Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $

Question

Compute $$ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $$

I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.

Answer 1

This answer is from my old calculation.

First, assume we are well aware of the following famous result.

$$\zeta(2) =\frac{\pi^{2}}{6}, \quad \zeta(4) =\frac{\pi^{4}}{90}$$

Next, by a simple calculation we obtain

$$ H_{n} := \sum_{k=1}^{n} \frac{1}{k} =\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt. $$

and

$$ \frac{\log (1-x)}{1-x}\ =\ -\sum_{n=1}^{\infty}H_{n}x^{n}. $$

Finally, define the polylogarithm as

$$ \mathrm{Li}_{s}(x) :=\sum_{n=1}^{\infty} \frac{x^n}{n^s}, $$

so that it satisfies the recurrence relation

$$ \mathrm{Li}_{1}(x) =-\log (1-x) , \quad \mathrm{Li}_{s+1}(x) =\int_{0}^{x}\frac{\mathrm{Li}_{s}(t)}{t}\, dt $$

and the identity

$$ \mathrm{Li}_{s}(1) =\zeta(s). $$

The the all-in-one straight calculation goes as follows:

\begin{align*} \int_{0}^{1}\frac{\log x\log^{2}(1-x)}{x}\, dx & = \int_{0}^{1}\frac{\log (1-x)\log^{2}x}{1-x}\, dx = -\sum_{n=1}^{\infty}H_{n}\int_{0}^{1}x^{n}\log^{2}x\, dx\\ & = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{3}}\\ & = 2\sum_{n=1}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right] = 2\sum_{n=0}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right]\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt = 2\zeta(4)-2\int_{0}^{1}\frac{\zeta(3)-\mathrm{Li}_{3}(t)}{1-t}\, dt\\ & = 2\zeta(4)+\left[2 (\zeta(3)-\mathrm{Li}_{3}(t))\log (1-t)\right]_{0}^{1}+2\int_{0}^{1}\frac{\mathrm{Li}_{2}(t)\log (1-t)}{t}\, dt\\ & = 2\zeta(4)-2\int_{0}^{1}\mathrm{Li}_{2}(t)\frac{d\mathrm{Li}_{2}(t)}{dt}\, dt\\ & = 2\zeta(4)-\left[\mathrm{Li}_{2}^{2}(t)\right]_{0}^{1} = 2\zeta(4)-\zeta(2)^{2} = \frac{\pi^{4}}{45}-\frac{\pi^{4}}{36} = -\frac{\pi^{4}}{180}\\ & = -\frac{1}{2}\zeta(4). \end{align*}

Answer 2

The Taylor expansion approach gives you $-2 \sum_{k=1}^\infty H_k/(k+1)^3$ where $H_k = \sum_{n=1}^k 1/n$. Wolfram Alpha says this is $-\pi^4/180$, but I don't know how it gets that.

Answer 3

@Chri's sister: see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=353720&p=1921474&hilit=Borwein#p1921474

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Huge\left. a\right)}$ \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = \left.\partiald[2]{}{\mu}\int_{0}^{1}\ln\pars{x} \,{\pars{1 - x}^{\mu} -1 \over x}\,\dd x\,\right\vert_{\large\ \mu\ =\ 0^{+}} \\[5mm] & = \left.\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{\mu \choose n}\pars{-1}^{n}\ \int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x\,\right\vert_{\large\ \mu\ =\ 0^{+}} = \left.-\,\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{\mu \choose n} {\pars{-1}^{n} \over n^{2}}\,\right\vert_{\large\ \mu\ =\ 0^{+}} \\[5mm] & = \left.-\,\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{1 \over n!}\, {\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}}\, {\pars{-1}^{n} \over n^{2}}\,\right\vert_{\large\ \mu\ =\ 0^{+}}\label{1}\tag{1} \end{align}

