📝 题目
例 5 设 $f\left( x\right) \in C\left\lbrack {0,1}\right\rbrack$ ,证明:
(1) $\displaystyle{\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{\frac{\pi }{2}}\sin {\varphi f}\left( {\sin \theta \sin \varphi }\right) \mathrm{d}\theta \mathrm{d}\varphi = \frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\sin {\varphi f}\left( {\cos \varphi }\right) \mathrm{d}\varphi$ ;
(2)计算重积分 $\displaystyle{I = {\iint }_{0 \leq \frac{\theta }{\varphi } \leq \frac{\pi }{2}}\sin \varphi {\mathrm{e}}^{\sin \theta \sin \varphi }\mathrm{d}\theta \mathrm{d}\varphi}$ .
💡 答案与解析
证 (1) 令 $S$ 为 ${x}^{2} + {y}^{2} + {z}^{2} = 1,x \geq 0,y \geq 0,z \geq 0$ . 由对称性显然可得
$$ {\iint }_{S}f\left( y\right) \mathrm{d}S = {\iint }_{S}f\left( z\right) \mathrm{d}S, $$
而
$$ {\iint }_{S}f\left( y\right) \mathrm{d}S = {\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{\frac{\pi }{2}}f\left( {\sin \theta \sin \varphi }\right) \sin \varphi \mathrm{d}\theta \mathrm{d}\varphi , $$
$$ {\iint }_{S}f\left( z\right) \mathrm{d}S = {\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{\frac{\pi }{2}}f\left( {\cos \varphi }\right) \sin \varphi \mathrm{d}\theta \mathrm{d}\varphi $$
$$ = \frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\sin {\varphi f}\left( {\cos \varphi }\right) \mathrm{d}\varphi , $$
所以
$$ {\int }_{0}^{\frac{\pi }{2}}{\int }_{0}^{\frac{\pi }{2}}\sin {\varphi f}\left( {\sin \theta \sin \varphi }\right) \mathrm{d}\theta \mathrm{d}\varphi = \frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\sin {\varphi f}\left( {\cos \varphi }\right) \mathrm{d}\varphi . $$
(2)利用(1)的结果得
$$ I = {\iint }_{0 \leq \frac{\theta }{\varphi } \leq \frac{\pi }{2}}\sin \varphi {\mathrm{e}}^{\sin \theta \sin \varphi }\mathrm{d}\theta \mathrm{d}\varphi = \frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\sin \varphi {\mathrm{e}}^{\cos \varphi }\mathrm{d}\varphi $$
$$ = - {\left. \frac{\pi }{2}{\mathrm{e}}^{\cos \varphi }\right| }_{0}^{\frac{\pi }{2}} = \frac{\pi }{2}\left( {\mathrm{e} - 1}\right) . $$