📝 题目
例 32 设 ${f}^{\prime }\left( x\right) \in C\left\lbrack {0,1}\right\rbrack$ ,求证:
$$ {\int }_{0}^{1}{x}^{n}f\left( x\right) \mathrm{d}x = \frac{f\left( 1\right) }{n} + o\left( \frac{1}{n}\right) \;\left( {n \rightarrow \infty }\right) . $$
💡 答案与解析
证 用分部积分法.
$$ {\int }_{0}^{1}{x}^{n}f\left( x\right) \mathrm{d}x = \frac{1}{n + 1}{\int }_{0}^{1}f\left( x\right) \mathrm{d}\left( {x}^{n + 1}\right) $$
$$ = \frac{f\left( 1\right) }{n + 1} - \frac{1}{n + 1}{\int }_{0}^{1}{x}^{n + 1}{f}^{\prime }\left( x\right) \mathrm{d}x $$
$$ = \frac{f\left( 1\right) }{n} + \frac{f\left( 1\right) }{n + 1} - \frac{f\left( 1\right) }{n} - \frac{1}{n + 1}{\int }_{0}^{1}{x}^{n + 1}{f}^{\prime }\left( x\right) \mathrm{d}x. $$
又
$$ \frac{f\left( 1\right) }{n + 1} - \frac{f\left( 1\right) }{n} = - \frac{f\left( 1\right) }{n\left( {n + 1}\right) } = o\left( \frac{1}{n}\right) $$
及 ${f}^{\prime }\left( x\right) \in C\left\lbrack {0,1}\right\rbrack$ ,推知存在 ${M}_{1} > 0$ ,使得 $\left| {{f}^{\prime }\left( x\right) }\right| \leq {M}_{1}(\forall x \in$ $\left\lbrack {0,1}\right\rbrack )$ ,故有
$$ \left| {\frac{1}{n + 1}{\int }_{0}^{1}{x}^{n + 1}{f}^{\prime }\left( x\right) \mathrm{d}x}\right| \leq \frac{{M}_{1}}{\left( {n + 1}\right) \left( {n + 2}\right) } $$
$$ \Rightarrow \frac{1}{n + 1}{\int }_{0}^{1}{x}^{n + 1}{f}^{\prime }\left( x\right) \mathrm{d}x = o\left( \frac{1}{n}\right) . $$
因此
$$ {\int }_{0}^{1}{x}^{n}f\left( x\right) \mathrm{d}x = \frac{f\left( 1\right) }{n} + o\left( \frac{1}{n}\right) \;\left( {n \rightarrow \infty }\right) . $$