📝 题目
例 2 设 $f\left( x\right) = {\mathrm{e}}^{{x}^{2}}{\int }_{x}^{+\infty }{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t$ ,求证: $f\left( x\right) \leq \sqrt{\pi }/2\left( {x \geq 0}\right)$ .
💡 答案与解析
证法 1 令 $u = t - x$ ,则当 $x \geq 0$ 时,
$$ f\left( x\right) = {\mathrm{e}}^{{x}^{2}}{\int }_{0}^{\infty }{\mathrm{e}}^{-{\left( u + x\right) }^{2}}\mathrm{\;d}u = {\int }_{0}^{\infty }{\mathrm{e}}^{-{u}^{2}} \cdot {\mathrm{e}}^{-{2ux}}\mathrm{\;d}u $$
$$ \leq {\int }_{0}^{\infty }{\mathrm{e}}^{-{u}^{2}}\mathrm{\;d}u = \frac{\sqrt{\pi }}{2}. $$
证法 2 令 $u = t - x,f\left( x\right) = {\int }_{0}^{\infty }{\mathrm{e}}^{-{u}^{2} - {2ux}}\mathrm{\;d}u$ ,利用广义参变量积分求导定理, 得
$$ {f}^{\prime }\left( x\right) = {\int }_{0}^{\infty }\left( {-{2u}}\right) {\mathrm{e}}^{-{u}^{2} - {2ux}}\mathrm{\;d}u < 0, $$
所以函数 $f\left( x\right)$ 在实轴严格递减. 特别当 $x \geq 0$ 时,有
$$ f\left( x\right) \leq f\left( 0\right) = \frac{\sqrt{\pi }}{2}. $$
证法 3 由
$$ {\left( {\int }_{x}^{\infty }{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t\right) }^{2} = {\int }_{x}^{\infty }{\int }_{x}^{\infty }{\mathrm{e}}^{-\left( {{u}^{2} + {v}^{2}}\right) }\mathrm{d}u\mathrm{\;d}v \leq {\iint }_{\begin{matrix} {{u}^{2} + {v}^{2} \geq 2{x}^{2}} \\ {u \geq 0,v \geq 0} \end{matrix}}{\mathrm{e}}^{-\left( {{u}^{2} + {v}^{2}}\right) }\mathrm{d}u\mathrm{\;d}v $$
$$ = {\int }_{\sqrt{2}x}^{\infty }r{\mathrm{e}}^{-{r}^{2}}\mathrm{\;d}r{\int }_{0}^{\frac{\pi }{2}}\mathrm{\;d}\theta = \frac{\pi }{4}{\mathrm{e}}^{-2{x}^{2}}\;\left( {x \geq 0}\right) , $$
两边开平方,即得 $f\left( x\right) \leq \sqrt{\pi }/2$ .