$$\int_{C_R}{\frac{\sin(z)}{z^2}e^{inz}\, dz}$$
With $C_R$ being a semicircle in the upper half plane. So
$$\begin{align}\left|\int_{0}^{\infty}{\frac{\sin(Re^{i\theta})}{(Re^{i\theta})^2}e^{inRe^{i\theta}}iRe^{i\theta}\, d\theta}\right|&\leq\frac{1}{R}\int_{0}^{\pi}{\left|\sin(Re^{i\theta})e^{inR(\cos\theta+i\sin\theta)}\right|\, d\theta}\\&\leq\frac{1}{R}\int_{0}^{\pi}{\left|\sin(Re^{i\theta})\right|\left|e^{-Rn\sin\theta}\right|\, d\theta}\end{align}$$
How can I continue? The exponential term is fine but I writing the $\sin$ is the problem, maybe $|\sin(Re^{i\theta})|=\frac{1}{2}|e^{iRe^{i\theta}}-e^{-iRe^{i\theta}}|$ but I can't see where to go with this.
Keep going
$$|\sin(Re^{i\theta})|=\frac{1}{2}|e^{iRe^{i\theta}}-e^{-iRe^{i\theta}}|\le \frac12|e^{iRe^{i\theta}}|+\frac12|e^{-iRe^{i\theta}}|=\frac12 e^{-R\sin\theta}+\frac12 e^{R\sin\theta}$$
hence
$$\begin{align}\left|\int_{0}^{\infty}{\frac{\sin(Re^{i\theta})}{(Re^{i\theta})^2}e^{inRe^{i\theta}}iRe^{i\theta}\, d\theta}\right|&\leq\frac{1}{2R}\int_{0}^{\pi}{\left(e^{-R\sin\theta}+e^{R\sin\theta}\right)\cdot e^{-Rn\sin\theta}\, d\theta}\end{align}$$
In the case of $n\ge1,$ above integral vanishes as $R\to\infty$