Integrals of the kind $$\int_0^1 \frac{\log(1-z)\log(1-z^n)}{z^2}dz$$ where $n\geq 1$ is an integer, arises from a natural way when one apply Möbius inversion to get identities realated to $\zeta(2)$ (see my appendix if there are no mistakes), but how
Question. Can you state a closed form for $$\int_0^1 \frac{\log(1-z)\log(1-z^3)}{z^2}dz?$$
I am asking for a detailed explanationabout how get the definite integral. Many thanks.
I don't know if the general integral was in the literature but my trials with Wolfram Alpha, seems that is very difficult to get particular cases. See this
Example. Other case of the indefinite integral, for example $n=5$ also is provided by Wolfram Alpha online calculator when one type
integrate (log(1-x)log(1-x^5))/x^2 dx
My idea was multiply both formulas of the first paragraph in page 77 from:
Benito, Navas, Varona, Möbius inversion from the point of view of flows, Proceedings of the Segundas Jornadas de Teoría de Números, Biblioteca de la Revista Matemática Iberoamericana (2008),
then integrating one gets $$\zeta(2)=\int_0^1\sum_{n=1}^\infty\frac{z^{n-1}}{n}dz=\int_0^1 \left(\frac{\log(1-z)}{z^2}\sum_{n=1}^\infty\frac{\mu(n)}{n}\log(1-z^n)\right)dz.$$ Here as you see $\zeta(s)$ is the Riemann's Zeta function and $\mu(n)$ is the Möbius function. And now from there one more time by absolute convergence, I've asked myself what's about these coefficients
$$a_n=\int_0^1 \frac{\log(1-z)\log(1-z^n)}{z^2}dz,$$ that for integers $n\geq 1$ satisfy $$\zeta(2)=\sum_{n=1}^\infty\frac{\mu(n)}{n}a_n.$$