How to evaluate $\int_{0}^{1}\frac{\log x}{1+x} dx$

Question

Today When I was computing measure theoretic entropy of Gauss map I encountered this integral. Then I check asked Questions but I couldn't find same question how to evaluate:
$$\int_{0}^{1}\frac{\log x}{1+x} dx$$

Thanks for any hint.

Answer 1

We perform the usual series expansion about $x = 0$ of $(1+x)^{-1}$ to obtain $$I = \int_{x=0}^1 \sum_{k=0}^\infty (-1)^k x^k \log x \, dx.$$ Interchanging the order of integration and summation gives $$I = \sum_{k=0}^\infty (-1)^k \int_{x=0}^1 x^k \log x \, dx,$$ and observing that $$x^k \log x = \frac{d}{dk} [x^k],$$ we have $$I = \sum_{k=0}^\infty (-1)^k \frac{d}{dk} \left[\int_{x=0}^1 x^k \, dx \right] = \sum_{k=0}^\infty (-1)^k \frac{d}{dk} \left[\frac{1}{k+1}\right] = \sum_{k=0}^\infty \frac{(-1)^{k+1}}{(k+1)^2}.$$ This of course is related to $$\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}:$$ a little algebra easily gives $$I = -\frac{\pi^2}{12}.$$