How to evaluate $\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx$ in an elegant way?

Question

How to prove, in an elegant way that

$$I=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}dx=\frac{11}{4}\zeta(4)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)\ ?$$

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First, let me show you how I did it

\begin{align} I&=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx\overset{1-x\ \mapsto x}{=}\int_0^1\frac{\ln(1-x)\ln^2x}{2-x}\ dx\\ &=\sum_{n=1}^\infty\frac1{2^n}\int_0^1x^{n-1}\ln^2x\ln(1-x)\ dx\\ &=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\int_0^1x^{n-1}\ln(1-x)\ dx\\ &=\sum_{n=1}^\infty\frac1{2^n}\frac{\partial^2}{\partial n^2}\left(-\frac{H_n}{n}\right)\\ &=\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2\zeta(2)}{n^2}+\frac{2\zeta(3)}{n}-\frac{2H_n}{n^32^n}-\frac{2H_n^{(2)}}{n^22^n}-\frac{2H_n^{(3)}}{n2^n}\right)\\ &=2\zeta(2)\operatorname{Li}_2\left(\frac12\right)+2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n} \end{align}

By substituting

$$S_1=\sum_{n=1}^\infty \frac{H_n}{n^32^n}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$

$$ S_2=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$

$$S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$

along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get the closed form on $I$. Note that $S_1$, $S_2$ and $S_3$ can be found here, here and here respectively.

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Now we can see how boring and tedious our calculations are as we used results of three harmonic series with powers of 2 in the denominator. A friend ( who proposed this problem ) suggested that the integral can be done without using harmonic series, so any idea how to do it that way?

Thanks

Answer 1

Integrate as follows \begin{align} I=&\int_0^1 \frac{\ln x\ln^2(1-x)}{1+x}dx\\ = &\int_0^1 \frac{\ln^2 x\ln(1-x)}{2-x}dx = \int_0^1 \ln(1-x)\> d\left( \int_1^x \frac{\ln^2 t}{2-t}dt\right) \\ =& \int_0^1 \frac1{1-x}\left( \int_0^x \frac{\ln^2 t}{2-t}\overset{t=xy}{dt}- \int_0^1 \frac{\ln^2 t}{2-t}dt \right) dx\\ =& \int_0^1 \left( \int_0^1 \frac{\ln^2 (xy)}{2-y} \left( \frac1{1-x}-\frac2{2-xy}\right){dy}- \frac1{1-x}\int_0^1 \frac{\ln^2 t}{2-t}dt \right) dx\\ =& \int_0^1 \int_0^1 \frac{\ln^2 x+2\ln x\ln y}{(1-x)(2-y)} dydx - \int_0^1 \int_0^1 \frac{2\ln^2(xy)}{(2-y)(2-xy)} \overset{t=xy}{dx}dy\\ =& \> 2\ln2 Li_3(1)+2Li_2(1)Li_2(\frac12) + \int_0^1 \overset{ibp}d\left(\ln \frac y{2-y}\right)\int_0^y\frac{\ln^2t}{2-t}dt\\ =& \>2\ln2 Li_3(1)+2Li_2(1)Li_2(\frac12)-6Li_4(\frac12)-K\tag1 \end{align} where \begin{align} K=&\int_0^1 \frac{\ln^2x \ln (2-x)}{2-x}dx\\ = &\>\ln2 \int_0^1 \frac{\ln^2x}{2-x}dx + \int_0^1 \frac{\ln^2x \ln\frac{2-x}2}{2-x} \overset{x\to 2x}{dx}\\ =&\>2\ln2Li_3(\frac12)+\ln2 \int_0^{\frac12} \frac{\ln(2x^2)\ln(1-x)}{1-x}\overset{ibp}{dx} + \int_0^{\frac12} \frac{\ln^2x\ln(1-x)}{1-x}dx\\ =&\>2\ln2Li_3(\frac12)+\ln2 \left( \frac12\ln^32 +\int_0^{\frac12} \frac{\ln^2(1-x)}{x}dx\right) -\frac14\ln^42+ \frac12J\\ =& \>2\ln2Li_3(1)-2\ln^22Li_2(\frac12)-\frac34\ln^42+\frac12J\tag2\\ \\ J=&\int_0^1 \frac{\ln^2x \ln (1-x)}{1-x}dx\\ = &\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {-x}{1-x y}dy\right) dx =\int_0^1 \frac{1}{1-y}\left(\int_0^1 \frac {\ln^2x}{1-yx}dx - 2Li_3(1)\right) dy \\ =& \>2 \int_0^1\frac{Li_3(y)}{y}dy + 2 \int_0^1\frac{Li_3(y)-Li_3(1)}{1-y}\>\overset{ibp}{dy} = 2Li_4(1) - Li_2^2(1) \end{align} Plug the result for $J$ into (2) and then $K$ into (1) to obtain \begin{align} I= &-6Li_4(\frac12) -Li_4(1)+2(Li_2(1)+\ln^22)Li_2(\frac12)+\frac12Li_2^2(1)+\frac34\ln^42\\ =& -6Li_4(\frac12)+\frac{11\pi^4}{360}-\frac14\ln^42 \end{align}

Answer 2

I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them.

