Proof
(related to Lemma: Riemann Integral of a Product of Continuously Differentiable Functions with Sine)
 By hypothesis, $[a,b]$ is a closed real interval and $f:[a,b]\to\mathbb R$ is a continuously differentiable function.
 We define for $y\in\mathbb R,$ the Riemann integral $F(y):=\int_{a}^{b}f(x)\sin(yx)dx.$
 Please note that the derivative of $\frac {d}{dx}\frac {\cos(yx)}{y}$ is $sin(xy).$
 With this result, if $y\neq 0,$ then the partial integration gives us $$F(y)=\underbrace{f(x)\frac{\cos(yx)}{y}\Rule{1px}{4ex}{2ex}^b_a}_{=:A}+\underbrace{\frac 1y\int_a^b f'(x)\cos(yx)dx}_{=:B}.\label{eq:E20241}\tag{1}$$
 Because $f$ and the derivative $f'$ are continuous and because $[a,b]$ is closed, both continuous functions are bounded.
 By definition of bounded functions there is a constant $M\ge 0$ such that for all $$f(x)\le M,\quad f'(x)\le M,\quad\text{for all }x\in[a,b].$$
 Since $\cos(yx)\le 1$ for all $x\in[a,b],$ the term $A$ in $(\ref{eq:E20241})$ can be approximated to $A\le \frac{2M}{y}$ and the term $B$ to $B\le \frac{(ba)M}{y},$ where $y$ is the absolute value of $y.$
 Since both upper bounds tend to zero as $y$ tends to infinity, we have $\lim_{y\to\infty}F(y)=0.$
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983