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One of the ways to approach it lies in the area of the dilogarithm, but is it possible to evaluate it
by other means of the real analysis (without using dilogarithm)?

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

EDIT: maybe you're aware of some easy way to do that. I'd appreciate it!
Some words on the generalization case (by means of the real analysis again)?

$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$

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    $\begingroup$ do you need a closed form? Otherwise $\log(1+x) \sim x - \frac{x^2}{2} + \frac{x^3}{3}$ is a good approximation $\endgroup$
    – Alex
    May 15, 2014 at 13:17
  • $\begingroup$ Yes, I need closed forms. $\endgroup$ May 15, 2014 at 13:44
  • $\begingroup$ Wolfram Alpha thinks the first two answers (for $n=1$ and $n=2$) are ${1\over2}\zeta(2)$ and ${1\over4}\zeta(3)$. After that it gets dilogarithmically messy. $\endgroup$ May 15, 2014 at 13:54
  • $\begingroup$ An antiderivative of $\frac{\log^{n}(1+x)}{x}$ in terms of polylogarithms can found by repeatedly integrating by parts. And if you're interested, joriki evaluated the case $n=2$ using contour integration. math.stackexchange.com/questions/316745/… $\endgroup$ May 15, 2014 at 14:09
  • $\begingroup$ @RandomVariable Yes, indeed (I missed a tricky series while doing things in a hurry). $\endgroup$ May 15, 2014 at 14:20

8 Answers 8

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Squaring the series for $\log(1+x)$ yields $$ \log(1+x)^2=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^kx^k}{j(k-j)} $$ Dividing by $x$ and integrating gives $$ \begin{align} \int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x &=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^k}{jk(k-j)}\\ &=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac{(-1)^k}{jk(k-j)}\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)}\\[9pt] &=\frac{\zeta(3)}{4} \end{align} $$ Using $(5)$ from this answer: $$ \sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n =-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn} $$ and $(6)$ from the same answer: $$ -\frac58\zeta(3) =\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n $$ we get $$ \sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)} =\frac{\zeta(3)}{4} $$

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  • $\begingroup$ Thanks, that's the way. (+1) $\endgroup$ May 15, 2014 at 14:23
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} =\int_{1}^{2}{\ln^{2}\pars{x} \over x - 1}\,\dd x =\int_{1}^{1/2}{\ln^{2}\pars{1/x} \over 1/x - 1}\,\pars{-\,{\dd x \over x^{2}}} =\int_{1/2}^{1}{\ln^{2}\pars{x} \over x\pars{1 - x}}\,\dd x \\[3mm]&=\int_{1/2}^{1}{\ln^{2}\pars{x} \over x}\,\dd x + \int_{1/2}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x ={1 \over 3}\,\ln^{3}\pars{2} +\sum_{n = 1}^{\infty}\color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} \qquad\qquad\pars{1} \end{align}

$$ \color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} =\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu}\int_{1/2}^{1}x^{\mu}\,\dd x =\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu} \bracks{{1 - \pars{1/2}^{\mu + 1} \over \mu + 1}} $$

$$ \color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} =-2\,{\pars{1/2}^{n} \over n^{3}}+ {2 \over n^{3}} -\ln^{2}\pars{2}\,{\pars{1/2}^{n} \over n} -2\ln\pars{2}\,{\pars{1/2}^{n} \over n^{2}} $$

By replacing in $\pars{1}$: \begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} \\[3mm]&={1 \over 3}\,\ln^{3}\pars{2} -2{\rm Li}_{3}\pars{\half} +2\zeta\pars{3} - \ln^{2}\pars{2}{\rm Li}_{1}\pars{\half} -2\ln\pars{2}{\rm Li}_{2}\pars{\half}\tag{2} \end{align}

You'll find values for the PolyLogarithm Function $\ds{{\rm Li}_{s}\pars{\half}\,,\ \pars{~s = 1,2,3~}\,,\ }$ in this page: \begin{align} {\rm Li}_{1}\pars{\half} &= \ln\pars{2} \\[3mm] {\rm Li}_{2}\pars{\half} &= {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[3mm] {\rm Li}_{3}\pars{\half} &= {1 \over 6}\,\ln^{3}\pars{2}- {\pi^{2} \over 12}\,\ln\pars{2} +{7 \over 8}\,\zeta\pars{3} \end{align}

