[裴礼文数学分析中的典型问题与方法习题参考解答]5.1.28

设 $k>0$, $a>0$. 证明:

(1). $\dps{\int_a^\infty \frac{\sin 2n\pi x\rd x}{x^k}}$ 收敛;

(2). $\dps{\vsm{n}\frac{1}{n}\int_a^\infty \frac{\sin 2n\pi x\rd x}{x^k}}$ 收敛.

证明:

(1). 由 Dirichlet 判别法即知结论成立.

(2). 由积分第二中值定理, $$\beex \bea \sev{\frac{1}{n}\int_a^A \frac{\sin 2n\pi x\rd x}{x^k}} &=\sev{\frac{1}{n}\sez{\frac{1}{a^k}\int_a^\xi \sin 2n\pi x\rd x +\frac{1}{A^k}\int_\xi^A \sin 2n\pi x\rd x}}\\ &\leq \frac{1}{n}\sex{\frac{1}{a^k}\frac{1}{n\pi}+\frac{1}{A^k}\frac{1}{n\pi}}\\ &=\sex{\frac{1}{a^k}+\frac{1}{A^k}}\frac{1}{n^2\pi}, \eea \eeex$$ 而 $$\bex \sev{\frac{1}{n}\int_a^\infty \frac{\sin 2n\pi x\rd x}{x^k}} \leq \frac{1}{a^k n^2\pi}. \eex$$ 故结论成立.