On $\int_0^1\frac{\log^2(x)}{1+x^3}dx$ and $\zeta(3)$

Question

When I was playing with Wolfram Alpha about the integral $$\int_0^1\frac{x^{s-1}}{1+x^2}dx$$ and its derivatives, since I know the relationship between the Apéry's constant and particular values of the polygamma function, and since I presume that this way will be known, I found playing with the code a closed-form for this $$\int_0^1\frac{\log^2(x)}{1+x^3}dx$$ see this code

integrate 1/(1+x^3)(log^2(x))dx, from x=0 to x=1

in the online calculator. I am not able to get easily the calculations for $$\int\frac{\log^2(x)}{1+x^3}dx$$ and after evaluate it as a definite integral. And you?

Question. This can be a good integral for this friday. Can you prove the closed-form for $$\int_0^1\frac{\log^2(x)}{1+x^3}dx?$$ Thanks in advance.

Answer 1

Claim: $$ \int_{0}^{1}\frac{\log^2(x)}{1+x^3}\,dx = \color{red}{\frac{5\pi^3}{81\sqrt{3}}+\frac{13\zeta(3)}{18}}.\tag{0}$$

Proof: since $$ \int_{0}^{1}\frac{x^s}{1+x^3}\,dx = \frac{1}{6}\left[\psi\left(\frac{s+4}{6}\right)-\psi\left(\frac{s+1}{6}\right)\right]\tag{1} $$ the answer just depends on $\psi''\left(\frac{2}{3}\right)-\psi''\left(\frac{1}{6}\right)$. In particular, $(0)$ follows by applying the differential operator $\frac{d^3}{dz^3}\log(\cdot)$ to the reflection formula and the duplication formula for the $\Gamma$ function, then evaluating at $z=\frac{1}{3}$. This appears to be a further generalization of the result shown here about the values of the $\eta$ function at odd positive integers.

Answer 2

By expanding the denominator and performing the integral

$$\int_0^1 dx \, x^{3 k} \log^2{x} $$

on each term in the sum, we find that the integral is equal to

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(3 k+1)^3} $$

This is as far as I will take this.