Show that: $\lim\limits_{r\to\infty}\int\limits_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$ I would like to show $\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$.
Now, of course, the integrand does not converge uniformly to $0$ on $\theta\in [0, \pi/2]$, since it has value $1$ at $\theta =0$ for all $r\in \mathbb{R}$. 
If $F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$, we can find the $j$th derivative $F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$, but I don't see how this is helping.
The function is strictly decreasing on $[0,\pi/2]$, since $\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$, which is strictly negative on $(0,\pi/2)$.
Any ideas?
 A: 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 
$$
\lim_{r \to \infty} \int_0^{{\pi/2}} e^{-r \sin \theta}\text d\theta =  \int_0^{{\pi/2}}\lim_{r \to \infty} e^{-r \sin \theta}\text d\theta = \int_0^{\pi/2}0 \text d\theta = 0.
$$
A: It's only enough to show that
$$ \int\limits_{0}^{\pi/2}{e^{-r\sin\theta}\text d\theta}\le \int\limits_{0}^{\pi/2}{e^{-r\frac{2}{\pi}\theta}\text d\theta}=\frac{\pi}{2r}\left(1-e^{-r}\right) \to 0 \quad (r \to +\infty)$$
A: 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. 
A: Note that, the integral converges uniformly, since
$$ e^{-r\sin(\theta)} \leq e^{-\sin(\theta)}, \quad r\geq 1, $$
which justifies changing the limit with the integral and the answer is $0$. 
A: This can also be handled by Monotone Convergence.
The functions $f_r(\theta)=1-e^{-r\sin(\theta)}$ monotonically converge to
$$
\lim_{r\to\infty}f_r(\theta)=f(\theta)=\left\{\begin{array}{}
1&\text{if }0\lt\theta\le\frac\pi2\\
0&\text{if }\theta=0
\end{array}\right.
$$
Thus,
$$
\begin{align}
\lim_{r\to\infty}\int_0^{\pi/2}e^{-r\sin(\theta)}\,\mathrm{d}\theta
&=\lim_{r\to\infty}\int_0^{\pi/2}(1-f_r(\theta))\,\mathrm{d}\theta\\
&=\frac\pi2-\lim_{r\to\infty}\int_0^{\pi/2}f_r(\theta)\,\mathrm{d}\theta\\
&=\frac\pi2-\int_0^{\pi/2}f(\theta)\,\mathrm{d}\theta\\[4pt]
&=\frac\pi2-\frac\pi2\\[9pt]
&=0
\end{align}
$$
A: Very simple trick:
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 
$$\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-R\sin\theta}d\theta\leq\lim_{R\to\infty}\int_0^{\frac{\pi}{2}} e^{-\frac{2R}{\pi}\theta}d\theta =\lim_{R\to\infty}\frac{\pi}{2R}(1-e^{-R}) =0$$