For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity: $$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$ $$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\underbrace{ \int_0^1 \frac{\ln x\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx}_{I_2}-\underbrace{\int_0^1 \frac{\ln x\ln^2(1+x)}{1+x}dx}_{I_3}$$

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For the first integral we will write the denominator as: $$\frac{1}{1+x}=\frac{1}{1-x^2}-\frac{x}{1-x^2}$$ $$\Rightarrow I_1=\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1-x^2}dx-{\int_0^1 \frac{x\ln x\ln^2(1-x^2)}{1-x^2}dx}$$ $$\overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}\frac{dx}{\sqrt x}-\frac14\int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}dx$$ Those two integral can be found using the second identity from here.

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Let's also take $I_2$ and substitute $\frac{1-x}{1+x}\to x$. $$\Rightarrow I_2=\underbrace{\int_0^1 \frac{\ln(1-x)\ln^2 x}{1+x}dx}_{P}-\underbrace{\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx}_{Q}$$ $$P-Q=I_2;\quad P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1+x}dx$$ And again with the same trick done for $I_1$, we have: $$P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1-x^2}dx-\int_0^1 \frac{x\ln(1-x^2)\ln^2 x}{1-x^2}dx$$ $$\overset{x^2\to x}=\frac18\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}\frac{dx}{\sqrt x}-\frac18 \int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx$$ Henceforth we can extract our second integral, $I_2$ as: $$I_2=P-Q=(P+Q)-2Q$$ Note that $P+Q$ can again be found using the second identity from here.
Finally, we only need to find $Q$.

$$Q=\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx=\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{n}\ln^2 x=2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^3}$$ So $Q$ is actually an Euler sum in disguise, but you nicely found it here.

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Also, $I_3$ is pretty easy, one just needs to use the same approach as done for $I_1$ in your following post. $$I_3=\int_0^1 \frac{\ln x \ln^2(1+x)}{1+x}dx\overset{IBP}=-\frac12\int_0^1 \frac{\ln^3(1+x)}{x}dx$$

Answer 3

Here is a solution using only integration and with a bonus

Let

$$P=\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx$$

$$Q=\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}\ dx$$

I am going to establish two relations and solve for $P$ and $Q$.

The first relation:

Use the identity

$$3ab^2-3a^2b=(a-b)^3-a^3+b^3$$

set $a=\ln x$ and $b=\ln(1-x)$ we have

$$3P-3Q=\int_0^1\frac{\ln^3\left(\frac{x}{1-x}\right)}{1+x}\ dx-\int_0^1\frac{\ln^3x}{1+x}\ dx+\underbrace{\int_0^1\frac{\ln^3(1-x)}{1+x}\ dx}_{1-x\to x}$$

Subbing $\frac{x}{1-x}\to x$ gives

$$\int_0^1\frac{\ln^3\left(\frac{x}{1-x}\right)}{1+x}\ dx=\int_0^\infty\frac{\ln^3x}{(1+x)(1+2x)}\ dx$$ $$=\int_0^1\frac{\ln^3x}{(1+x)(1+2x)}\ dx+\underbrace{\int_1^\infty\frac{\ln^3x}{(1+x)(1+2x)}\ dx}_{x\to 1/x}$$

$$=\int_0^1\frac{\ln^3x}{(1+x)(1+2x)}\ dx-\int_0^1\frac{\ln^3x}{(1+x)(2+x)}\ dx$$

$$=2\int_0^1\frac{\ln^3x}{1+2x}\ dx+\int_0^1\frac{\ln^3x}{2+x}\ dx-2\int_0^1\frac{\ln^3x}{1+x}\ dx$$

$$\Longrightarrow 3P-3Q=2\int_0^1\frac{\ln^3x}{1+2x}\ dx+\int_0^1\frac{\ln^3x}{2+x}+\int_0^1\frac{\ln^3x}{2-x}\ dx-3\int_0^1\frac{\ln^3x}{1+x}\ dx$$

$$P-Q=-2\operatorname{Li}_4\left(\frac12\right)+\frac{7}4\zeta(4)-\ln^22\zeta(2)-\frac1{12}\ln^42\tag1$$

