Definite integral for $\zeta(3)$

Question

By making use of Mathematica, I detected the following integral expression for zeta(3): $$\int_0^1\frac{\log(x)}{1+x}\log\left(\frac{2+x}{1+x}\right)dx=-\frac5{12}\zeta(3).$$ Any proof of it would be gratefully appreciated.

Answer 1

The proof is simple : the antiderivative exists (ask Mathematica).

The result is not so complicated (beside logarithms, there are just a few polylogarithm functions).

Use the bounds and simplify to get $$\text{Li}_3(-3)-2 \text{Li}_3(3)+i \pi \left(-\text{Li}_2(-3)+2 \text{Li}_2(3)+\log ^2(3)\right)-$$ $$\text{Li}_2(-3) \log (3)+\text{Li}_2(3) \log (9)+\frac{7 \zeta (3)}{4}-\frac{2 i \pi ^3}{3}-\pi ^2 \log (9)$$

This simplifies nicely (just a litle work - I did it by hand).

If you do not know what it is, ask for the numerical value $$I=-0.5008570429831642855832242$$ and the inverse symbolic calculator will tell you that $I$ satisfies the following Z-linear combination : $$ 12 I + 5\, \zeta(3)=0$$

You could have fun with $$J(a,b,c,d)=\int \frac{\log (x+a) \log \left(\frac{x+b}{x+c}\right)}{x+d}\,dx$$ which has a closed form antiderivative and get a bunch of identities.