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I was collecting some exercises for my students, and I found this one in a book: compute, if it exists, the limit $$ \lim_{x \to +\infty} \int_x^{2x} \sin \left( \frac{1}{t} \right) \, dt. $$ It seems to me that this limit exists by monotonicity. Moreover, since $\frac{2}{\pi}x \leq \sin x \leq x$ for $0 \leq x \leq \pi/2$, I could easily show that $$ \frac{2}{\pi} \log 2 \leq \lim_{x \to +\infty} \int_x^{2x} \sin \left( \frac{1}{t} \right) \, dt \leq \log 2. $$ WolframAlpha suggests a "closed" form for the integral, and by dominated convergence the limit turns out to be $\log 2$. However, passing to a limit in the $\operatorname{Ci}(\cdot)$ function is not really elementary. I wonder if there is a simpler approach that a student can understand at the end of a first course in mathematical analysis.

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Given $0<\delta<1$, for $t$ sufficiently large, $\sin(1/t)\ge (1-\delta)/t$. –  David Mitra Jan 19 at 11:24
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Substituting $u = 1/t$ and using $u - u^3/6 < \sin u < u$ for $0 < u < 1$ looks elementary to me. –  Daniel Fischer Jan 19 at 11:54
    
@Siminore: Sorry for asking a basic question but what is the hint for sinx $\leq$ x? Do you use the expansion of sine function? –  Abhimanyu Arora Jan 19 at 12:21
    
I take this for granted with my students. I suggest looking at the graphs. For a rigorous proof, you can indeed expand. –  Siminore Jan 19 at 12:23
    
When $x \to \infty$, just replace $\sin\left(1/t\right)$ by $1/t$. –  Felix Marin Jan 20 at 4:14

7 Answers 7

up vote 5 down vote accepted

Integrate by parts, with $u=\sin\left(\frac{1}{t}\right)$ and $dv=dt$. We get that this is equal to

$$ \left[t\sin\left(\frac{1}{t}\right)\right|_x^{2x}+\int_x^{2x}\frac{1}{t}\cos\left(\frac{1}{t}\right)dt $$

Now the first term equals

$$ \frac{\sin\left(\frac{1}{2x}\right)}{\frac{1}{2x}}-\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}} $$

which limits to zero. On the second term, for sufficiently large $t$, $1-\varepsilon\leq\cos\left(\frac{1}{t}\right)\leq 1$. So we get that the second term is between $(1-\varepsilon)\log2$ and $\log2$.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t&= \lim_{x \to \infty}\int_{1/x}^{1/2x}\sin\pars{t}\,\pars{-\,{\dd t \over t^{2}}} =\lim_{x \to \infty}\int_{1/x}^{1/2x}{\cos\pars{t} \over t}\,\dd t \\[3mm]&=\lim_{x \to \infty}\braces{% -\bracks{-\int_{1/x}^{\infty}{\cos\pars{t} \over t}\,\dd t} + \bracks{-\int_{1/2x}^{\infty}{\cos\pars{t} \over t}\,\dd t}} \\[3mm]&= \lim_{x \to \infty}\bracks{-{\rm ci}\pars{1 \over x} + {\rm ci}\pars{1 \over 2x}} \end{align} where ${\rm ci}\pars{z}$ is the $\it\mbox{Cosine Integral}$ which satisfies $\lim_{z \to 0}\bracks{{\rm ci}\pars{z} - \gamma - \ln\pars{z}} = 0$. $\gamma$ is the Euler-Mascheroni constant. Then,

\begin{align} &\lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t= \lim_{x \to \infty}\braces{% -\bracks{{\rm ci}\pars{1 \over x} - \gamma - \ln\pars{1 \over x}} +\bracks{{\rm ci}\pars{1 \over 2x} - \gamma - \ln\pars{1 \over 2x}} + \ln\pars{2}} \end{align}

$$\color{#00f}{\large% \lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t = \ln\pars{2}} $$

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$$\int_{x}^{2x}{\sin\left(\frac{1}{t}\right)dt}=\int_{x}^{2x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\cdot t^{-(2n+1)}\,dt= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!} \int_{x}^{2x}{t^{-(2n+1)} dt}$$

Since $$\lim_{x \rightarrow \infty} \int_{x}^{2x}{t^{-(2n+1)}dt}=\begin{cases}0,& n >0 \\ \log2,& n=0\end{cases} $$

We conclude that $$\lim_{x \rightarrow \infty} \int_{x}^{2x}{\sin\left(\frac{1}{t}\right)\,dt}= \log2.$$

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Yes, I had devised this proof, but I should prove that the series and the integral can be swapped. This is not taught in a first course in analysis. –  Siminore Jan 19 at 18:08

If $t$ approaches infinity, $sin(1/t)$ will approach $0$, so you can just use the Taylor expansion, which we know to be absolutely convergent (if the students have had Taylor series in their first analysis course):

$$ \begin{align*}& lim_{x\to\infty} \int_x^{2x} sin(1/t) dx\\ = & lim_{x\to\infty} \int_x^{2x} (1/t) + O((1/t)^3) dx\\ = & lim_{x\to\infty} \left[ln(t) + O((1/t)^2)\right]_x^{2x}\\ = & lim_{x\to\infty} \left[ln(2) + O((1/x)^2)\right]\\ = & log(2) \end{align*} $$

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In the continuation of what Gaffney wrote, Cos[1/t] can be developed for large values of t as

1 - 1/(2 t^2) + 1/(24 t^4) - 1/(720 t^6) + ....

So, the value of Gaffney's last integral between $x$ and $2x$ is

Log[2] - 3/(16 x^2) + 5/(512 x^4) + ...

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$$ \int_x^{2x}\sin\left(\frac{1}{t} \right)\,dt=\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{\sin(u)}{u^2}\,du, $$ $$ \sin(u)=u+o(1)u, $$ if $u$ is around $0$, (here we use the elemetary $\lim_{u\to0}\frac{\sin(u)}{u}=1$) so $$ \int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{\sin(u)}{u^2}\,du=\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{1}{u}(1+o(1))\,du, $$ where $o(1)\to 0$ as $u\to0$, which is the case, because $x\to+\infty$.

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Hint: $$\int_{x}^{2x}\sin\left(\frac{1}{t}\right)\,dt=\frac{1}{2}\int_{2x}^{4x}\sin\left(\frac{2}{t}\right)\,dt=\int_{2x}^{4x}\sin\left(\frac{1}{t}\right)\cos\left(\frac{1}{t}\right)\,dt<\int_{2x}^{4x}\sin\left(\frac{1}{t}\right)\,dt$$

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