一个和取整函数有关的积分

计算积分

\[\int_{0}^{1}x\left[\frac{1}{x}\right]dx=\frac{\pi^{2}}{12}\]

 

解：

\begin{align*}\int_{0}^{1}x\left[\frac{1}{x}\right]dx&=-\sum_{n=1}^{\infty}\int_{\frac{1}{n}}^{\frac{1}{n+1}}x\left[\frac{1}{x}\right]dx \\&=\sum_{n=1}^{\infty}\frac{n}{2}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]\\&=\frac{1}{2} \sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{(n+1)^{2}})\\&=\frac{\pi ^{2}}{12}\end{align*}

 

对于 $s>1$，同样的方法可以计算
$$\zeta(s)=s\int_{1}^{\infty} \frac{[ x ]}{x^{s+1}} dx$$