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The second relation:

By integration by parts we have

$$P=2\underbrace{\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{1-x}\ dx}_{Y}-\underbrace{\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx}_{Z}\tag2$$

For $Y$, use the identity $4ab=(a+b)^2-(a-b)^2$

set $a=\ln(1-x)$ and $b=\ln(1+x)$ we have

$$4Y=\underbrace{\int_0^1\frac{\ln x\ln^2(1-x^2)}{1-x}\ dx}_{Y_1}-\underbrace{\int_0^1\frac{\ln x\ln^2\left(\frac{1-x}{1+x}\right)}{1-x}\ dx}_{Y_2}$$

For $Y_1$, using @Zacky's technique above $\frac1{1-x}=\frac{1+x}{1-x^2}=\frac{1}{1-x^2}+\frac{x}{1-x^2}$

$$Y_1=\int_0^1\frac{\ln x\ln^2(1-x^2)}{1-x^2}\ dx+\int_0^1\frac{x\ln x\ln^2(1-x^2)}{1-x^2}\ dx$$

$$\overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}\frac{dx}{\sqrt x}+\frac14\int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}dx$$

$$Y_1=-\frac{17}{4}\zeta(4)+7\ln2\zeta(3)-3\ln^22\zeta(2)$$

where the last result follows from using Beta function.

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For $Y_2$, set $\frac{1-x}{1+x}\to x$

$$Y_2=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^2x}{x(1+x)}\ dx$$ $$=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^2x}{x}\ dx-\underbrace{\int_0^1\frac{\ln(1-x)\ln^2x}{1+x}\ dx}_{Q}+\int_0^1\frac{\ln(1+x)\ln^2x}{1+x}\ dx$$

where

$$\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^2x}{x}\ dx=\sum_{n=0}^\infty\frac{-2}{2n+1}\int_0^1 x^{2n}\ln^2x \ dx=\sum_{n=0}^\infty\frac{-4}{(2n+1)^4}=-\frac{15}{4}\zeta(4)$$

and we proved here

$$\int_0^1\frac{\ln(1+x)\ln^2x}{1+x}\ dx=4\operatorname{Li_4}\left(\frac12\right)-\frac{15}4\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac{1}{6}\ln^42$$

$$\Longrightarrow Y_2=4\operatorname{Li_4}\left(\frac12\right)-\frac{15}2\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac{1}{6}\ln^42-Q$$

Collect the results of $Y_1$ and $Y_2$ we get

$$4Y=-4\operatorname{Li_4}\left(\frac12\right)+\frac{13}4\zeta(4)+\frac72\ln2\zeta(3)-2\ln^22\zeta(2)-\frac{1}{6}\ln^42+Q\tag3$$

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For the integral $Z$, its common to use algebraic identities for such integrals

$$a^2b=\frac16(a+b)^3-\frac16(a-b)^3-\frac13b^3$$

$$Z=\frac16\underbrace{\int_0^1\frac{\ln^3(1-x^2)}{x}\ dx}_{x^2\to x}-\frac16\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}=x}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx$$

$$=\frac1{12}\underbrace{\int_0^1\frac{\ln^3(1-x)}{x}\ dx}_{-6\zeta(4)}-\frac13\underbrace{\int_0^1\frac{\ln^3x}{1-x^2}\ dx}_{-\frac{45}{8}\zeta(4)}-\frac13\int_0^1\frac{\ln^3(1+x)}{x}\ dx$$

The last integral can be calculated using the generalization

$$\int_0^1\frac{\ln^n(1+x)}{x}\ dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

$$\Longrightarrow\int_0^1\frac{\ln^3(1+x)}{x}\ dx=-6\operatorname{Li_4}\left(\frac12\right)+6\zeta(4)-\frac{21}{4}\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac{1}{4}\ln^42$$

Therefore

$$Z=2\operatorname{Li_4}\left(\frac12\right)-\frac58\zeta(4)+\frac{7}{4}\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42\tag4$$

Plug $(3)$ and $(4)$ in $(2)$ we get

$$2P-Q=-8\operatorname{Li_4}\left(\frac12\right)+\frac92\zeta(4)-\ln^22\zeta(2)-\frac{1}{3}\ln^42\tag5$$

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Now solve $(1)$ and $(5)$ as a system of equations,

$$P=\frac{11}{4}\zeta(4)-\frac14\ln^42-6\operatorname{Li}_4\left(\frac12\right)$$

$$Q=\zeta(4)+\ln^22\zeta(2)-\frac16\ln^42-4\operatorname{Li}_4\left(\frac12\right)$$

