Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$

Solution 1:

This integral is equal to $$ -4\big( \zeta(-3,-1,-1,-1) +\zeta(-3,-1,1,-1) +\zeta(-3,1,-1,1) +\zeta(3,-1,-1,-1) +\zeta(3,-1,1,-1) +\zeta(3,1,-1,1) \big) $$ in terms of the multiple zeta function, which can also be simplified to $$ 2\zeta(-5,-1)-2\zeta(-5,1)+2\zeta(5,-1)+{\textstyle\frac32}\zeta(5,1)+4\zeta(-3,1,1,1), $$ of which only $$ \begin{aligned} \zeta(5,1) &= {\textstyle\frac34}\zeta(6)-{\textstyle\frac12}\zeta(3)^2 \\ \zeta(5,-1) &= {\textstyle\frac{111}{64}} \zeta (6)-{\textstyle\frac{9}{32}} \zeta (3)^2-{\textstyle\frac{31}{16}} \zeta (5) \log (2) \end{aligned} $$ have a known closed form (see also this article about Euler sums, and also Euler Sums and Contour Integral Representations by Philippe Flajolet and Bruno Salvy).

Update (by editor): Based on MZV reduction of weight $6$, expression above is furtherly simplified to: $$-2 \zeta(\bar5,1)+8 \text{Li}_6\left(\frac{1}{2}\right)+4 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+8 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{13 \zeta (3)^2}{16}+\frac{7}{6} \zeta (3) \log ^3(2)-\frac{221 \pi ^6}{30240}+\frac{\log ^6(2)}{9}-\frac{1}{12} \pi ^2 \log ^4(2)$$

Solution 2:

There is no closed form for this integral as the answer involves $\sum_{n=1}^\infty\frac{H_n}{n^52^n}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ which have no known closed form and here is how I found them:

Let $I$ denotes our integral $\displaystyle \int_0^1\frac{\ln x\ln^2(1-x)\ln^2(1+x)}{x}\ dx$

Using the algebraic identity

$$12a^2b^2=(a+b)^4+(a-b)^4-2a^4-2b^4$$ and by letting $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write our integral :

$$\small{12I=\underbrace{\int_0^1\frac{\ln x\ln^4(1-x^2)}{x}}_{1-x^2\mapsto x}+\underbrace{\int_0^1\frac{\ln x\ln^4\left(\frac{1-x}{1+x}\right)}{x}}_{\frac{1-x}{1+x}\mapsto x}-2\underbrace{\int_0^1\frac{\ln x\ln^4(1-x)}{x}}_{1-x\mapsto x}\ dx-2\int_0^1\frac{\ln x\ln^4(1+x)}{x}\ dx}$$

$$12I=-\frac74\underbrace{\int_0^1\frac{\ln(1-x)\ln^4x}{1-x}\ dx}_{K}+2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^4x}{1-x^2}\ dx}_{J}-2\underbrace{\int_0^1\frac{\ln x\ln^4(1+x)}{x}\ dx}_{M}$$

$$K=\int_0^1\frac{\ln(1-x)\ln^4x}{1-x}\ dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^4x\ dx\\ =-24\sum_{n=1}^\infty\frac{H_n}{(n+1)^5}=-24\sum_{n=1}^\infty\frac{H_n}{n^5}+24\zeta(6)=\boxed{12\zeta^2(3)-18\zeta(6)}$$

To evaluate $J$ we are going to use the identity

$$\frac{1}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)x^{2n-1}$$

$$J=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^4x}{1-x^2}\ dx=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)\int_0^1x^{2n-1}\ln^4x\ dx\\ \sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)\left(\frac{3}{4n^5}\right)=-\frac{93}{4}\sum_{n=1}^\infty\frac{H_n}{n^5}-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}\\ =\boxed{\frac{93}{8}\zeta^2(3)-\frac{651}{16}\zeta(6)-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}}$$

I managed to simplify $M$ here

$$M=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}$$

Combining the results of $K$, $J$ and $M$ we get

$$I=20\operatorname{Li}_6\left(\frac12\right)+12\ln2\operatorname{Li}_5\left(\frac12\right)+4\ln^22\operatorname{Li}_4\left(\frac12\right)-\frac{549}{32}\zeta(6) -\frac18\ln2\zeta(5)+\frac14\ln^22\zeta(4)\\ +\frac12\ln^32\zeta(3)-\frac13\ln^42\zeta(2)-\frac{29}{16}\zeta^2(3)+2\ln2\zeta(2)\zeta(3)\\ +\frac{17}{180}\ln^62-4\sum_{n=1}^\infty\frac{H_n}{n^52^n}-4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$

and here we see the two sums appeared and because their numerical values (given by wolfram) are different so unfortunately they don't cancel each other out. So the integral $I$ has no closed form.