Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to integrate $\;\displaystyle \frac{\log(x)}{1−x}\;$ from $0$ to $1$.

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!

share|improve this question
    
Can anyone tell me how to do it via contour integration please? –  Ansari Jun 16 '11 at 21:02

1 Answer 1

up vote 3 down vote accepted

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 $\log(1-x)$.

share|improve this answer
    
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
1  
Chandru: use \text{d}x instead of \text{dx}. –  Did Jun 16 '11 at 19:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.