Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ and do a comparision with $\zeta(3)$?

Question

I know from Wolfram Alpha that $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=1.20206$$ and in the other hand, too from this online tool that $$\int\frac{\log(x)\log(1-x)}{x}dx=\mathrm{Li}_3(x)-\mathrm{Li}_2(x)\log(x)+constant.$$

Question. I would like made a comparision, and need obtain $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx,$$ more precisely than $1.20206$. I believe that could be $\zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.

Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ to discard that this value is $\zeta(3)$, Apéry constant, or claim that the equality $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=\zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?

This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.

I've searched in this site about this integral $\int\frac{\log(x)\log(1-x)}{x}dx$, and in Wikipedia about a possible identity between $\zeta(3)$ and particular values of logarithmic integrals $\mathrm{Li}_2(x)$ and $\mathrm{Li}_3(x)$.

References:

[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant

Answer 1

$$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,.$$

Therefore, since $\log(1)=0$, we have:

$$\operatorname{Li}_3(1) = \sum_{k=1}^\infty {1 \over k^3} = \zeta(3)$$

$$\operatorname{Li}_3(1)-\log(1)\operatorname{Li}_2(1) = \zeta(3)$$

It remains to show that $$\lim_{x\to0} \operatorname{Li}_3(x)-\log(x)\operatorname{Li}_2(x)=0$$

Note that if $z<1$,

$$\operatorname{Li}_2(z) = \sum_{k=1}^\infty {z^k \over k^2} = z + {z^2 \over 2^2} + {z^3 \over 3^2} + \cdots \\< z + {z \over 2^2} + {z \over 2^2}+ {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 8^2}+\cdots \leq z\sum_{k=0}^\infty {1 \over 2^k} = 2z$$

Since $\log(z)<z$, for $z>1$, we also have $\log(z^2)<2z$ thus $\log(x)<2\sqrt{x}$. Thus $\log(\frac1x)>-2\sqrt{x}$, thus $\log(u)>-2\sqrt{\frac1u}$, thus $0>\log(u)\operatorname{Li}_2(u)>-2\sqrt{u}$.

We may use the Squeeze Theorem to finish the result.

Answer 2

An informal argument: $$\log (1-x) = - \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}, \quad |x| < 1.$$ Then $$\frac{\log x \log(1-x)}{x} = -\sum_{k=0}^\infty \frac{x^k \log x}{k+1},$$ and integrating term by term gives $$\int_{x=0}^1 x^k \log x \, dx = \left[ \frac{x^{k+1} \log x}{k+1} \right]_{x=0}^1 - \int_{x=0}^1 \frac{x^k}{k+1} \, dx = - \frac{1}{(k+1)^2}.$$ Therefore, $$\int_{x=0}^1 \frac{\log x \log(1-x)}{x} \, dx = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3).$$

Answer 3

Let $x = e^{-y}$, we have $$\int_0^1 \frac{\log x\log(1-x)}{x} dx = \int_0^1 \frac{(-\log x)}{x} \sum_{n=1}^\infty \frac{x^n}{n} dx = \sum_{n=1}^\infty \frac{1}{n}\int_0^1 (-\log x) x^{n-1} dx\\ = \sum_{n=1}^\infty \frac{1}{n}\int_0^\infty y e^{-ny} dy = \sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3) $$ Please note that we can switch the order of summation and integration because all the individual terms are non-negative.

Answer 4

The Beta function and Feynman's trick are another way to go:

$$I=\int_{0}^{1}\frac{\log(x)\log(1-x)}{x}\,dx =\left.\frac{\partial^2}{\partial a \partial b}\int_{0}^{1}x^{a-1}(1-x)^{b}\,dx\,\right|_{\alpha,\beta=0^+}\tag{1} $$ hence: $$ I = \left.\frac{\partial^2}{\partial a \partial b}\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}\,\right|_{\alpha,\beta=0^+}\tag{2} $$ and by exploiting $\Gamma'(z) = \Gamma(z)\cdot\psi(z)$ we get: $$ I = -\frac{1}{2}\psi''(2)=\sum_{n\geq 0}\frac{1}{(n+1)^3}=\color{red}{\zeta(3)}\tag{3} $$ as wanted.

Answer 5

Note

$$\zeta(3)= \mathrm{Li}_3(1)=\int_0^1\frac{\mathrm{Li}_2(x)}xdx =\int_0^1\left(- \int_0^x\frac{\ln (1-t)}tdt \right)d(\ln x) = \int_0^1\frac{\ln x\ln(1-x)}{x}dx $$