Ryfujk

I am curious to know how to solve this problem; I know the answer comes out to be $1.20205$ (Apéry's constant).

$$int_0^1 frac{ln(1-x)ln(x)}{x} dx= ? $$

There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them

I: Taylor Series Expansion of $log(1-x)$

As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain

begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
&=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
&=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
&=sum_{n=1}^inftyfrac1{n^3}\
&=zeta(3)
end{align*}

This might be the most straightforward approach possible.

II: Integration By Parts

Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives

begin{align*}
int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
&=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
&=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
&=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
&=sum_{n=1}^inftyfrac1{n^3}\
&=zeta(3)
end{align*}

Again, we utilized a series expansion, this time the one of the geometric series.

III: Integral Representation of the Zeta Function

To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find

begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
&=-int_0^infty xlog(1-e^{-x})mathrm dx\
&=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
&=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
&=zeta(3)
end{align*}

Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.

IV: The Trilogarithm $operatorname{Li}_3(1)$

Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get

begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
&=[operatorname{Li}_3(x)]_0^1\
&=zeta(3)
end{align*}

A quick look at the series representation of the Trilogarithm verifies the last line.

begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
&=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
&=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
&=sum_{n=1}^infty frac{1}{n^3}\
&=zeta(3)
end{align}