Show that: $\lim\limits_{r\to\infty}\int\limits_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$

Solution 1:

It's only enough to show that

$$ \int\limits_{0}^{\pi/2}{e^{-r\sin\theta}\text d\theta}\le \int\limits_{0}^{\pi/2}{e^{-r\frac{2}{\pi}\theta}\text d\theta}=\frac{\pi}{2r}\left(1-e^{-r}\right) \to 0 \quad (r \to +\infty)$$

Solution 2:

On $[0,\pi/2]$ the sine is nonnegative and so $|e^{-r\sin \theta}| \leq 1$ for $r \geq 0$. It follows by dominated convergence that $$ \lim_{r \to \infty} \int_0^{{\pi/2}} e^{-r \sin \theta}\text d\theta = \int_0^{{\pi/2}}\lim_{r \to \infty} e^{-r \sin \theta}\text d\theta = \int_0^{\pi/2}0 \text d\theta = 0. $$

Solution 3:

Split it into two pieces: one integral over $[0,\epsilon]$ and another over $[\epsilon, \pi/2]$. Since the integrand is bounded on the first piece, you can make it arbitrarily small by choosing $\epsilon$ small. On the other piece, it converges uniformly to zero. I think you can take it from there.