方企勤 第六章 多元函数积分学 第4题
📝 题目
例 4 假设可测图形 $\Omega$ 关于超平面 ${x}_{m} = 0$ 对称,即若 $\left( {{x}_{1},\cdots }\right.$ , $\left. {{x}_{m - 1},{x}_{m}}\right) \in \Omega$ ,必有 $\left( {{x}_{1},\cdots ,{x}_{m - 1}, - {x}_{m}}\right) \in \Omega$ . 又函数 $f\left( {{x}_{1},\cdots ,{x}_{m}}\right)$ 关于 ${x}_{m}$ 是奇函数,且 $f \in R\left( \Omega \right)$ ,则 $\displaystyle{\int }_{\Omega }f\left( \mathbf{x}\right) \mathrm{d}V = 0$ .
💡 答案解析
证 设 $\Omega$ 包含在正方形 $\left\{ {\left( {{x}_{1},\cdots ,{x}_{m}}\right) \left| \right| {x}_{i} \mid \leq M\left( {i = 1,\cdots ,m}\right) }\right\}$ 内. 考虑集合
$$ \Omega \cap \left\{ {\left( {{x}_{1},\cdots ,{x}_{m - 1},{x}_{m}}\right) \left| {{x}_{m} \geq 0,}\right| {x}_{i} \mid \leq M\left( {i = 1,\cdots ,m - 1}\right) }\right\} . $$
它们是两个可测图形的交集,故为可测图形. 对此可测图形作一剖分 $\left\{ {{\Omega }_{1},\cdots ,{\Omega }_{n}}\right\}$ ,并取 ${\xi }_{i} \in {\Omega }_{i}\left( {i = 1,\cdots ,n}\right)$ .
令 ${\Omega }_{n + i} = \left\{ {\left( {{x}_{1},\cdots ,{x}_{m - 1},{x}_{m}}\right) \mid \left( {{x}_{1},\cdots ,{x}_{m - 1}, - {x}_{m}}\right) \in {\Omega }_{i}}\right\} (i = 1$ , $\cdots ,n)$ ,显然它是可测图形. 再令 ${\xi }_{n + i}$ 为 ${\xi }_{i}$ 分量中第 $m$ 个分量变号所得的点,显然 ${\xi }_{n + i} \in {\Omega }_{n + i},f\left( {\xi }_{n + i}\right) = - f\left( {\xi }_{i}\right) \left( {i = 1,\cdots ,n}\right)$ .
这样我们得到 $\Omega$ 的一个剖分 $\Delta = \left\{ {{\Omega }_{1},\cdots ,{\Omega }_{n},{\Omega }_{n + 1},\cdots ,{\Omega }_{2n}}\right\}$ ,相应的黎曼和为
$$ \mathop{\sum }\limits_{{i = 1}}^{n}\left\lbrack {f\left( {\xi }_{i}\right) V\left( {\Omega }_{i}\right) + f\left( {\xi }_{n + i}\right) V\left( {\Omega }_{n + i}\right) }\right\rbrack $$
$$ = \mathop{\sum }\limits_{{i = 1}}^{n}\left\lbrack {f\left( {\xi }_{i}\right) + f\left( {\xi }_{n + i}\right) }\right\rbrack V\left( {\Omega }_{i}\right) = 0. $$
令 $\parallel \Delta \parallel \rightarrow 0$ ,即得 $\displaystyle{\int }_{\Omega }f\left( \mathbf{x}\right) \mathrm{d}V = 0$ .