With these identities and result $\pars{2}$: \begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} \\[3mm]&=\color{#00f}{{1 \over 3}\,\ln^{3}\pars{2}} +\ \overbrace{\bracks{\color{#00f}{-\,{1 \over 3}\,\ln^{3}\pars{2}} + \color{magenta}{{\pi^{2} \over 6}\,\ln\pars{2}} {\large -{7 \over 4}\,\zeta\pars{3}}}}^{\ds{-2{\rm Li}_{3}\pars{\half}}}\ +\ {\large 2\zeta\pars{3}} \\[3mm]&+\ \underbrace{\bracks{\color{#990099}{-\ln^{3}\pars{2}}}} _{\ds{-\ln^{2}\pars{2}{\rm Li}_{1}\pars{\half}}}\ +\ \underbrace{\bracks{\color{magenta}{-\,{\pi^{2} \over 6}\,\ln\pars{2}} +\color{#990099}{\ln^{3}\pars{2}}}}_{\ds{-2\ln\pars{2}{\rm Li}_{2}\pars{\half}}}\ =\ \pars{2 - {7 \over 4}}\zeta\pars{3} \end{align}

$$ \color{#66f}{\large% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x = {\zeta\pars{3} \over 4}} \approx 0.3005 $$

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  • $\begingroup$ Good job there! :-) (+1) $\endgroup$ Jun 30, 2014 at 19:38
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The following new solution to the classical harmonic series result, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{5}{8}\zeta(3)$, is proposed by Cornel Ioan Valean, using the powerful identity, $$\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)}=\frac{1}{2(2n-1)}\left(2H_{2n}-H_n-2\log(2)\right),\tag1$$ found and proved in $(6.289)$ in the book (Almost) Impossible Integrals, Sums, and Series.

If we multiply both sides of $(1)$ by $1/(2n-1)$, consider the sum from $n=1$ to $\infty$ and then reindex, we have for the right-hand side that $$\sum_{n=1}^{\infty} \frac{H_{2n}}{(2n-1)^2}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n-1)^2}-\log(2)\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$ $$=-\frac{3}{4}\log(2)\zeta(2)+\sum_{n=1}^{\infty} \frac{H_{2n-1}}{(2n-1)^2}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n+1)^2}$$ $$=-\frac{7}{8}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_n}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{1}{8}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}.\tag2$$

On the other hand, based on $(1)$, we have for the left-hand side that $$\sum _{n=1}^{\infty}\left(\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)}\right)=\sum _{k=1}^{\infty}\left(\sum _{n=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)}\right)$$ $$=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\sum_{n=1}^k \frac{1}{2n-1}=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\left(H_{2k}-\frac{1}{2}H_k\right)=\sum _{k=1}^{\infty}\frac{H_{2k}}{(2k)^2}-\frac{1}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}$$ $$=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\sum_{n=1}^k \frac{1}{2n-1}=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\left(H_{2k}-\frac{1}{2}H_k\right)=\sum _{k=1}^{\infty}\frac{H_{2k}}{(2k)^2}-\frac{1}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}$$ $$=\frac{3}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}-\frac{1}{2}\sum _{k=1}^{\infty}(-1)^{k-1}\frac{H_k}{k^2}=\frac{3}{4}\zeta(3)-\frac{1}{2}\sum _{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}.\tag3$$

By combining $(2)$ and $(3)$, we obtain that

$$\sum _{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{5}{8}\zeta(3).$$

In the calculations we needed particular cases of the generalizations, \begin{equation*} 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\ge2, \end{equation*} and \begin{equation*} \sum _{k=1}^{\infty}\frac{H_k}{(2k+1)^{2m}}=2m\left(1-\frac{1}{2^{2m+1}}\right)\zeta(2m+1)-2\log(2)\left(1-\frac{1}{2^{2m}}\right)\zeta(2m) \end{equation*} \begin{equation*} -\frac{1}{2^{2m}}\sum_{i=1}^{m-1}(1-2^{i+1})(1-2^{2m-i})\zeta(1+i)\zeta(2m-i), \end{equation*} proved in https://math.stackexchange.com/q/3268851. Cornel's solution to the case, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}=\frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3)$, may be found in https://math.stackexchange.com/q/3269815, and the present technique may be easily extended to calculate the generalization, $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n^{2m}}$.

