📝 题目
例 33 设 $f\left( x\right)$ 在 $\left\lbrack {0,{2\pi }}\right\rbrack$ 上单调. 求证:
$$ \mathop{\lim }\limits_{{\lambda \rightarrow \infty }}{\int }_{0}^{2\pi }f\left( x\right) \sin {\lambda x}\mathrm{\;d}x = 0. $$
💡 答案与解析
证 由 $f\left( x\right)$ 在 $\left\lbrack {0,{2\pi }}\right\rbrack$ 上单调,应用积分第二中值定理,有
$$ {\int }_{0}^{2\pi }f\left( x\right) \sin {\lambda x}\mathrm{\;d}x = f\left( 0\right) {\int }_{0}^{\xi }\sin {\lambda x}\mathrm{\;d}x + f\left( {2\pi }\right) {\int }_{\xi }^{2\pi }\sin {\lambda x}\mathrm{\;d}x $$
$$ = f\left( 0\right) \frac{1 - \cos {\lambda \xi }}{\lambda } - f\left( {2\pi }\right) \frac{1 - \cos {\lambda \xi }}{\lambda }, $$
因此
$$ \left| {{\int }_{0}^{2\pi }f\left( x\right) \sin {\lambda x}\mathrm{\;d}x}\right| \leq \max \{ \left| {f\left( 0\right) }\right| ,\left| {f\left( {2\pi }\right) }\right| \} \frac{2}{\lambda } $$
$$ \Rightarrow \mathop{\lim }\limits_{{\lambda \rightarrow \infty }}{\int }_{0}^{2\pi }f\left( x\right) \sin {\lambda x}\mathrm{\;d}x = 0. $$