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How to show that:

$$\lim_{x\rightarrow 0} \frac{1}{x}\int_0^x |\sin(1/y)| \mathrm{d} y \not=0$$

It seems like a easy example of illustrating 0 is not in the Lebesgue set of $g(x)$ where $g(x)=\sin(1/x)$ if $x\neq 0$ and $g(0)=0$. But I fail to see why the above integral is true.

I tried looking at the intervals such that $\sin(1/y)$ is greater or equal to some constant (for example, $\left[\frac{1}{k\pi+\pi/6}, \frac{1}{k\pi+5\pi/6}\right]$ such that $\sin(1/y)\geq \frac{1}{2}$), however, $$\sum_{k \text{ large}} \left(\frac{1}{k\pi+\pi/6}-\frac{1}{k\pi+5\pi/6}\right)$$ converges, which is not strong enough to prove the claim. Any thoughts? Thanks in advance.

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up vote 2 down vote accepted

Let $y= 1/t$. We then get that $$I(x) = \dfrac1x \int_0^x \left \vert \sin(1/y) \right \vert dy = \dfrac1x \int_{\infty}^{1/x} \left \vert \sin(t) \right \vert \dfrac{-dt}{t^2} = \dfrac1x \int_{1/x}^{\infty} \dfrac{\left \vert \sin(t) \right \vert}{t^2} dt$$ Let $x=\dfrac1{n \pi}$. Hence, \begin{align} I_n & = n \pi \int_{n \pi}^{\infty} \dfrac{\vert \sin(t) \vert}{t^2} dt\\ & =n \pi \left(\sum_{k=n}^{\infty} \int_{k \pi}^{(k+1) \pi} \dfrac{\vert \sin(t) \vert}{t^2} dt \right)\\ & \geq n \pi \left(\sum_{k=n}^{\infty} \int_{k \pi}^{(k+1) \pi} \dfrac{\vert \sin(t) \vert}{(k+1)^2 \pi^2} dt \right)\\ & = n \pi \sum_{k=n}^{\infty} \left(\dfrac{1}{(k+1)^2 \pi^2}\displaystyle \int_{k \pi}^{(k+1) \pi} \vert \sin(t) \vert dt \right)\\ & = n \pi \sum_{k=n}^{\infty} \dfrac2{(k+1)^2 \pi^2}\\ & = \dfrac{2n}{\pi} \sum_{k=n}^{\infty} \dfrac1{(k+1)^2}\\ & > \dfrac{2n}{\pi} \int_{n+1}^{\infty} \dfrac{dt}{t^2}\\ & = \dfrac{2n}{\pi(n+1)} \end{align} Hence, we if let $x = \dfrac1{n \pi}$, then we get that $$I\left( \dfrac1{n \pi} \right) = I_n \geq \dfrac{2n}{\pi(n+1)}$$ Hence, $$\lim_{n \to \infty} I\left( \dfrac1{n \pi} \right) \geq \dfrac{2}{\pi}$$

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Wouldn't the limit be exactly $\frac{2}{\pi}$? – susan Nov 28 '12 at 8:04
@susan: no, notice the > sign on the second to last line of the long list of steps. – Willie Wong Nov 28 '12 at 8:20
@willie wong: Yes, but what if we use the lower limit of the integral, namely $k\pi$ to obtain the first inequality(the reverse in this case). Then, wouldn't the integral be bounded by $2/\pi$ above as well? – solea Nov 28 '12 at 8:35
@susan The error induced in the second inequality goes as $\dfrac1n$ and hence the error goes to $0$ as $n \to \infty$ i.e. $$\lim_{n \to \infty} n \sum_{k=n+1}^{\infty} \dfrac1{k^2} = \lim_{n \to \infty}(1+\mathcal{O}(1/n)) = 1$$ However, there is also the first inequality, whose error must also be proved to go to $0$ as $n \to \infty$. – user17762 Nov 28 '12 at 8:47

You had the right idea: $$\int_{1/((k+1) \pi)}^{1/(k\pi)} |\sin(1/y)| \ dy \ge \frac{1}{2\pi} \left(\frac{1}{k+1/6} - \frac{1}{k+5/6}\right) \ge \frac{C}{(k+1)(k+2)} = \frac{C}{k+1} - \frac{C}{k+2}$$ for some positive constant $C$, so if $n \pi \le 1/x < (n+1)\pi$ $$ \int_{0}^{x} |\sin(1/y)|\ dy \ge \int_{0}^{1/(n \pi)} |\sin(1/y)|\ dy \ge \sum_{k=n}^\infty \frac{C}{k+1} - \frac{C}{k+2} = \frac{C}{n+1}$$ and $$\frac{1}{x} \int_0^x |\sin(1/y)|\ dy \ge \ldots$$

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