Note that

\begin{align} {\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}} & = {\Gamma\pars{\mu + 1} \over \pi/\braces{\Gamma\pars{n - \mu}\sin\pars{\pi\bracks{n - \mu}}}} = {\pars{-1}^{n + 1} \over \pi}\,\sin\pars{\pi\mu} \Gamma\pars{\mu + 1}\Gamma\pars{n - \mu} \\[5mm] & = \pars{-1}^{n + 1}\,\mu \braces{\Gamma\pars{1}\Gamma\pars{n} + \bracks{\Gamma'\pars{1}\Gamma\pars{n} - \Gamma\pars{1}\Gamma'\pars{n}}\mu} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu + \pars{-1}^{n + 1}\bracks{-\gamma\pars{n - 1}! -\pars{n - 1}!\,\Psi\pars{n}}\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu - \pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}\,\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \end{align}

such that $\ds{\left.\partiald[2]{}{\mu}{\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}}\right\vert_{\ \mu\ =\ 0^{+}} = -2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}$

Expression \eqref{1} becomes \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = -\sum_{n = 1}^{\infty}{1 \over n!}\, \bracks{-2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}\, {\pars{-1}^{n} \over n^{2}} = -2\sum_{n = 1}^{\infty}{H_{n - 1} \over n^{3}} \\[5mm] & = -2\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} + 2\sum_{n = 1}^{\infty}{1 \over n^{4}} = -2\pars{\pi^{4} \over 72} + 2\zeta\pars{4} = -\,{5 \over 2}\,\zeta\pars{4} + 2\zeta\pars{4} = \bbx{-\,{1 \over 2}\,\zeta\pars{4}} \end{align}

Note that $\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} = {\pi^{4} \over 72} = {5 \over 4}\,\zeta\pars{4}}$ is a well known identity. See expression $\pars{19}$ in this page.

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$\ds{\Huge\left. b\right)}$ \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = {1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x} \\[5mm] + &\ {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x\ +\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 - x} \over x}\,\dd x} _{\ds{-\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x}}\label{2}\tag{2} \end{align}

Note that

\begin{align} {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x = {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x = \int_{0}^{1}\ln\pars{1 - x}{\ln^{2}\pars{x} \over x}\,\dd x = -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \end{align} such that \eqref{2} is reduced to \begin{align} \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x & = \underbrace{{1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x}} _{\ds{\mc{I}_{1}}} \\[5mm] & -\ \underbrace{2\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x} _{\ds{\mc{I}_{2}}}\ =\ \mc{I}_{1} - \,\mc{I}_{2}\label{3}\tag{3} \end{align}

Hereafter, I'll evaluate $\ds{\,\mc{I}_{1}}$ and $\ds{\,\mc{I}_{2}}$.

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$\ds{\large\quad\mc{I}_{1}:\ ?}$. \begin{align} \mc{I}_{1} & \equiv {1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\bracks{% {1 \over 3}\int_{\epsilon}^{1}\ln^{3}\pars{x \over 1 - x}\,{\dd x \over x} - {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \end{align} In the RHS first integral, lets make the change $\ds{{x \over 1 - x} \mapsto x}$: \begin{align} \mc{I}_{1} & = \lim_{\epsilon \to 0^{+}}\bracks{% {1 \over 3}\int_{\epsilon/\pars{1 - \epsilon}}^{\infty} {\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x - {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\left[% -\,{1 \over 3}\int_{\epsilon}^{\epsilon/\pars{1 - \epsilon}} {\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x + {1 \over 3}\int_{\epsilon}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}} \,\dd x\right. \\[5mm] & \left.\phantom{= \lim_{\epsilon \to 0^{+}}\left[\right]}- {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x\right] \end{align} The RHS first integral vanishes out when $\ds{\epsilon \to 0^{+}}$: \begin{align} \mc{I}_{1} & = {1 \over 3}\int_{0}^{1}\ln^{3}\pars{x} \bracks{{1 \over x\pars{1 + x}} - {1 \over x}}\,\dd x + {1 \over 3}\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x \\[5mm] & = -\,{1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x + {1 \over 3}\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x \end{align} In the second integral, lets $\ds{x \mapsto 1/x}$: \begin{align} \mc{I}_{1} & = -\,{1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x + {1 \over 3}\int_{1}^{0}{\ln^{3}\pars{1/x} \over \pars{1/x}\pars{1 + 1/x}} \pars{-\,{1 \over x^{2}}}\,\dd x \\[5mm] & = -\,{2 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x = {2 \over 3}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x = 2\int_{0}^{-1}\ln\pars{1 - x}\,{\ln^{2}\pars{-x} \over x}\,\dd x \\[5mm] & = -2\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x = 4\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\ln\pars{-x}\,\dd x = -4\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & = -4\,\underbrace{\mrm{Li}_{4}\pars{-1}}_{\ds{-\,{7 \over 8}\,\zeta\pars{4}}} \implies \bbx{\,\mc{I}_{1} = {7 \over 2}\,\zeta\pars{4}}\label{4}\tag{4} \end{align}

