# Proving that $\lim\limits_{R\to\infty}\int_0^\pi e^{-R\sin\theta}d\theta=0$ [duplicate]

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.

## marked as duplicate by Guy Fsone, Lord Shark the Unknown, Claude Leibovici integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 1 '17 at 7:23

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

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 $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$