📝 题目
例 3 求证: $f\left( x\right) = x{\mathrm{e}}^{-{x}^{2}}{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t$ 在 $\left( {-\infty ,\infty }\right)$ 上有界.
💡 答案与解析
证法 1 因为 $f\left( {-x}\right) = f\left( x\right)$ ,即 $f\left( x\right)$ 是偶函数,所以只需证明 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上有界.
用洛必达法则:
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t}{\frac{1}{x}{\mathrm{e}}^{{x}^{2}}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\mathrm{e}}^{{x}^{2}}}{2{\mathrm{e}}^{{x}^{2}} - \frac{1}{{x}^{2}}{\mathrm{e}}^{{x}^{2}}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{2 - \frac{1}{{x}^{2}}} = \frac{1}{2}. $$
因此 $\exists A > 0$ ,使得 $0 < f\left( x\right) < 1\left( {\forall x > A}\right)$ . 又由 $f\left( x\right)$ 在 $\left\lbrack {0,A}\right\rbrack$ 上连续,故 $\exists {M}_{1} > 0$ ,使得 $0 \leq f\left( x\right) \leq {M}_{1}\left( {\forall x \in \left\lbrack {0,A}\right\rbrack }\right)$ . 取
$$ M = \max \left( {1,{M}_{1}}\right) , $$
则有 $0 \leq f\left( x\right) \leq M\left( {\forall x \in \lbrack 0, + \infty }\right)$ ).
证法 2 由证法 1 知只需证 $f\left( x\right)$ 在 $\lbrack 0, + \infty )$ 上有界,对 $\forall x >$ 1,根据柯西中值定理, $\exists \xi \in \left( {1,x}\right)$ ,使得
$$ \frac{{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t - {\int }_{0}^{1}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t}{\frac{1}{x}{\mathrm{e}}^{{x}^{2}} - \mathrm{e}} = \frac{{\mathrm{e}}^{{\xi }^{2}}}{-\frac{1}{{\xi }^{2}}{\mathrm{e}}^{{\xi }^{2}} + 2{\mathrm{e}}^{{\xi }^{2}}} = \frac{1}{2 - \frac{1}{{\xi }^{2}}} \leq 1. \tag{7.15} $$
又当 $x > 1$ 时, ${\mathrm{e}}^{{x}^{2} - 1} > {x}^{2} > x$ ,即得 $\frac{1}{x}{\mathrm{e}}^{{x}^{2}} > \mathrm{e}$ . 于是由 (7.15) 式得
$$ {\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t \leq \frac{1}{x}{\mathrm{e}}^{{x}^{2}} - \mathrm{e} + {\int }_{0}^{1}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t \leq \frac{1}{x}{\mathrm{e}}^{{x}^{2}}, $$
即得 $0 \leq f\left( x\right) \leq 1\left( {\forall x > 1}\right)$ . 又当 $x \in \left\lbrack {0,1}\right\rbrack$ 时,
$$ 0 \leq f\left( x\right) \leq {\int }_{0}^{1}{\mathrm{e}}^{t}\mathrm{\;d}t \cdot \mathop{\max }\limits_{{0 \leq x \leq 1}}x{\mathrm{e}}^{-{x}^{2}} = \frac{\mathrm{e} - 1}{\sqrt{2\mathrm{e}}} < \sqrt{\frac{2}{\mathrm{e}}} < 1. $$
故有
$$ 0 \leq f\left( x\right) \leq 1\;\left( {\forall x \in \lbrack 0, + \infty }\right) ). $$
证法 3 利用幂级数逐项积分.
$$ {\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t = {\int }_{0}^{x}\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{t}^{2n}}{n!}\mathrm{d}t = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{x}^{{2n} + 1}}{\left( {{2n} + 1}\right) n!}, $$
即得
$$ x{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{x}^{{2n} + 2}}{\left( {{2n} + 1}\right) n!}; $$
又
$$ {\mathrm{e}}^{{x}^{2}} = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{x}^{2n}}{n!} = 1 + \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{x}^{{2n} + 2}}{\left( {n + 1}\right) !}. $$
由于 $\left( {{2n} + 1}\right) n! \geq \left( {n + 1}\right) !\left( {n \geq 0}\right)$ ,得到
$$ 0 \leq x{\int }_{0}^{x}{\mathrm{e}}^{{t}^{2}}\mathrm{\;d}t < {\mathrm{e}}^{{x}^{2}}\;\left( {\forall x \in \mathbf{R}}\right) , $$
即
$$ 0 \leq f\left( x\right) < 1\;\left( {\forall x \in \mathbf{R}}\right) . $$