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$\ds{\large\quad\mc{I}_{2}:\ ?}$. \begin{align} \mc{I}_{2} & \equiv 2\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x = -4\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln\pars{x}\,\dd x = 4\int_{0}^{1}\mrm{Li}_{4}\pars{x}\,\dd x = 4\mrm{Li}_{4}\pars{1} \\[5mm] & \implies \bbx{\,\mc{I}_{2} = 4\zeta\pars{4}}\label{5}\tag{5} \end{align}

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With \eqref{3}, \eqref{4} and \eqref{5}: \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = {7 \over 2}\,\zeta\pars{4} - 4\zeta\pars{4} \end{align} $$ \bbox[#ffe,15px,border:1px dotted navy]{\ds{% \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = -\,{1 \over 2}\,\zeta\pars{4}}} $$

Answer 5

In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be $$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$$ and the well-known Euler sum of $$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$ several proofs for which can be found here.

From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as \begin{align*} \int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx. \end{align*} The integral that appears to the right can be readily found by parts. The result is $$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$$ Thus $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$$

From properties of harmonic numbers, since $$H_n = H_{n - 1} + \frac{1}{n},$$ the integral becomes $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$$ As $$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$ we have $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$$ as expected.

Answer 6

\begin{align} &\int_0^1 \frac{\ln x\ln^2(1-x)}{x}\overset{x\to 1-x}{dx}\\ =& \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-y x}dy\right) dx\\ =&\int_0^1 \frac{1}{1-y}\left(\int_0^1 \frac {\ln^2x}{1-yx}dx -2\text{Li}_3(1)\right) dy\\ =&\>2 \int_0^1\frac{\text{Li}_3(y)}{y}dy + 2 \int_0^1\frac{\text{Li}_3(y)-\text{Li}_3(1)}{1-y}\>\overset{ibp}{dy}\\ = &\>2 \text{Li}_4(1) - \text{Li}_2^2(1)= 2\left(\frac{\pi^2}{90}\right)-\left(\frac{\pi^2}6\right)^2= -\frac{\pi^4}{180}\\ \end{align}

Answer 7

\begin{align}J&=\int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx\\ &\overset{\text{IBP}}=\int_0^1\frac{\ln^2 x\ln(1-x)}{1-x}dx\\ &\overset{\text{IBP}}=-\left[\left(\int_0^x\frac{\ln^2 t}{1-t}dt-\int_0^1\frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1+\\&\int_0^1 \frac{1}{1-x}\left(\int_0^1 \frac{x\ln^2(tx)}{1-tx}dt-\int_0^1\frac{\ln^2 t}{1-t}dt\right)dx\\ &=\int_0^1\int_0^1 \left(\frac{\ln(tx)^2}{(1-t)(1-x)}-\frac{\ln(tx)^2}{(1-t)(1-tx)}-\frac{\ln^2 t}{(1-t)(1-x)}\right)dtdx\\ &\overset{\text{Fubini}}=2\zeta(2)^2+\int_0^1 \int_0^1\left(\frac{\ln^2 x}{(1-t)(1-x)}-\frac{\ln^2(tx)}{(1-t)(1-tx)}\right)dtdx\\ &=2\zeta(2)^2+\int_0^1\left(\frac{1}{1-t}\left(\int_t^1\frac{\ln^2 x}{1-x}dx\right)-\frac{1}{t}\left(\int_0^t\frac{\ln^2 x}{1-x}dx\right)\right)dt\\ &\overset{\text{IBP}}=2\zeta(2)^2-J-6\zeta(4)\\ J&=\boxed{\zeta(2)^2-3\zeta(4)=-\frac{\pi^4}{180}} \end{align}

