How to evaluate the limit $\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta (as \quad R \rightarrow \infty)$ 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 
$$\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta  \rightarrow ? \; (as \quad R \rightarrow \infty)$$
I have no idea how to calculate, I even don't know if the limit exists.
 A: 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. 
A: This is a classic integral in any textbook. $$\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta\le\int_{0}^{\frac{\pi}{2}}Re^{-R\frac{2\theta}{\pi}}d\theta=\frac{1}{2}\pi(1-e^{-R})\rightarrow\frac{\pi}{2}\; (as \quad R \rightarrow \infty).$$
A: $$I_R=\int_{0}^{\frac{\pi}{2}}R\,e^{-R\sin(\theta)}\,d\theta=\frac{\pi  R}{2}  \,(I_0(R)-\pmb{L}_0(R))$$ where appear Bessel and Struve functions.
The result tends (quite slowly) to $1$ when $R$ becomes larger and larger as shown in the table below
$$\left(
\begin{array}{cc}
R & I_R \\
  10 & 1.01126 \\
 11 & 1.00909 \\
 12 & 1.00750 \\
 13 & 1.00630 \\
 14 & 1.00538 \\
 15 & 1.00465 \\
 16 & 1.00406 \\
 17 & 1.00358 \\
 18 & 1.00318 \\
 19 & 1.00284 \\
 20 & 1.00256
\end{array}
\right)$$
Edit
If you look here, at the bottom of the page, you will find the very interesting asymprotic relation
$$\pmb{L}_n(x)=I_{-n}(x)-\frac{x^{n-1}}{2^{n-1}\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right)}$$ Use $n=0$ to get the value of $1$.
A: This is a straightforward application of Laplace's Method. Since $\sin(\theta)$ is strictly increasing on $[0,\pi/2]$ we have that the main contribution of the integral comes from near $\theta$ around zero. So by applying Laplace's method, we get:
\begin{align}
\int^{\pi/2}_0 e^{-R \sin\theta}\,d\theta = \frac{1}{R} + o\left( \frac 1 R\right)
\end{align}
as $R\to\infty$ and therefore:
\begin{align} 
\lim_{R\to\infty} \int^{\pi/2}_0 Re^{-R \sin\theta}\,d\theta = 1
\end{align}
A: Put
\begin{equation*}
I(R)=\int_{0}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta= \int_{0}^{\pi/4}Re^{-R\sin \theta}\,\mathrm{d}\theta+\int_{\pi/4}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta .
\end{equation*}
Integration by parts yields
\begin{gather*}
\int_{0}^{\pi/4}Re^{-R\sin \theta}\,\mathrm{d}\theta = \left[\dfrac{-1}{\cos\theta}e^{-R\sin \theta}\right]_{0}^{\pi/4}+ \int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta=1-\sqrt{2}e^{-R/\sqrt{2}}+\\[2ex]\int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta .
\end{gather*}
But according to Lebesgue's dominated convergence theorem
\begin{equation*}
\int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta \to 0, \quad R\to\infty
\end{equation*}
and
\begin{equation*}
\int_{\pi/4}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta \to 0, \quad R\to\infty.
\end{equation*}
Consequently
\begin{equation*}
\lim_{R\to\infty}I(R)=1.
\end{equation*}
