$f:(0,1)\rightarrow \mathbb{R}$ given by $f(x)=\frac{1}{x}\sin(\frac{1}{x})$ is not Lebesgue integrable.

Question

I'm trying to show that $f:(0,1)\rightarrow \mathbb{R}$ given by $f(x)=\frac{1}{x}\sin(\frac{1}{x})$ is not Lebesgue integrable.

My initial thought is coming up with a smaller function $g(x)\leq |f(x)|$ and showing that $\int_{(0,1)}g(x)=\infty$.

I was thinking $g(x)=|\sin(\frac{1}{x})|$, then, $\int_{(0,1)}|\sin(\frac{1}{x})|=[x^2\cos(\frac{1}{x})]_0^1=1$ which doesn't help.

Can anyone help me?

Answer 1

At an intuitive level, $\sin(\frac1x)$ fluctuates near $x\approx0$, such that roughly one third of the time, $\sin(\frac1x)>\frac12$. So the integral diverges at least as badly as $\frac{1}{6x}$. You can turn this into a precise argument by computing the region where $\sin(\frac1x)>\frac12$ and comparing the integral of $\frac1x$ there with the harmonic series.

Answer 2

Hint: For $n=1,2,\dots, $ let $a_n=1/(2n\pi + \pi/2),$ $b_n=1/(2n\pi + \pi/4).$ On the intervals $[a_n,b_n],$ which are disjoint, $\sin(1/x)\ge 1/\sqrt 2.$