数学

一个定积分

建议将帖子移至题目互答

$S=\int_0^1\frac{x\ln{(x^{2n-1})}}{1+x^2} dx=(2n-1)\int_0^1\frac{x\ln{x}}{1+x^2} dx$

$则\frac{S}{2n-1}=\frac{1}{4}\int_0^1\frac{\ln{u}}{1+u} du$

$则\frac{4S}{2n-1}=\int_0^1 \ln{u}\cdot\sum_{i=0}^{\infty}(-1)^i u^i du=\sum_{i=0}^{\infty}(-1)^i\int_0^1 u^i\ln{u} du$

$~~~~~~~=-\sum_{i=0}^{\infty}(-1)^i \frac{1}{(i+1)^2}=-\frac{\pi^2}{12}$

$所以说S=-\frac{\pi^2(2n-1)}{48}$