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I need to evaluate $\int_0^1 \frac{\log(x)}{1−x}\;dx$

I know I need to use contour integration and I read the chapter in Churchill but I'm still running into issues doing it properly.

I also know the answer is $\displaystyle\;\; −\frac{\pi^2}{6},$ but I'd like to know how to arrive at that answer.

Thanks!

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    Can anyone tell me how to do it via contour integration please?2011-06-16

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We know that $\int\limits_{0}^{a} f(x) \: \text{d}x = \int\limits_{0}^{a} f(a-x) \: \text{dx}$ Using this fact $\int\limits_{0}^{1} \frac{\log(x)}{1-x} \ \mathrm{d}x = \int\limits_{0}^{1} \frac{\log(1-x)}{x} \ \mathrm{d}x$ and then use the expansion of $\log(1-x)$.

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    @Kahen: Ok, now lets, remove all of our comments, as I feel it shouldn't be here. I shall remove this comment in some time as well.2011-06-17