For \(0\le\theta\le\pi/2\), prove \[ \sin\theta\ge \frac{2}{\pi}\theta. \]
Using part 1 or otherwise, prove that if \(\lambda<1\), then \[ \lim_{R\to\infty}R^\lambda\int_0^{\pi/2}e^{-R\sin\theta}\,d\theta=0. \]
Solution
1. For \(0\le\theta\le\pi/2\),\[ \sin\theta \ge \frac{2}{\pi}\theta. \]
On \([0,\pi/2]\), \[ \frac{d^2}{d\theta^2}\sin\theta = -\sin\theta \le 0, \] so \(\sin\theta\) is concave. A concave function lies above the chord joining any two points of its graph. The chord joining \[ (0,0) \qquad\text{and}\qquad (\pi/2,1) \] has equation \[ y=\frac{2}{\pi}\theta. \] Thus the displayed inequality holds.
2. For every \(R>0\),\[ 0 \le \int_0^{\pi/2}e^{-R\sin\theta}\,d\theta \le \frac{\pi}{2R}. \]
By step 1, \[ e^{-R\sin\theta} \le e^{-2R\theta/\pi}. \] Therefore \[ \begin{aligned} \int_0^{\pi/2}e^{-R\sin\theta}\,d\theta &\le \int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta\\ &= \frac{\pi}{2R}(1-e^{-R})\\ &\le \frac{\pi}{2R}. \end{aligned} \] The lower bound is immediate from positivity of the integrand.
3. If \(\lambda<1\), then\[ \boxed{ \lim_{R\to\infty} R^\lambda \int_0^{\pi/2}e^{-R\sin\theta}\,d\theta = 0 }. \]
Multiplying step 2 by \(R^\lambda\) gives \[ 0 \le R^\lambda \int_0^{\pi/2}e^{-R\sin\theta}\,d\theta \le \frac\pi2 R^{\lambda-1}. \] Since \(\lambda-1<0\), the right-hand side tends to \(0\). The squeeze theorem gives the displayed limit.
4. Q.E.D.
Step 1 proves part 1, and step 3 proves part 2.