# Lebesgue integrable sin(1/x)

How can I prove that the function

$$f:(0, 1) \to \mathbb{R}$$
$$f(x) = \frac{|\sin(\frac{1}{x})|}{x}$$

is not Lebesgue integrable?

Should I try the definition?

Because I need to find a function $$f:A \to \mathbb{R}$$ such that $$A$$ is open and bounded and $$f$$ is continuous, not bounded and not Lebesgue integrable, and I think that $$f$$ works, right?

• You could just use $f(x) = 1/x^2$ on the same domain.
– user169852
Oct 29, 2019 at 19:31

Observe that if $$x \in \left[ \frac{(6n+1)\pi}{6}, \frac{(6n+5)\pi}6 \right] = \left[ n\pi + \frac \pi 6, (n+1)\pi - \frac \pi 6 \right]$$ then $$|\sin x| \ge \dfrac 12$$. Define $$E_n = \left[ \frac{6}{(6n+5)\pi},\frac{6}{(6n+1)\pi} \right].$$ It follows that $$x \in E_n \implies \frac 1x \ge \frac{(6n+1)\pi}{6} \ge n\pi \quad \text{and} \quad \left|\sin\left( \frac 1x \right) \right| \ge \frac 12$$ and consequently $$x \in E_n \implies f(x) = \frac{|\sin(\frac 1x)|}{x} \ge \frac{n\pi}2 \ge n.$$ It is simple to verify that the intervals $$\{E_n\}_{n=1}^\infty$$ are pairwise disjoint and that each $$E_n \subset (0,1)$$. This implies that for any $$N$$ you have $$\sum_{n=1}^N n \chi_{E_n}(x) \le f(x)$$ on $$(0,1)$$. Since $$f \ge 0$$ the definition of the Lebesgue integral gives you $$\sum_{n=1}^N n \ell(E_n) \le \int_{(0,1)} f.$$ It appears that $$\ell(E_n) \ge \dfrac 1{5(n+1)^2}$$ so that $$\frac 15 \sum_{n=1}^N \frac{n}{(n+1)^2} \le \int_{(0,1)} f.$$ This is true for any natural number $$N$$---the divergence of the harmonic series tells you that the left-hand-side is unbounded as $$N \to \infty$$, so $$f$$ cannot have finite integral.