方企勤 第三章 一元函数积分学 第9题
📝 题目
例 9 求证: (1) 极限 $\mathop{\lim }\limits_{{b \rightarrow 1}}{\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x\left( {0 < b < 1}\right)$ 存在;
(2) $\displaystyle{\mathop{\lim }\limits_{{b \rightarrow 1}}{\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x > 1}$ .
💡 答案解析
证(1)考虑函数 $f\left( b\right) \overset{\text{ 定义 }}{ = }{\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x\left( {0 < b < 1}\right)$ . 当 $0 < {b}_{1}$ $< {b}_{2} < 1$ 时,
$$ f\left( {b}_{2}\right) = {\int }_{0}^{{b}_{2}}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x = {\int }_{0}^{{b}_{1}}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x + {\int }_{{b}_{1}}^{{b}_{2}}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x $$
$$ \geq {\int }_{0}^{{b}_{1}}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x = f\left( {b}_{1}\right) , $$
即 $f\left( b\right)$ 在(0,1)上单调递增. 又
$$ f\left( b\right) \leq {\int }_{0}^{b}\frac{1}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x = \arcsin b \leq \frac{\pi }{2}\;\left( {0 < b < 1}\right) , $$
即 $f\left( b\right)$ 在(0,1)上有上界. 因此 $\mathop{\lim }\limits_{{b \rightarrow 1 - 0}}f\left( b\right)$ 存在.
(2)证法 1 因为 $\cos x = 1 - 2{\sin }^{2}\frac{x}{2} \geq 1 - \frac{{x}^{2}}{2} \geq \sqrt{1 - {x}^{2}}(0 \leq x$ $\leq b < 1$ ),所以
$$ {\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x \geq {\int }_{0}^{b}\mathrm{\;d}x = b\left( {0 < b < 1}\right) $$
$$ \Rightarrow \mathop{\lim }\limits_{{b \rightarrow 1}}{\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x \geq \mathop{\lim }\limits_{{b \rightarrow 1}}b = 1. $$
证法 2 因为 $\cos x \geq 1 - \frac{{x}^{2}}{2} = \frac{1 - {x}^{2}}{2} + \frac{1}{2}\left( {0 \leq x \leq b < 1}\right)$ ,所以
$$ {\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x \geq \frac{1}{2}{\int }_{0}^{b}\sqrt{1 - {x}^{2}}\mathrm{\;d}x + \frac{1}{2}{\int }_{0}^{b}\frac{1}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x $$
$$ = \frac{1}{4}b\sqrt{1 - {b}^{2}} + \frac{3}{4}\arcsin b. $$
因此
$$ \mathop{\lim }\limits_{{b \rightarrow 1}}{\int }_{0}^{b}\frac{\cos x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x \geq \frac{3}{4}\arcsin 1 = \frac{3\pi }{8} > 1. $$