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.
Proving that $\lim\limits_{R\to\infty}\int_0^\pi e^{-R\sin\theta}d\theta=0$
1
$\begingroup$
calculus
integration
complex-analysis
analysis
limits
-
1I think you can use the Dominated Convergence Theorem to pass the limit through the integral sign. – 2017-01-12
-
1there is a really famous inequality for the sine function.. – 2017-01-12
-
0I missed the very easy point. Thanks, especially for @tired . – 2017-01-12
1 Answers
3
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 $$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$