Answer 4

A fancy way of calculating the integral

Here is a fancy way proposed by Cornel (it's pretty amazing for the mathematical connections involved). Let's briefly start with recalling and using Dilogarithm reflection formula, $$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x),$$ where if we multiply both sides by $\displaystyle\frac{\log(1-x)}{1+x}$ and then consider to integrate from $x=0$ to $x=1$, we may express our integral as follows $$\int_0^1\frac{\log(x)\log^2(1-x)}{1+x}\textrm{d}x$$ $$=\zeta(2)\underbrace{\int_0^1\frac{\log(1-x)}{1+x}\textrm{d}x}_{\displaystyle \text{Trivial}}-\underbrace{\int_0^1\frac{\log(1-x)\operatorname{Li}_2(x)}{1+x}\textrm{d}x}_{\displaystyle I}-\underbrace{\int_0^1\frac{\log(1-x)\operatorname{Li}_2(1-x)}{1+x}\textrm{d}x}_{\displaystyle J}.$$ Let the party begin ...

By Landen's Identity, the integral $I$ may be connected to the integral $$\int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\log ^2(2) \zeta (2) -\frac{1}{8} \log ^4(2),$$ which appears in (Almost) Impossible Integrals, Sums, and Serie, page $17$, with a nice solution, and therefore we immediately obtain the desired value of $I$. A different solution than the one presented in the book may be found here.

The last integral (the integral $J$) is also a very pleasant and unexpected game! The magic will happen by letting the variable change $x\mapsto 1-x$, and then connecting the form of the integral to the generalization $$ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a},$$ which is given in the article A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by C. I. Valean, and the solution is straightforward if we expand the integral in series and then use the Cauchy product of $(\operatorname{Li_2}(x))^2$. In other words, we have that $$J=\int_0^1\frac{\log(1-x)\operatorname{Li}_2(1-x)}{1+x}\textrm{d}x=\frac{1}{2}\int_0^1\frac{\log(x)\operatorname{Li}_2(x)}{1-x/2}\textrm{d}x$$ $$=\frac{1}{2}\left(\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}\right) \biggr|_{a=1/2}.$$ End of party (story)

An important note: the necessity of calculating advanced alternating harmonic series or advanced harmonic series with powers of $2$ in the denominator is completely removed by the actual procedure. In fact, by carefully checking the development of the solution to the integral $\int_0^1 \frac{\log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x$ in the book (Almost) Impossible Integrals, Sums, and Series, one may observe that reaching the point with harmonic series may be completely avoided if necessary and the calculation can be accomplished with the use of integrals only (to be clearer, I'm talking about the famous Au-Yeung series).

Answer 5

This does indeed have a closed form. I have to go to work now, but I will be back later unless someone else gets around to it first.

You are aiming for $$\displaystyle \zeta(4)+\zeta(2)\ln^{2}(2)-1/6\ln^{4}(2)-4Li_{4}(1/2)$$

Perhaps begin with the representation:

$$\frac{\psi(k+1/2)-\psi(1/2)}{(2k+1)^{3}}=1/2\int_{0}^{\infty}t^{2}e^{-(2k+1)t}dt\int_{0}^{1}\frac{(1-u^{k})u^{-1/2}}{1-u}du$$

Or, expand $\displaystyle \frac{\ln(1-x)}{1+x}$ in a power series:

$$\frac{\ln(1-x)}{1+x}=\sum_{k=1}^{\infty}\left[(-1)^{k}\ln(2)-(\psi(k/2+1)-\psi(k+1)+\ln(2))\right]x^{k}$$

multiply by $\displaystyle \ln^{2}(x)$, then integrate. The resulting series will be in terms of digamma sums.

It leads to: $$2\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}}{(k+1)^{3}}-2\sum_{k=1}^{\infty}\frac{\psi(k/2+1)}{(k+1)^{3}}+2\sum_{k=1}^{\infty}\frac{\psi(k+1)}{(k+1)^{3}}-2\ln(2)\sum_{k=1}^{\infty}\frac{1}{(k+1)^{3}}$$

Of course, the two sums without digamma are simply zeta series:

$$2\ln(2)\sum_{k=1}^{\infty}\frac{1}{(k+1)^{3}}=2\ln(2)(\zeta(3)-1)$$

$$2\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}}{(k+1)^{3}}=2\ln(2)(3/4\zeta(3)-1)$$

and using the known:

$$\sum_{k=1}^{\infty}\frac{\psi(k+1)}{(k+1)^{3}}=\frac{\pi^{4}}{360}-\gamma\zeta(3)+\gamma$$.

and

$$\sum_{k=1}^{\infty}\frac{\psi(k/2+1)}{(k+1)^{3}}=\frac{-\pi^{4}}{360}-\gamma\zeta(3)+\gamma-1/4\ln(2)\zeta(3)-\frac{\pi^{2}}{12}\ln^{2}(2)+1/12\ln^{4}(2)+2Li_{4}(1/2)$$.

