Yhyjyk

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

$$int_0^1 fracln(1-x)ln(x)x dx= ? $$

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

beginalign*
int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
&=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
&=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
&=sum_n=1^inftyfrac1n^3\
&=zeta(3)
endalign*

This might be the most straightforward approach possible.

II: Integration By Parts

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

beginalign*
int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm 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)^3right]\
&=sum_n=1^inftyfrac1n^3\
&=zeta(3)
endalign*

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

beginalign*
int_0^1fraclog(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\
&=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
&=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
&=zeta(3)
endalign*

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 $operatornameLi_3(1)$

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 $mathrm dv=fraclog(1-x)x$ to get

beginalign*
int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
&=[operatornameLi_3(x)]_0^1\
&=zeta(3)
endalign*

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

beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
&=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
&=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
&=sum_n=1^infty frac1n^3\
&=zeta(3)
endalign