$\ln(1-x)$泰勒级数

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Brilliant wiki - Interchanging the summation and integral sign

First, use the taylor series of $\ln(1-x)$:
\[\int_0^1 \dfrac{\ln(1-x)}{x} dx=\int_0^1 \sum_{n=1}^\infty \dfrac{x^{n-1}}{n} dx.\]

积分的结果写的是$\frac{π^2}6$, 但是当$0<x<1$时, $\ln(1-x)<0$, 所以integrand是负的. 离谱啊.
看了一下,应该是$\displaystyle\frac{\ln(1-x)}x=-\sum_{n=1}^∞\frac{x^{n-1}}n$吧, 缺少负号?
所以, 答案应该是\[\int_0^1 \dfrac{\ln(1-x)}{x} dx=-\frac{π^2}6\]