integration - How to evaluate the limit $int_{0}^{frac{pi}{2}}Re^{-Rsintheta}dtheta (as quad R rightarrow infty)$

While doing a mathematical exercise(stein Complex Analysis chapter2,exercise 3),
I managed to reduce the problem to the following one:

$$\int_{0}^{\omega}Re^{-R\cos\theta}d\theta \rightarrow 0 \; (as \quad R \rightarrow \infty)$$ where $0\le \omega <\frac{\pi}{2}$.

I can prove this without much difficulty:
$$\int_{0}^{\omega}Re^{-R\cos\theta}d\theta \le \int_{0}^{\omega}Re^{-R\cos\omega}d\theta =\omega Re^{-R\cos\omega} \rightarrow 0 \; (as \quad R \rightarrow \infty)$$
It is crucial that $\omega$ is strictly less than $\frac{\pi}{2}$. This lead me to raise another interesting problem: what the limit will be if we replace $\omega$ by $\frac{\pi}{2}$. After changing $\cos\theta$ to $\sin\theta$( this doesn't matter), now my question is

I have no idea how to calculate, I even don't know if the limit exists.

Answer

Put $I(R)$ your integral and $J(R)=\int_{0}^{\pi/2}R\cos(\theta)^2\exp(-R\sin(\theta))d\theta$, $K(R)=\int_{0}^{\pi/2}R\sin(\theta)^2\exp(-R\sin(\theta))d\theta$. We have $I(R)=J(R)+K(R)$; Note that the function $u\exp(-u)$ is positive and bounded on $[0,+\infty[$, say by $M$.

a) For $K(R)$, we have $R\sin(\theta)^2\exp(-R\sin(\theta))\leq M$ for all $\theta$, and this function goes to $0$ everywhere if $R\to +\infty$. By the Dominated convergence theorem, $K(R)\to 0$ as $R\to +\infty$.

b) For $J(R)$, we integrate by parts:
$$J(R)=[(\cos(\theta)(-\exp(-R\sin(\theta))]_0^{\pi/2}-\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$$
We have hence $J(R)=1-\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$. Now apply the dominated convergence theorem to $\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$, and you are done.