📝 题目
例 20 假设函数 $f\left( x\right)$ 在闭区间 $\left\lbrack {0,b}\right\rbrack$ 上单调增加,函数 $g\left( x\right)$ 使得广义积分 $\displaystyle{\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x$ 收敛. 求证:
$$ \mathop{\lim }\limits_{{p \rightarrow + \infty }}{\int }_{0}^{b}f\left( x\right) \frac{g\left( {px}\right) }{x}\mathrm{\;d}x = f\left( {0}^{ + }\right) {\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x. $$
💡 答案与解析
证 分解
$$ {\int }_{0}^{b}f\left( x\right) \frac{g\left( {px}\right) }{x}\mathrm{\;d}x = f\left( {0}^{ + }\right) {\int }_{0}^{b}\frac{g\left( {px}\right) }{x}\mathrm{\;d}x $$
$$ + {\int }_{0}^{b}\left\lbrack {f\left( x\right) - f\left( {0}^{ + }\right) }\right\rbrack \frac{g\left( {px}\right) }{x}\mathrm{\;d}x. $$
因为
$$ \mathop{\lim }\limits_{{p \rightarrow + \infty }}f\left( {0}^{ + }\right) {\int }_{0}^{b}\frac{g\left( {px}\right) }{x}\mathrm{\;d}x = \mathop{\lim }\limits_{{p \rightarrow + \infty }}f\left( {0}^{ + }\right) {\int }_{0}^{bp}\frac{g\left( t\right) }{t}\mathrm{\;d}t $$
$$ = f\left( {0}^{ + }\right) {\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x, $$
所以为了结论成立, 只要证明
$$ \mathop{\lim }\limits_{{p \rightarrow + \infty }}{\int }_{0}^{b}\left\lbrack {f\left( x\right) - f\left( {0}^{ + }\right) }\right\rbrack \frac{g\left( {px}\right) }{x}\mathrm{\;d}x = 0. \tag{7.36} $$
由 $\displaystyle{\int }_{0}^{+\infty }\frac{g\left( x\right) }{x}\mathrm{\;d}x$ 收敛,可设 $\left| {{\int }_{0}^{x}\frac{g\left( x\right) }{x}\mathrm{\;d}x}\right| \leq M\left( {\forall X > 0}\right)$ . 又 $\forall \varepsilon > 0$ , $\exists 0 < \delta < b$ ,使得 $0 \leq f\left( x\right) - f\left( {0}^{ + }\right) < \varepsilon \left( {0 \leq x \leq \delta }\right)$ . 于是由积分第二中值定理, $\exists 0 < \xi < \delta$ 及 $\delta < \eta < b$ ,使得
$$ \left| {{\int }_{0}^{\delta }\left\lbrack {f\left( x\right) - f\left( {0}^{ + }\right) }\right\rbrack \frac{g\left( {px}\right) }{x}\mathrm{\;d}x}\right| $$
$$ = \left| {f\left( \delta \right) - f\left( {0}^{ + }\right) }\right| \left| {{\int }_{\xi }^{\delta }\frac{g\left( {px}\right) }{x}\mathrm{\;d}x}\right| $$
$$ = \left| {f\left( \delta \right) - f\left( {0}^{ + }\right) }\right| \left| {{\int }_{p\xi }^{p\delta }\frac{g\left( x\right) }{x}\mathrm{\;d}x}\right| < {2M\varepsilon }, \tag{7.37} $$
及
$$ \left| {{\int }_{\delta }^{b}\left\lbrack {f\left( x\right) - f\left( {0}^{ + }\right) }\right\rbrack \frac{g\left( {px}\right) }{x}\mathrm{\;d}x}\right| $$
$$ = \left| {f\left( b\right) - f\left( {0}^{ + }\right) }\right| \left| {{\int }_{\eta }^{b}\frac{g\left( {px}\right) }{x}\mathrm{\;d}x}\right| $$
$$ = \left| {f\left( b\right) - f\left( {0}^{ + }\right) }\right| \left| {{\int }_{p\eta }^{pb}\frac{g\left( x\right) }{x}\mathrm{\;d}x}\right| . \tag{7.38} $$
又由柯西收敛准则, $\exists P > 0$ ,使得
$$ \left| {{\int }_{p\eta }^{pb}\frac{g\left( x\right) }{x}\mathrm{\;d}x}\right| < \varepsilon \;\left( {p > P}\right) . \tag{7.39} $$
联立 (7.37) $\sim \left( {7.39}\right)$ 式,当 $p > P$ 时,有
$$ \left| {{\int }_{0}^{b}\left\lbrack {f\left( x\right) - f\left( {0}^{ + }\right) }\right\rbrack \frac{g\left( {px}\right) }{x}\mathrm{\;d}x}\right| < \left( {{2M} + \left| {f\left( b\right) - f\left( {0}^{ + }\right) }\right| }\right) \varepsilon . $$
由 $\varepsilon$ 的任意性,(7.36)式得证.