Putting these pieces together gives the required result.

Answer 6

we start with the following result: $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right) = \frac{7}{4}\zeta \left( 3 \right)\ln 2 - \frac{{{\pi ^4}}}{{288}}$$ which can be found in the paper "Euler Sums and Contour Integral Representations",written by Philippe Flajolet and Bruno Salvy,then we can get the value of this integral after expanding it into a series: $$\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right){{\ln }^2}x}}{{1 - x}}dx} = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}\left( {\sum\limits_{k = n + 1}^\infty {\frac{1}{{{k^3}}}} } \right)}$$ by Bailey's Transform,it is $$= 2\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} - 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}} = - \frac{{19{\pi ^4}}}{{720}} + \frac{7}{2}\zeta \left( 3 \right)\ln 2$$....................$(1)$ and,we have to caculate another one: $$\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right)\ln \left( {1 - x} \right)\ln x}}{x}dx} = - \frac{{3{\pi ^4}}}{{160}} - \frac{{{\pi ^2}}}{{12}}{\ln ^2}2 + \frac{{{{\ln }^4}2}}{{12}} + 2L{i_4}\left( {\frac{1}{2}} \right) + \frac{7}{4}\zeta \left( 3 \right)\ln 2$$....................$(2)$ which comes from $$\frac{1}{4}\int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 - x} \right)\ln x}}{x}dx} = \int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 - x} \right)\ln x}}{x}dx} + \int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 + x} \right)\ln x}}{x}dx} + 2\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right)\ln \left( {1 - x} \right)\ln x}}{x}dx}$$ and $$\int\limits_1^\infty {\frac{{{{\ln }^3}\left( {1 + x} \right) - {{\ln }^3}x}}{x}dx} = \int\limits_0^1 {\frac{{{{\ln }^3}\left( {1 + x} \right) - 3{{\ln }^2}\left( {1 + x} \right)\ln x + 3\ln \left( {1 + x} \right){{\ln }^2}x}}{x}dx}$$, now,with some replacements of variables and using the definition of polylogarithm function,it turns true,anyone who is intrested can make up the details. now,back to the question above, integrate $$\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right){{\ln }^2}x}}{{1 - x}}dx}$$ by parts,we get $$\int\limits_0^1 {\frac{{\ln \left( {1 - x} \right){{\ln }^2}x}}{{1 + x}}} dx = \zeta \left( 4 \right) + \zeta \left( 2 \right){\ln ^2}2 - \frac{{{{\ln }^4}2}}{6} - 4L{i_4}\left( {\frac{1}{2}} \right)$$

Answer 7

there are many of this kind associated with EulerSums or Alernating EulerSums,they can be proved by combination of the methods of Contour integration、skills of Translating integrals into series or Translating series into integrals、 bailey transform and something else, I'd like to share some with you. $(1)$. $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} = \frac{{{\pi ^4}}}{{60}} + \frac{{{\pi ^2}}}{{12}}{\ln ^2}2 - \frac{1}{{12}}{\ln ^4}2 - 2L{i_4}\left( {\frac{1}{2}} \right)$$

$(2)$.$$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = - \frac{{17{\pi ^4}}}{{480}} - \frac{{{\pi ^2}}}{6}{\ln ^2}2 + \frac{1}{6}{\ln ^4}2 + 4L{i_4}\left( {\frac{1}{2}} \right) + \frac{7}{2}\zeta \left( 3 \right)\ln 2$$

$(3)$.$${\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} ^2} = - \frac{{61{\pi ^4}}}{{1440}} + \frac{{{\pi ^2}}}{3}{\ln ^2}2 + \frac{1}{6}{\ln ^4}2 + 4L{i_4}\left( {\frac{1}{2}} \right) + \frac{7}{4}\zeta \left( 3 \right)\ln 2$$

$(4)$.$$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} = \frac{{29{\pi ^4}}}{{1440}} + \frac{{{\pi ^2}}}{8}{\ln ^2}2 - \frac{1}{8}{\ln ^4}2 - 3L{i_4}\left( {\frac{1}{2}} \right)$$

