pi}dt\,\frac{|\sin(\pi t)|}{t}=\int_0^{y}dt\,\frac{|\sin(\pi t)|}{t}$ (we introduced $y:=x/\pi$). The integrand $t \mapsto \frac{|\sin(\pi t)|}{t}$ is squeezed between $h(t):=\sum_{k=1}^{\infty}\chi_{[k,k+1)}(t)\frac{1}{k+1}|\sin(t)|$ below and $j(t):=\chi_{[0,1)}(t)\pi+\sum_{k=1}^{\infty}\chi_{[k,k+1)}(t)\frac{1}{k}|\sin(t)|$ above. Hence $$\frac{2}{\pi}\sum_{k=1}^{\lfloor y \rfloor-1} \frac{1}{k+1}=\int_0^{\lfloor y \rfloor}dt\,h(t)\leq \int_0^ydt\,\frac{|\sin(\pi t)|}{t} \leq \int_0^{\lceil y\rceil}dt\,j(t) =\pi+\frac{2}{\pi}\sum_{k=1}^{\lceil y\rceil-1}\frac{1}{k}$$

Concerning the numerator, note that $\ln(x)=\ln(x/\pi)+\ln(\pi)=\int_1^y \frac{dt}{t}+\ln(\pi)$. The integrand function $t \mapsto \frac{1}{t}\chi_{(1,y)}(t)$ is squeezed between the step function $f(t):=\sum_{k=1}^{\infty}\chi_{[k,k+1)}(t)\frac{1}{k+1}$ below and the step function $g(t):=\sum_{k=1}^{\infty}\chi_{[k,k+1)}(t)\frac{1}{k}$ above. Hence $$\sum_{k=1}^{\lfloor y \rfloor-1} \frac{1}{k+1}=\int_1^{\lfloor y \rfloor}dt\,f(t)\leq \int_1^y\frac{dt}{t} \leq \int_1^{\lceil y\rceil}dt\,g(t) =\sum_{k=1}^{\lceil y\rceil-1}\frac{1}{k}$$ Gathering all our results, we find (recall that y=x/\pi) $$\frac{\left(\sum_{k=1}^{\lceil y\rceil-1}\frac{1}{k}\right)-1+\ln(\pi)}{\frac{2}{\pi}\sum_{k=1}^{\lceil y \rceil-1} \frac{1}{k}}=\frac{\left(\sum_{k=1}^{\lfloor y\rfloor-1}\frac{1}{k+1}\right)+\ln(\pi)}{\frac{2}{\pi}\sum_{k=1}^{\lceil y \rceil-1} \frac{1}{k}}\leq \frac{\ln(x)}{\int_0^xdt\,\frac{|\sin(t)|}{t}}\leq \frac{\left(\sum_{k=1}^{\lceil y\rceil-1}\frac{1}{k}\right)+\ln(\pi)}{\frac{2}{\pi}\sum_{k=1}^{\lfloor y \rfloor-1} \frac{1}{k+1}}=\frac{\left(\sum_{k=1}^{\lceil y\rceil-1}\frac{1}{k}\right)+\ln(\pi)}{\frac{2}{\pi}\left(\sum_{k=1}^{\lceil y \rceil-1} \frac{1}{k}\right)-\frac{2}{\pi}}$$ since the expression on the left and on the right converge to $\pi/2$ as we let $y=x/\pi\to \infty$, so does the middle expression.

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