Evaluating $\lim_{r \to \infty} \int_0^\frac{\pi}{2} e^{-r\sin\theta} d\theta$ Evaluating $$\lim_{r \to \infty} \int_0^\frac{\pi}{2} e^{-r\sin\theta} d\theta$$
First for all, I need to show that $e^{-r\sin\theta}$ converges uniformly to a function $F(r)$ Then I can easily take the limit inside, since $e^{-r\sin\theta}$ is continuous.
How can I show it is uniformly continuous- my approach was:
$$|e^{-r\sin\theta}|\le |e^{-\delta \sin\theta}| \ for \ \delta \le r$$ 
Then If that $\int_0^\frac{\pi}{2}e^{-\delta \sin\theta}$ exists, in the sense that converges LHD is uniformly continuous so I can take the limit inside and arrive at the soulution. However I couldn't take the integration, any hints? 

If we set
$$ I(r)=\int_{0}^{\pi/2}e^{-r\sin\theta}\,d\theta $$
we have that $I(r)$ is a moment. Since $e^{-\sin\theta}$ is a positive and decreasing function on $(0,\pi/2)$, $I(r)$ is a positive, continuous and log-convex function on $\mathbb{R}^+$, due to the Cauchy-Schwarz inequality. $I(r)$ is also decreasing, in particular $\lim_{r\to +\infty}I(r)$ is a finite, non-negative number $L$. On the other hand, for any $r$ large enough
$$ I(r) = \int_{0}^{\arcsin\frac{1}{\sqrt{r}}}e^{-r\sin\theta}\,d\theta+\int_{0}^{\pi/2}e^{-r\sin\theta}\,d\theta \leq \arcsin\frac{1}{\sqrt{r}}+\frac{\pi}{2}e^{-\sqrt{r}} $$
hence the only chance is $L=0$.