$(5)$.$$\int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 - x} \right)\ln \left( {1 + x} \right)}}{x}} dx = - \frac{{{\pi ^4}}}{{144}} - \frac{{{\pi ^2}}}{{12}}{\ln ^2}2 + \frac{1}{{12}}{\ln ^4}2 + 2L{i_4}\left( {\frac{1}{2}} \right) + \frac{7}{4}\zeta \left( 3 \right)\ln 2$$

$(6)$.$$\int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 + x} \right)\ln \left( {1 - x} \right)}}{x}} dx = - \frac{{{\pi ^4}}}{{240}}$$

$$(7)$.let $I = \int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 + x} \right)\ln \left( {1 - x} \right)}}{x}} dx$ and $J = \int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 - x} \right)\ln \left( {1 + x} \right)}}{x}} dx$$,then $$I = \frac{7}{{48}}{\pi ^2}\zeta \left( 3 \right) - \frac{{25}}{{16}}\zeta \left( 5 \right)$$ $$J = - \frac{1}{9}{\pi ^2}{\ln ^3}2 + \frac{2}{{15}}{\ln ^5}2 + 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 + 4L{i_5}\left( {\frac{1}{2}} \right) - \frac{1}{{16}}{\pi ^2}\zeta \left( 3 \right) + \frac{7}{4}\zeta \left( 3 \right){\ln ^2}2 - \frac{7}{2}\zeta \left( 5 \right)$$ proof(brief,detailed proof is very long): notice the fact $$\int\limits_1^\infty {\frac{{{{\ln }^4}\left( {1 + x} \right) - {{\ln }^4}x}}{x}} dx = \int\limits_0^1 {\frac{{{{\ln }^4}\left( {1 + x} \right)}}{x}} dx - 4\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 + x} \right)}}{x}} dx + 6\int\limits_0^1 {\frac{{{{\ln }^2}x{{\ln }^2}\left( {1 + x} \right)}}{x}} dx - 4\int\limits_0^1 {\frac{{{{\ln }^3}x\ln \left( {1 + x} \right)}}{x}} dx$$, the left integral,and the 1st、4th integral of right side are esay to be caculated.and the 3rd integral can be translated into series,the hard term of which is $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^4}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}$$ and has been studied in "Euler Sums and Contour Integral Representations",therefore,we finally get $$\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 + x} \right)}}{x}} dx = \frac{1}{3}{\pi ^2}{\ln ^3}2 - \frac{2}{5}{\ln ^5}2 - 12L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 12L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{2}{\pi ^2}\zeta \left( 3 \right) - \frac{{21}}{4}\zeta \left( 3 \right){\ln ^2}2 + \frac{{99}}{{16}}\zeta \left( 5 \right)$$...............$(1)$ on the other hand,we consider $$\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 - x} \right)}}{x}} dx - 3\int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 - x} \right)\ln \left( {1 + x} \right)}}{x}} dx + 3\int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 + x} \right)\ln \left( {1 - x} \right)}}{x}} dx - \int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 + x} \right)}}{x}} dx$$ $$ = \int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {\frac{{1 - x}}{{1 + x}}} \right)}}{x}} dx$$ $$ = 2\int\limits_0^1 {\frac{{\ln \left( {1 - x} \right){{\ln }^3}x}}{{1 - {x^2}}}} dx - 2\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right){{\ln }^3}x}}{{1 - {x^2}}}} dx$$ $$ = \int\limits_0^1 {\frac{{\ln \left( {1 - x} \right){{\ln }^3}x}}{{1 - x}}} dx + \int\limits_0^1 {\frac{{\ln \left( {1 - x} \right){{\ln }^3}x}}{{1 + x}}} dx - \int\limits_0^1 {\frac{{\ln \left( {1 + x} \right){{\ln }^3}x}}{{1 - x}}} dx - \int\limits_0^1 {\frac{{\ln \left( {1 + x} \right){{\ln }^3}x}}{{1 + x}}} dx$$ the 1st、4th integrals are easy to be worked out,and all integrals of right side can be translated into EulerSums,the results of which have been gotten in "Euler Sums and Contour Integral Representations",so we arrive at $$3I - 3J = \frac{1}{3}{\pi ^2}{\ln ^3}2 - \frac{2}{5}{\ln ^5}2 - 12L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 12L{i_5}\left( {\frac{1}{2}} \right) + \frac{5}{8}{\pi ^2}\zeta \left( 3 \right) - \frac{{21}}{4}\zeta \left( 3 \right){\ln ^2}2 + \frac{{93}}{{16}}\zeta \left( 5 \right)$$...............$(2)$ another relation comes from $$\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 - x} \right)}}{x}} dx = 4\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 - {x^2}} \right)}}{x}} dx = 4\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 - x} \right)}}{x}} dx + 4\int\limits_0^1 {\frac{{\ln x{{\ln }^3}\left( {1 + x} \right)}}{x}} dx + 12\int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 - x} \right)\ln \left( {1 + x} \right)}}{x}} dx + 12\int\limits_0^1 {\frac{{\ln x{{\ln }^2}\left( {1 + x} \right)\ln \left( {1 - x} \right)}}{x}} dx$$ with $(1)$, after simplified,the above is $$12I + 12J = - \frac{4}{3}{\pi ^2}{\ln ^3}2 + \frac{8}{5}{\ln ^5}2 + 48L{i_4}\left( {\frac{1}{2}} \right)\ln 2 + 48L{i_5}\left( {\frac{1}{2}} \right) + {\pi ^2}\zeta \left( 3 \right) + 21\zeta \left( 3 \right){\ln ^2}2 - \frac{{243}}{4}\zeta \left( 5 \right)$$...............