Improper Integral Involving Logarithmic Function

Question

The integral I was facing trouble with is the following: $$\int^1_0 \frac{\log x}{1+x}\,dx$$ I substituted, $$x=\tan^2\theta$$ This is what I got: $$2\int^{\frac{\pi}{4}}_0 \frac{\log(\tan \theta)}{\sec^2\theta} \cdot 2\tan\theta\sec^2\theta \, d\theta\\ =4\int^{\frac{\pi}{4}}_0 \tan\theta\cdot\log(\tan\theta)\,d\theta$$ I can't proceed any further. What do I do next? Is this the right approach?

Answer 1

Expanding the series for $\frac{1}{1+x}$, your integral is $$\sum_{k=0}^\infty (-1)^k \int_0^1 x^k\log x \, dx.$$ Using integration by parts, with $U=\log x$ and $dV=x^k\,dx$, we have that $$\int_0^1 x^k \log x \, dx=-\frac{1}{(k+1)^2},$$ and so $$\int_0^1 \frac{\log x}{1+x}=-\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}$$ $$=-\eta(2)=-\frac{\pi^2}{12},$$ where $\eta(s)$ is the Dirichlet Eta Function. Evaluating $\eta(2)$ is exactly as difficult as evaluating $\zeta(2)$ (it's not hard to show that $\eta(2)=\zeta(2)/2$) so there will be no "easy" integration by parts solution to this integral. See this answer for more details on how to evaluate $\zeta(2)$: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$.