If $ 0 \le t \lt 1$ then, $\lim\limits_{R \uparrow \infty}\int_{0}^{\pi}e^{-R\sin\theta}R^t\text d\theta = 0$

Question

If $ 0 \le t \lt 1$ then, $$\lim\limits_{R \uparrow \infty}\int_{0}^{\pi}e^{-R\sin\theta}R^t\,\text d\theta = 0$$

I'm not sure how to solve this, I thought about starting to show that $ 0 \le \sin\theta \le \pi/2$, then $ 0 \le \sin\theta \le 2\theta/\pi$. But I'm still not sure how to continue from there.

Answer 1

Using the fact that $\sin(\pi -\theta)=\sin \,\theta$ you can see that $\int_{\pi /2}^{\pi} e^{-R\sin\, \theta} R^{t}\text d\theta =\int_{0}^{\pi /2} e^{-R\sin\, \theta} R^{t}\text d\theta$. On $(0, \pi/2)$ we have $\sin\, \theta \geq \frac 2 {\pi} \theta$. Can you complete the argument now?