Since the given integral easily reduces to the calculations of $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}$, the solution is finalized.

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Different approach:

We have

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

Divide by $x$ then integrate to get

\begin{align} \int_0^1\frac{\ln^2(1+x)}{x}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{n+1}\int_0^1x^n\ dx\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n}H_{n-1}}{n^2}\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n}}{n^3}\\ &=2\left(-\frac58\zeta(3)\right)-2\operatorname{Li}_3(-1)\\ &=-\frac54\zeta(3)-2\left(-\frac34\zeta(3)\right)\\ &=\boxed{\frac14\zeta(3)} \end{align}


Note:

We have the generating identity

$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

and by setting $x=-1$ and considering only the real parts we have

$$\Re\sum_{n=1}^\infty (-1)^n\frac{H_n}{n^2}=\operatorname{Li}_3(-1)-\Re\operatorname{Li}_3(2)+\Re\ln2\operatorname{Li}_2(2)+\frac12\underbrace{\Re\ln(-1)\ln^22}_{0}+\zeta(3)\tag{1}$$

Using the trilogarithmic identity

$$\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\operatorname{Li}_3\left(\frac{x-1}{x}\right)=\frac16\ln^3x+\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)+\zeta(3)$$

set $x=-1$ and take the real parts to have

$$ \boxed{\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)}$$

also Landen's identity gives

$$ \boxed{\Re\operatorname{Li}_2(2)=\frac32\zeta(2)}$$

Plugging the boxed results along with $\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ in (1) we have

$$\Re\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}=-\frac58\zeta(3)$$

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You can find a nice generalization for $\int_0^1\frac{\ln^n(1+x)}{x}dx$ in lemma $2.2$ in this article and I am going to type it here with little more details.

Start with subbing $\frac{1}{1+x}=y$

$$I_n=\int_0^1\frac{\ln^n(1+x)}{x}dx=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy$$

$$=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y}dy+(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{1-y}dy$$

$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\int_{0}^1\frac{\ln^n(y)}{1-y}dy-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$

$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\left[(-1)^n n!\zeta(n+1)\right]-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$

$$=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\tag1$$

By using

$$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$

or $$(x-y)^n=(-1)^n(y-x)^n=(-1)^n \sum_{k=0}^n{n\choose k}y^{n-k}(-x)^k=\sum_{k=0}^n{n\choose k}(-y)^{n-k}x^k\tag2$$

we get

$$\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\overset{2y=x}{=}-\int_0^1\frac{(\ln(x)-\ln(2))^n}{2-x}dx$$

$$\overset{(2)}{=}-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\int_0^1\frac{\ln^k(x)}{2-x}dx\right)$$

$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\sum_{i=1}^\infty\frac1{2^i} \int_0^1 x^{i-1}\ln^k(x)dx\right)$$

$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left((-1)^k k!\sum_{i=1}^\infty\frac1{2^i i^{k+1}}\right)$$

$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)\tag3$$

Plug $(3)$ in $(1)$ we get

$$I_n=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$ or $$(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

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On the path of Felix Marin, \begin{align}J&=\int_0^1 \frac{\ln(1+x)^2}{x}\\ &\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x(1-x)}\,dx\\ &=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x}\,dx+\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{3}\left(\ln^3 (1)-\ln^3\left(\frac{1}{2}\right)\right)+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &\overset{y=\frac{x}{1-x},\text{the 2nd integral}}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx+2\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\ &\overset{IBP}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx-J\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{2x\ln^2 x}{1-x}\,dx-\frac{1}{3}\ln^3 2-J\\ &\overset{y=x^2}=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx-J\\ J&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{8}\times 2\zeta(3)\\ &=\boxed{\frac{1}{4}\zeta(3)} \end{align} NB: i assume that, \begin{align}\int_0^1 \frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align} (proof: Taylor expansion)

