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I want to show that the limit below is zero. $$\lim_{R\to\infty}\int_0^\pi e^{-R\sin\theta}d\theta$$ Wolframalpha and my intuition say that the limit is truly zero but I cannot approach.

Any hints will be appreciated. Thanks.

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marked as duplicate by Guy Fsone, Lord Shark the Unknown, Claude Leibovici integration Nov 1 '17 at 7:23

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    $\begingroup$ I think you can use the Dominated Convergence Theorem to pass the limit through the integral sign. $\endgroup$ – Thoth Jan 12 '17 at 10:58
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    $\begingroup$ there is a really famous inequality for the sine function.. $\endgroup$ – tired Jan 12 '17 at 11:05
  • $\begingroup$ I missed the very easy point. Thanks, especially for @tired . $\endgroup$ – Jinmu You Jan 12 '17 at 17:45
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This is a very nice question.

First split the integral involves at $\theta=\frac{\pi}{2}$ then take the change of variable $\theta'=\pi -\theta$

$$ \int_0^\pi e^{-R\sin\theta}d\theta =2\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta$$

afterward shows by studying the function $[0,\frac{\pi}{2}]\ni\theta \mapsto\frac{\sin\theta}{\theta}$ that

$$ \color{blue}{\sin\theta \geq \frac{2}{\pi}\theta ~~ \forall \theta\in [0,\frac{\pi}{2}] } $$ therefore we get that $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$

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