Closed form for $\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx$

Question

I am trying to get this to equal $\displaystyle-\frac {\pi^a}{b}$ for some positive integers $a$ and $b$ . My efforts so far give:

$\displaystyle \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx=A-B+C$

where $A=\displaystyle \sum_{n=1}^\infty\frac{2\zeta(3)}{n^3}=2\zeta(3)^2$

and $C=\displaystyle \sum_{n=1}^\infty\frac{\pi^2}{3n^4}=\frac{\pi^6}{270}$

and $B=\displaystyle \sum_{n=1}^\infty\frac{f(n)}{n^6}$ with f(1)=6, f(2)=34, f(3)=393/4, f(4)=5750/27, f(5)=339059/864, f(6)=325493/500, f(7)=36107863/36000, ... ,f(11)=29071954407257/8001504000, ...

I can not figure out what $f(n)$ might be. The sequences of numerators and denominators are unrecognized by OEIS. Maybe I am off the track.

Perhaps a different approach from the beginning would be better. The results for A and C were interesting anyway.

Answer 1

Hint. One may use some MZVs algebra.

For $n\geq1$, set $$ I_n:=\frac1{n^2}\int_0^1 x^{n-1}\ln^2(x)\ln(1-x)\:dx, $$ then, differentiating the Euler beta integral three times, one gets $$ I_n=-\frac4{n^6}-2\frac{H_{n-1}}{n^5}+2\frac{\frac{\pi^2}6-H_{n-1,2}}{n^4}+2\frac{\zeta(3)-H_{n-1,3}}{n^3} $$ summing with respect to $n$ gives $$ \sum_{n=1}^\infty I_n=-4\zeta(6)-2\zeta(5,1)+\frac{\pi^2}3\zeta(4)-2\zeta(4,2)+2\zeta(3)^2-2\zeta(3,3), $$ using $$ \begin{align} \zeta(5,1)&=-\frac12\zeta(3)^2 +\frac{\pi^6}{1260} \\\\ \zeta(4,2)&= \zeta(3)^2 -\frac{4\pi^6}{2835} \\\\ \zeta(3,3)&=\frac12\zeta(3)^2 -\frac{\pi^6}{1890} \end{align} $$ leads to

$$ \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2}=-\frac{\pi^6}{2835} $$

or

$$ \int_0^1 \frac{\text{Li}_2(x)\ln^2(x)\ln(1-x)}x=-\frac{\pi^6}{2835}. $$

Probably there is a direct path using $\rm{Li}_2(\cdot)$ properties.

Answer 2

Only an intermediate result, not a proof.

The following GNU Maxima 5.36.1 script results in zeros, so $D(n)$ should be tried for calculating the integrals.

D(n) :=
    -1/n*psi[2](n+1)
    +2/n^2*psi[1](n+1)
    -2/n^3*psi[0](n+1)
    -2/n^3*%gamma;

for n : 20 thru 50 do block(
    print(factor(integrate(x^(n-1)*log(x)*log(x)*log(1-x),x,0,1)-D(n)))
);

Answer 3

With @jim's help, I finally got it to be $\displaystyle -\frac{\pi^6}{2835}$ .I will try to figure out MathJax Details soon.

Of course, this means my $B=2\zeta(3)^2+\frac{23\pi^6}{5670}$