$(3)$ from $(2)$、$(3)$,we get I、J. $(8)$.let$$x = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}{{\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}^2}}$$,$$y = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)}$$,$$z = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)}$$,$$w = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)}$$,then $$x = - \frac{1}{9}{\pi ^2}{\ln ^3}2 + \frac{2}{{15}}{\ln ^5}2 + 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 + 4L{i_5}\left( {\frac{1}{2}} \right) - \frac{{11}}{{48}}{\pi ^2}\zeta \left( 3 \right) + \frac{7}{2}\zeta \left( 3 \right){\ln ^2}2 - \frac{{19}}{{32}}\zeta \left( 5 \right)$$ $$y = \frac{1}{9}{\pi ^2}{\ln ^3}2 - \frac{2}{{15}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 4L{i_5}\left( {\frac{1}{2}} \right) + \frac{5}{{32}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{4}\zeta \left( 3 \right){\ln ^2}2 + \frac{{23}}{8}\zeta \left( 5 \right)$ $z = - \frac{{13}}{{48}}{\pi ^2}\zeta \left( 3 \right) + \frac{{125}}{{32}}\zeta \left( 5 \right)$$ $$w = \frac{2}{{45}}{\pi ^4}\ln 2 + \frac{1}{{36}}{\pi ^2}{\ln ^3}2 - \frac{1}{{60}}{\ln ^5}2 + 2L{i_5}\left( {\frac{1}{2}} \right) - \frac{1}{{48}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{{37}}{{16}}\zeta \left( 5 \right)$$ brief proof: we transform the integrals of I、J of $(7)$ into series of EulerSums: $$x + y = - \frac{7}{{96}}{\pi ^2}\zeta \left( 3 \right) + \frac{{73}}{{32}}\zeta \left( 5 \right)$$...................$(1)$ $$z + w = \frac{2}{{45}}{\pi ^4}\ln 2 + \frac{1}{{36}}{\pi ^2}{\ln ^3}2 - \frac{1}{{60}}{\ln ^5}2 + 2L{i_5}\left( {\frac{1}{2}} \right) - \frac{7}{{24}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{8}\zeta \left( 3 \right){\ln ^2}2 + \frac{{51}}{{32}}\zeta \left( 5 \right)$$...............$(2)$ and we consider two contour integrals $$\int_C {\frac{1}{{{z^2}}}} {\mathop \psi \limits^{\_\_} ^{(1)}}\left( z \right){\left( {\psi \left( z \right) + \gamma } \right)^2}dz$,$\int_C {\frac{1}{{{z^3}}}} \mathop \psi \limits^{\_\_} \left( z \right){\left( {\psi \left( z \right) + \gamma } \right)^2}dz$$ where $$\psi \left( z \right) + \gamma = \frac{d}{{dz}}\ln \Gamma \left( z \right) = - \frac{1}{z} + \sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + z}}} \right)}$$,$$\mathop \psi \limits^{\_\_} \left( z \right) = \frac{1}{2}\psi \left( {\frac{{z + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{z}{2}} \right)$,${\mathop \psi \limits^{\_\_} ^{(1)}}\left( z \right) = \frac{d}{{dz}}\mathop \psi \limits^{\_\_} \left( z \right)$$,just same as the the authors did in the paper " Euler Sums and Contour Integral Representations", we get $$x + y + z = - \frac{{11}}{{32}}{\pi ^2}\zeta \left( 3 \right) + \frac{{99}}{{16}}\zeta \left( 5 \right)$$...............$(3)$ $$x - 2w = - \frac{4}{{45}}{\pi ^4}\ln 2 - \frac{1}{6}{\pi ^2}{\ln ^3}2 + \frac{1}{6}{\ln ^5}2 + 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - \frac{3}{{16}}{\pi ^2}\zeta \left( 3 \right) + \frac{7}{2}\zeta \left( 3 \right){\ln ^2}2 + \frac{{129}}{{32}}\zeta \left( 5 \right)$$...............$(4)$ solve equations $(1)$,$(2)$,$(3)$,$(4)$,we complete the proof. $(9)$.here are some other EluerSums,on the base of above results,they can be proved. $$\int\limits_0^1 {\frac{{\ln \left( {1 + x} \right)\ln \left( {1 - x} \right)}}{x}} L{i_2}\left( { - x} \right)dx = \frac{1}{9}{\pi ^2}{\ln ^3}2 - \frac{2}{{15}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 4L{i_5}\left( {\frac{1}{2}} \right) + \frac{5}{{96}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{4}\zeta \left( 3 \right){\ln ^2}2 + \frac{{123}}{{32}}\zeta \left( 5 \right)$$ $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} = \frac{2}{{45}}{\pi ^4}\ln 2 + \frac{1}{{36}}{\pi ^2}{\ln ^3}2 - \frac{1}{{60}}{\ln ^5}2 + 2L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{16}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{{193}}{{64}}\zeta \left( 5 \right)$$ $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}{{\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}^3}} = - \frac{1}{6}{\pi ^2}{\ln ^3}2 + \frac{1}{5}{\ln ^5}2 + 6L{i_4}\left( {\frac{1}{2}} \right)\ln 2 + 6L{i_5}\left( {\frac{1}{2}} \right) - \frac{9}{{32}}{\pi ^2}\zeta \left( 3 \right) + \frac{{21}}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{9}{4}\zeta \left( 5 \right)$$ $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}{{\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)}^3}} = - \frac{{29}}{{160}}{\pi ^4}\ln 2 + \frac{{11}}{{12}}{\pi ^2}{\ln ^3}2 - \frac{1}{{20}}{\ln ^5}2 - 6L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 24L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{367}}{{16}}\zeta \left( 5 \right)$$ $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)} = \frac{{49}}{{720}}{\pi ^4}\ln 2 - \frac{1}{{18}}{\pi ^2}{\ln ^3}2 + \frac{1}{{10}}{\ln ^5}2 + 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 + 8L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{96}}{\pi ^2}\zeta \left( 3 \right) - \frac{{35}}{4}\zeta \left( 5 \right)$$ $$(10)$.$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = - \frac{{101}}{{45360}}{\pi ^6} + \frac{5}{2}{\zeta ^2}\left( 3 \right)$$ $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}{{\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}^3} = } \frac{{31}}{{5040}}{\pi ^6} - \frac{5}{2}{\zeta ^2}\left( 3 \right)$$

