微积分(A)随缘一题[31]

若 \(f(x) \in C[0,\pi]\)，求证：\(\lim_{n \to \infty}\int_0^\pi f(x)|\sin nx|dx=\frac{2}{\pi}\int_0^\pi f(x)dx\)

\[\begin{aligned} \int_0^\pi f(x) |\sin nx|dx \xlongequal{x_i=\frac{i\pi}{n}} &\sum_{i=1}^{n}\int_{x_{i-1}}^{x_i}f(x)|\sin nx|dx \\ =&\sum_{i=1}^{n}f(\zeta_i)\int_{x_{i-1}}^{x_i}|\sin nx|dx \\ =&\sum_{i=1}^{n}f(\zeta_i)\int_{(i-1)\pi}^{i\pi}|\sin x|d\frac{x}{n} \\ =&\sum_{i=1}^{n}f(\zeta_i) \frac{2}{n} \\ =&\frac{2}{\pi}\sum_{i=1}^{n}\frac{\pi}{n}f(\zeta_i) \\ =&\frac{2}{\pi}\int_0^{\pi}f(x)dx \end{aligned} \]