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I'm asked to calculate

$$\int_{0}^{+\infty} \frac{\sin nx \sin^n x}{x^{n+1}}\text{d}x$$ by integrating $\int_{\Gamma} \frac{e^{niz}\sin^n z}{z^{n+1}}\text{d}z$ on a semi-arc. $(n\in \mathbb{N}; n\geqslant 1)$

I'm having troubles proving:

$$\left|\int_{\Gamma} \frac{e^{niz}\sin^n z}{z^{n+1}}\text{d}z \right| \to 0 \qquad \text{ if } R\to \infty$$

Using the parameterization $z(\theta) = Re^{i\theta}\quad \theta: 0\to \pi$ I always end up with something like this:

$$\left|\int_{\Gamma} \frac{e^{niz}\sin^n z}{z^{n+1}}\text{d}z \right|\leqslant \frac{(1+\color{red}{e^{R}})^n}{R^{n+1}}\cdot M$$

With $M$ some constant, so this obviously does not $\to 0$ if $R\to \infty$

The $\color{red}{e^R}$ originates from $|\sin^n z|$. The problem would be situated here, how can I bound this non-exponential?

How could I tame this integral?

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$$e^{iz}\sin z = \frac{e^{2iz}-1}{2i},$$ hence, if $z=Re^{i\theta}$ with $R>0$ and $\theta\in[0,\pi]$: $$\|e^{iz}\sin z\,\|=\frac{1}{2}\left\|e^{-2R\sin\theta} e^{2Ri\cos\theta}-1\right\|\leq\frac{e^{-2R\sin\theta}+1}{2}\leq 1.$$

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