设 k>0, a>0. 证明:
(1). \dps∫∞asin2nπx\rdxxk 收敛;
(2). \dps\vsmn1n∫∞asin2nπx\rdxxk 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\sev1n∫∞asin2nπx\rdxxk≤1akn2π.\eex 故结论成立.