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

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 $$ \sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)} =\frac{\zeta(3)}{4} $$

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

\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}

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

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.