Answer 8

Let $I$ denote the integral in the question. Then we can write

\begin{align*} I &= \int_0^1\frac{\log^2 (x) \log(1-x^2)-\log^2(x)\log(1+x)}{1+x}dx \\ &= \int_0^1\frac{\log^2 (x) \log(1-x^2)}{1+x}dx-\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x}dx \quad \cdots (1) \end{align*}

The first integral can be manipulated as follows:

\begin{align*} \int_0^1\frac{\log^2 (x) \log(1-x^2)}{1+x}dx &= \frac{1}{8}\int_0^1 \frac{\log^2(x) \log(1-x)}{(1+\sqrt{x})\sqrt{x}}dx \\ &= \frac{1}{8} \left( \int_0^1 \frac{\log^2(x)\log(1-x)}{(1-x)\sqrt{x}}dx-\int_0^1 \frac{\log^2(x)\log(1-x)}{1-x}dx\right) \end{align*}

Calculation of these two integrals is done by differentiating the beta function. The end result is

$$\displaystyle \int_0^1\frac{\log^2 (x) \log(1-x^2)}{1+x}dx = \frac{7}{2} \zeta (3) \log (2)-\frac{11 \pi ^4}{360} \quad ... (2)$$

Now we turn our attention to the second integral of equation (1): \begin{align*} \int_0^1 \frac{\log^2(x)\log(1+x)}{1+x}dx &= \sum_{n=1}^\infty (-1)^{n+1} H_n \int_0^1 x^n \log^2(x)\; dx \\ &= 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^3} \\ &= 4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{2} \zeta (3) \log (2)-\frac{\pi ^4}{24}+\frac{\log ^4(2)}{6}-\frac{1}{6} \pi ^2 \log ^2(2) \quad ...(3) \end{align*}

Substitute everything in equation (1) to get

$$\displaystyle I=\boxed{\displaystyle -4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{\log ^4(2)}{6}+\frac{1}{6} \pi ^2 \log ^2(2)}$$