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Is $g(x) = sin(1/x) \forall x \in \mathbb{R} \setminus \{0\}$ and $g(0)=0$ Lebesgue integrable?

karnan
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2 Answers2

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We have $\lim_{x\to \pm \infty} \frac{g(x)}{\frac{1}{x}}= \lim_{t\to 0}\frac{\sin t}{t} = 1$, and $\frac{1}{x}$ is not integrable.

orangeskid
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  • How can I conclude from that, that $g(x)$ is not lebesgue integrable? – karnan Oct 22 '20 at 08:02
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    @reij: For instance, $g(x)\ge \frac{1}{2} \frac{1}{x}$ on some interval $[x_0, \infty)$. Since $\int_{x_0}^\infty \frac{1}{x} dx = \infty$, we also have $\int_{x_0}^\infty g(x) dx = \infty$, so $g$ is not integrable on $[x_0, \infty)$, so not integrable on $\mathbb{R}$. – orangeskid Oct 22 '20 at 23:45
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No. Since $\sin(y)\geq \frac{2}{\pi}|y|$ for all $y\in [0,\frac{\pi}{2}]$, we have $g(x)\geq \frac{2}{\pi}\frac{1}{x}$ for all $x\in I:=[\frac{2}{\pi}, \infty)$. As $x\mapsto\frac{1}{x}$ is not integrable on $I$, it follows that $g$ is also not integrable on $I$. In particular, $g$ is not integrable on $\mathbb{R}\supset I$.

ym94
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