integral of monotonic increasing function

Question

I have to show that if we have a monotonic increasing function $ g :[0,1] \to \mathbb{R} $ such that $\lim_{x\to0^+}g(x)=0$, then $$\lim_{n\to\infty}\int_{0}^1\frac {g(x) \sin (nx)}{x}dx=0$$ Because of the limit of $g(x)$ as $x$ goes to $0$, it implies that $\forall\varepsilon\gt0$, $\exists \delta\gt 0$ , $0\le g(x)\lt\varepsilon$, if $0\lt x\lt\delta$. We can now split the integral in two parts: $$\lim_{n\to\infty}\int_{0}^1\frac {g(x) \sin (nx)}{x}dx=\lim_{n\to\infty}\int_{0}^\delta\frac {g(x) \sin (nx)}{x}dx+\lim_{n\to\infty}\int_{\delta}^1\frac {g(x) \sin (nx)}{x}dx=$$$$\lim_{n\to\infty}\int_{0}^\delta\frac {g(x) \sin (nx)}{x}dx$$ The second part vanishes as n goes to infinity because of the Riemann lemma. So, my problem is to evaluate the first part.

Answer 1

HINT:

From Bonnet's Mean Value Theorem, since $g(x)$ is monotonic and $\frac{\sin(nx)}{x}$ is integrable, there exists a number $\xi\in(0,\delta)$ such that

$$\begin{align} \int_0^\delta \frac{g(x)}{x}\sin(nx)\,dx&= g(0^+)\int_0^\xi \frac{\sin(nx)}{x}\,dx+g(\delta)\int_\xi^\delta \frac{\sin(nx)}{x}\,dx\\\\ &=g(\delta)\int_{n\xi}^{n\delta}\frac{\sin(x)}{x}\,dx \end{align}$$

Can you finish?