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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 $|B|\le \frac{(b-a)M}{|y|},$ 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

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983