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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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    Thank you! I should have seen that. OK but then don't I need to prove convergence? Integral of sum is sum of integrals? Also I may have a bunch of similar integrals, so knowing the contour way of doing it would be a big help2011-06-16
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    Alright thanks very much! :)2011-06-16
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    (minor edit - not to nitpick, but in the final integral it should be simply $x$ not $-x$)2011-06-16
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    Chandru: use \text{d}x instead of \text{dx}.2011-06-16
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    You can use the Monotone Convergence Theorem for the convergence. I don't see a way of doing it using contours.2011-06-17
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    I deleted my previous comment because the example was terrible. But my point was that you should use `\mathrm` for when you want roman typeface in mathmode and save `\text` for when you actually want text2011-06-17
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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