📝 题目
例 16 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上可微, ${f}^{\prime }\left( x\right)$ 单调上升并满足 $\left| {{f}^{\prime }\left( x\right) }\right|$ $\geq m > 0$ . 求证:
$$ \left| {{\int }_{a}^{b}\cos f\left( x\right) \mathrm{d}x}\right| \leq \frac{2}{m}. $$
💡 答案与解析
证法 1 根据达布定理, ${f}^{\prime }\left( x\right)$ 的单调性条件蕴含 ${f}^{\prime }\left( x\right)$ 的连续性,又因为 $\left| {{f}^{\prime }\left( x\right) }\right| \geq m > 0$ ,所以 ${f}^{\prime }\left( x\right)$ 不变号,不妨设 ${f}^{\prime }\left( x\right) > 0$ ,则有 $f\left( x\right)$ 严格单调增加. 根据积分第二中值定理,
$$ {\int }_{a}^{b}\cos f\left( x\right) \mathrm{d}x = {\int }_{a}^{b}\frac{{f}^{\prime }\left( x\right) \cos f\left( x\right) }{{f}^{\prime }\left( x\right) }\mathrm{d}x = \frac{1}{{f}^{\prime }\left( a\right) }{\int }_{a}^{\xi }\cos f\left( x\right) \mathrm{d}f\left( x\right) , $$
从而
$$ \left| {{\int }_{a}^{b}\cos f\left( x\right) \mathrm{d}x}\right| \leq \frac{1}{{f}^{\prime }\left( a\right) }\left| {\sin f\left( \xi \right) - \sin f\left( a\right) }\right| \leq \frac{2}{m}. $$
证法 2 不妨设 ${f}^{\prime }\left( x\right) > 0$ (否则考虑 $- f\left( x\right)$ ),则有 $f\left( x\right)$ 严格单调增加. 设 $f$ 的反函数为 $g : \left\lbrack {f\left( a\right) ,f\left( b\right) }\right\rbrack \rightarrow \left\lbrack {a,b}\right\rbrack$ ,则
$$ \frac{\mathrm{d}g}{\mathrm{\;d}y} = \frac{1}{{f}^{\prime }\left( x\right) } \leq \frac{1}{m}\;\left( {\forall y \in \left\lbrack {f\left( a\right) ,f\left( b\right) }\right\rbrack }\right) . $$
对要估计的积分作反函数换元, 再利用积分第二中值定理, 得到
$$ \left| {{\int }_{a}^{b}\cos f\left( x\right) \mathrm{d}x}\right| = \left| {{\int }_{f\left( a\right) }^{f\left( b\right) }\cos y \cdot {g}^{\prime }\left( y\right) \mathrm{d}y}\right| = \left| {{g}^{\prime }\left( {f\left( a\right) }\right) {\int }_{f\left( a\right) }^{\xi }\cos y\mathrm{\;d}y}\right| $$
$$ \leq \frac{1}{m}\left| {\sin \xi - \sin f\left( a\right) }\right| \leq \frac{2}{m}. $$