新讲 第10章 广义积分 第6题
📝 题目
解 因为
$$ \mathop{\lim }\limits_{{x \rightarrow 1 - }}{\left( 1 - x\right) }^{0}\left| \frac{\ln x}{1 - {x}^{2}}\right| = \mathop{\lim }\limits_{{x \rightarrow 1 - }}\frac{-\ln x}{1 - {x}^{2}} = \mathop{\lim }\limits_{{x \rightarrow 1 - }}\frac{\frac{1}{x}}{2x} = \frac{1}{2}, $$
$$ \mathop{\lim }\limits_{{x \rightarrow 0 + }}{x}^{\frac{1}{2}}\left| \frac{\ln x}{1 - {x}^{2}}\right| = \mathop{\lim }\limits_{{x \rightarrow 0 + }}\frac{-{x}^{\frac{1}{2}}\ln x}{1 - {x}^{2}} = 0, $$
所以积分 $\displaystyle{\int }_{0}^{1}\frac{\ln x}{1 - {x}^{2}}\mathrm{\;d}x}$ 是绝对收敛的.
💡 答案解析
解 因为
$$ \mathop{\lim }\limits_{{x \rightarrow 1 - }}{\left( 1 - x\right) }^{0}\left| \frac{\ln x}{1 - {x}^{2}}\right| = \mathop{\lim }\limits_{{x \rightarrow 1 - }}\frac{-\ln x}{1 - {x}^{2}} = \mathop{\lim }\limits_{{x \rightarrow 1 - }}\frac{\frac{1}{x}}{2x} = \frac{1}{2}, $$
$$ \mathop{\lim }\limits_{{x \rightarrow 0 + }}{x}^{\frac{1}{2}}\left| \frac{\ln x}{1 - {x}^{2}}\right| = \mathop{\lim }\limits_{{x \rightarrow 0 + }}\frac{-{x}^{\frac{1}{2}}\ln x}{1 - {x}^{2}} = 0, $$
所以积分 $\displaystyle{\int }_{0}^{1}\frac{\ln x}{1 - {x}^{2}}\mathrm{\;d}x}$ 是绝对收敛的.