📝 题目
例 8 试证明
$$ {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \sqrt{\pi },\;{\int }_{0}^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \frac{\sqrt{\pi }}{2}. $$
💡 答案与解析
解 因为 ${\mathrm{e}}^{{x}^{2}} \geq 1 + {x}^{2}$ ,
$$ {\mathrm{e}}^{-{x}^{2}} \leq \frac{1}{1 + {x}^{2}}, $$
所以两积分都收敛. 又, 显然有
$$ {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = 2{\int }_{0}^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x. $$
所以只需计算其中任何一个就可以了. 下面, 我们来计算第一个积分. 考察
$$ I\left( a\right) = {\int }_{-a}^{a}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x. $$
我们有
$$ {\left( I\left( a\right) \right) }^{2} = {\int }_{-a}^{a}{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x{\int }_{-a}^{a}{\mathrm{e}}^{-{y}^{2}}\mathrm{\;d}y $$
$$ = {\iint }_{\left\lbrack {-a,a}\right\rbrack \times \left\lbrack {-a,a}\right\rbrack }{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}\left( {x,y}\right) . $$
显然有 (参看图 13-15):
$$ {\iint }_{{x}^{2} + {y}^{2} \leq {a}^{2}}{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}\left( {x,y}\right) $$
$$ \leq {\left( I\left( a\right) \right) }^{2} \leq {\iint }_{{x}^{2} + {y}^{2} \leq 2{a}^{2}}{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}\left( {x,y}\right) . $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/069.jpg} \end{center} \hspace*{3em}
图 13-15
换极坐标计算可得
$$ {\iint }_{{x}^{2} + {y}^{2} \leq {a}^{2}}{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}\left( {x,y}\right) = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{a}{\mathrm{e}}^{-{r}^{2}}r\mathrm{\;d}r $$
$$ = \pi \left( {1 - {\mathrm{e}}^{-{a}^{2}}}\right) . $$
同样可得
$$ {\iint }_{{x}^{2} + {y}^{2} \leq 2{a}^{2}}{\mathrm{e}}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}\left( {x,y}\right) = \pi \left( {1 - {\mathrm{e}}^{-2{a}^{2}}}\right) . $$
于是有
$$ \pi \left( {1 - {\mathrm{e}}^{-{a}^{2}}}\right) \leq {\left( I\left( a\right) \right) }^{2} \leq \pi \left( {1 - {\mathrm{e}}^{-2{a}^{2}}}\right) . $$
由此可得
$$ \mathop{\lim }\limits_{{a \rightarrow + \infty }}{\left( I\left( a\right) \right) }^{2} = \pi , $$
$$ {\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x = \mathop{\lim }\limits_{{a \rightarrow + \infty }}I\left( a\right) $$
$$ = \sqrt{\pi }\text{ . } $$