Riemann-Lebesgue Lemma And Integral Having $\sin(\cdot)$

Question

Let $f$ be continuous and absolutely integrable. The formulation of the Riemann-Lebesgue lemma which I am given is $$\lim_{\omega \to \infty}\int_a^bf(t)e^{i\omega t}dt=0.$$ It is taken from 'Integral Transforms and Their Applications' by Brian Davies. Later on, he claims that $$\lim_{\omega \to \infty} \int_a^b f(t)\frac{\sin(\omega t)}{t}dt=\lim_{\omega \to \infty} \int_b^\infty f(t)\frac{\sin(\omega t)}{t}=0$$ follows from the lemma ($b>a>0$). However, I fail to see why this is the case. I'd be even more glad, if there is a elementary way to show this without using the lemma.

Answer 1

Hint: Set $\frac{f(t)}{t} = g(t)$ and observe that $$\lim_{w \to \infty}\int_{a}^{b} g(t)e^{i w t} = 0.$$