📝 题目
例 31 利用级数收敛性证明无穷积分 $\displaystyle{\int }_{0}^{+\infty }{\left( -1\right) }^{\left\lbrack {x}^{2}\right\rbrack }$ 收敛,其中 $\left\lbrack {x}^{2}\right\rbrack$ 表示不超过 ${x}^{2}$ 的最大整数.
💡 答案与解析
证 因为当 $\sqrt{n} \leq x < \sqrt{n + 1}$ 时, $\left\lbrack {x}^{2}\right\rbrack = n\left( {n = 0,1,2,\cdots }\right)$ ,所以
$$ {\int }_{0}^{+\infty }{\left( -1\right) }^{\left\lbrack {x}^{2}\right\rbrack } = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{\sqrt{n}}^{\sqrt{n + 1}}{\left( -1\right) }^{\left\lbrack {x}^{2}\right\rbrack }\mathrm{d}x $$
$$ = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\left( {\sqrt{n + 1} - \sqrt{n}}\right) . $$
于是由于上式右端的无穷级数收敛, 得知左端的无穷积分收敛.