证明: 当 $\lm<1$ 时, $\dps{\lim_{R\to+\infty} R^\lm\int_0^{\pi/2} e^{-R\sin\tt}\rd \tt=0}$.

证明: 由 $$\beex \bea 0\leq R^\lm\int_0^{\pi/2} e^{-R\sin\tt}\rd \tt &\leq R^\lm \int_0^{\pi/2} e^{-R \frac{2}{\pi}\tt}\rd \tt\\ &=R^\lm \sex{-\frac{\pi}{2R}e^{-R\frac{2}{\pi}\tt}}^{\frac{\pi}{2}}_0\\ &=R^\lm \cfrac{\pi}{2R} \sex{1-e^{-R}}\\ &\leq \cfrac{\pi}{2} R^{\lm-1} \eea \eeex$$ 即知结论.