# simple(?) contour integral

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? – Ansari Jun 16 '11 at 21:02

## 1 Answer

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 help – Ansari Jun 16 '11 at 18:13
Alright thanks very much! :) – Ansari Jun 16 '11 at 18:27
(minor edit - not to nitpick, but in the final integral it should be simply $x$ not $-x$) – Ansari Jun 16 '11 at 18:45
thanks for helping out – user9413 Jun 16 '11 at 18:54
Chandru: use \text{d}x instead of \text{dx}. – Did Jun 16 '11 at 19:37