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Using the algebraic identity

$$b^2=\frac12(a-b)^2+\frac12(a+b)^2-a^2$$

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

$$\int_0^1\frac{\ln^2(1+x)}{x}\ dx=\frac12\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}=y}+\frac12\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}\ dx}_{1-x^2=y}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{1-x=y}\\=\int_0^1\frac{\ln^2y}{1-y^2}\ dy+\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy-\int_0^1\frac{\ln^2y}{1-y}\ dy\\=\frac12\int_0^1\frac{\ln^2y}{1+y}\ dy-\frac14\int_0^1\frac{\ln^2y}{1-y}\ dy=\frac12\left(\frac32\zeta(3)\right)-\frac14(2\zeta(3))=\boxed{\frac14\zeta(3)}$$

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  • $\begingroup$ For the last approach see math.stackexchange.com/a/2143161/186817 $\endgroup$
    – FDP
    Nov 17, 2019 at 16:10
  • $\begingroup$ Yes same idea. Using algebraic identity for logarithmic integrals is very common technique. $\endgroup$ Nov 17, 2019 at 16:40
  • $\begingroup$ The proof of Felix Marin can be much more shorter. $\endgroup$
    – FDP
    Nov 17, 2019 at 17:02
  • $\begingroup$ yes but there you need to use the special values of $\operatorname{Li}_2(1/2)$ and $\operatorname{Li}_3(1/2)$ which is not a big deal. $\endgroup$ Nov 17, 2019 at 17:08
  • $\begingroup$ Maybe using not directly such values. $\endgroup$
    – FDP
    Nov 17, 2019 at 17:11
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Here is a solution by finding the closed form of $\int \frac{\ln^2(1-x)}{x}dx$ then letting $x\mapsto -x$:

$$\int \frac{\ln^2(1-x)}{x}dx=\int \frac{\ln(1-x)\ln(1-x)}{x}dx\overset{IBP}{=}-\operatorname{Li}_2(x)\ln(1-x)-\int\frac{\operatorname{Li}_2(x)}{1-x}dx$$

For the last integral, set $1-x=y$ then use the reflection formula: $$\operatorname{Li}_2(1-y)=\zeta(2)-\ln(y)\ln(1-y)-\operatorname{Li}_2(y)$$

We obtain that

$$\int\frac{\operatorname{Li}_2(x)}{1-x}dx=-\int\frac{\operatorname{Li}_2(1-y)}{y}dy$$

$$=-\zeta(2)\int\frac{dy}y+\int\frac{\ln(y)\ln(1-y)}{y}dy+\int\frac{\operatorname{Li}_2(y)}{y}dy$$

$$=-\zeta(2)\ln(y)+\left[-\operatorname{Li}_2(y)\ln(y)+\int\frac{\operatorname{Li}_2(y)}{y}dy\right]+\int\frac{\operatorname{Li}_2(y)}{y}dy$$

$$=-\zeta(2)\ln(y)-\operatorname{Li}_2(y)\ln(y)+2\operatorname{Li}_3(y)$$

$$=-\zeta(2)\ln(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)+2\operatorname{Li}_3(1-x)$$

Then

$$\int\frac{\ln^2(1-x)}{x}dx=\ln(1-x)\left[\operatorname{Li}_2(1-x)-\operatorname{Li}_2(x)+\zeta(2)\right]-2\operatorname{Li}_3(1-x)$$

Now consider the integral boundaries $(0,a)$,

$$\int_0^a\frac{\ln^2(1-x)}{x}dx=\ln(1-a)\left[\operatorname{Li}_2(1-a)-\operatorname{Li}_2(a)+\zeta(2)\right]-2\operatorname{Li}_3(1-a)+2\zeta(3)$$

Therefore

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

$$=\ln(2)\left[\operatorname{Li}_2(2)-\operatorname{Li}_2(-1)+\zeta(2)\right]-2\operatorname{Li}_3(2)+2\zeta(3)$$

substitute $\Re\operatorname{Li}_2(2)=\frac32\zeta(2)$ and $\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$, the closed form follows.

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