Let $g(x)$ be a continuous function on the interval $(0,1]$ such that:
- $x^2\, g(x)$ is increasing on $(0,1]$,
- $\lim\limits_{x\to 0^+} x^2g(x) = 0$,
- $g'(x)$ is continuous on $(0,1]$.
Show that $\int_0^1 g(x) \sin \left( \frac{1}{x} \right)\, dx$ converges.
I have a proof for this problem, but it doesn't use the fact that $g'(x)$ is continuous on $(0,1]$. For this reason, I would appreciate if someone point out where my mistake is. Thank you.
Proof. Since $h(x) = -\frac{1}{x}$ is differentiable and increasing on $[0,1]$, $h'$ is continuous on $(0, 1]$, and $f(x) = g(x)\, \sin \left(\frac{1}{x}\right)$ is continuous on every closed interval $[-a, -1]\subset (-\infty,1]$, we have that (by change of variables) $$ \int_0^1 g(x)\, \sin \left(\frac{1}{x}\right)\, dx = \int_{-\infty}^{-1} \frac{1}{u^2} g \left( -\frac{1}{u} \right) \sin (-u)\, du .$$ Since $k(u) = -u$ is differentiable and increasing on $[0,1]$, $h'$ is continuous, and $l(u) = -u^2 g \left( -\frac{1}{u} \right) \sin (-u)$ is continuous on every closed interval $[1, b]\subset [1, \infty)$, we have that $$ \int_0^1 g(x)\, \sin \left(\frac{1}{x}\right)\, dx = \int_1^{\infty} \left( \frac{1}{v} \right)^2 g\left( \frac{1}{v} \right) \sin{v}\, dv .$$
Because $\left| \int_a^b \sin{v}\, dv\right|\le 2$ for all $b>1$ and the function $v\mapsto \left( \frac{1}{v} \right)^2 g\left( \frac{1}{v} \right)$ monotonically tends to $0$ as $x\to \infty$, the integral $$ \int_1^{\infty} \left( \frac{1}{v} \right)^2 g\left( \frac{1}{v} \right) \sin{v}\, dv \;\mbox{ diverges} $$ by Dirichlet's test.