Answer 8

For $\left| z \right|<1$ we have that $S=\sum\limits_{j=1}^{+\infty }{H_{j}z^{j}}=-\frac{\ln \left( 1-z \right)}{1-z}$

Proof: Expanding the logarithm and the geometric series $$S=-\frac{\ln \left( 1-z \right)}{1-z}=\frac{1}{1-z}\sum\limits_{j=1}^{+\infty }{\frac{z^{j}}{j}}=\frac{1}{1-z}\sum\limits_{j=1}^{+\infty }{\frac{z^{j}}{j}}=\frac{z}{1-z}+\frac{1}{2}\frac{z^{2}}{1-z}+\frac{1}{3}\frac{z^{3}}{1-z}+\frac{1}{4}\frac{z^{4}}{1-z}+...$$

$$=\frac{1}{1-z}\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)=\left( 1+z+z^{2}+... \right)\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)$$

$$=\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)+z\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)+z^{2}\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)+...$$

$$=\left( z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\frac{z^{4}}{4}+... \right)+\left( z^{2}+\frac{z^{3}}{2}+\frac{z^{4}}{3}+\frac{z^{5}}{4}+... \right)+\left( z^{3}+\frac{z^{4}}{2}+\frac{z^{5}}{3}+\frac{z^{6}}{4}+.. \right)+...$$

$$=z+\left( 1+\frac{1}{2} \right)z^{2}+\left( 1+\frac{1}{2}+\frac{1}{3} \right)z^{3}+\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \right)z^{4}+\left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} \right)z^{5}+...$$

$$=H_{1}z+H_{2}z^{2}+H_{3}z^{3}+H_{4}z^{4}+H_{5}z^{5}+...=\sum\limits_{j=1}^{+\infty }{H_{j}z^{j}}$$

Is the first identity in Harmonic numbers http://en.wikipedia.org/wiki/List_of_ma ... cal_series

Aso we know that $$H_{n}=1+\frac{1}{2}+...+\frac{1}{n}=\int\limits_{0}^{1}{\left( 1+x+...+x^{n-1} \right)dx}=\int\limits_{0}^{1}{\frac{1-x^{n}}{1-x}dx}$ with $H_{0}=0$$ by convention. Consider the polylogarithm function http://mathworld.wolfram.com/Polylogarithm.html We know that $\frac{d}{dx}Li_{n}\left( x \right)=\frac{Li_{n-1}\left( x \right)}{x}$

Then $$I=\int\limits_{0}^{1}{\ln x\ln ^{2}\left( 1-x \right)\frac{dx}{x}}=\int\limits_{0}^{1}{\frac{\ln \left( 1-x \right)}{1-x}\ln ^{2}xdx}=-\int\limits_{0}^{1}{\sum\limits_{j=1}^{+\infty }{H_{j}x^{j}}\ln ^{2}xdx}$$

$$=-\sum\limits_{j=1}^{+\infty }{H_{j}\int\limits_{0}^{1}{x^{j}\ln ^{2}xdx}}=-2\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}$$

as $$K\left( j \right)=\int\limits_{0}^{1}{x^{j}dx}=\left( \frac{x^{j+1}}{j+1} \right)\left| _{0}^{1} \right.=\frac{1}{j+1}\Rightarrow K''\left( j \right)=\int\limits_{0}^{1}{x^{j}\ln ^{2}xdx}=-\frac{1}{\left( j+1 \right)^{2}}=\frac{2}{\left( j+1 \right)^{3}}$$

Denote S the sum $S=\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}$ bringing the 0 $$S=\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}=\sum\limits_{j=1}^{+\infty }{\left( \frac{\frac{1}{j+1}-\frac{1}{j+1}+H_{j}}{\left( j+1 \right)^{3}} \right)}=\sum\limits_{j=1}^{+\infty }{\left( \frac{H_{j+1}-\frac{1}{j+1}}{\left( j+1 \right)^{3}} \right)}=\sum\limits_{j=1}^{+\infty }{\left( \frac{H_{j+1}}{\left( j+1 \right)^{3}}-\frac{1}{\left( j+1 \right)^{4}} \right)}$$

$$=\sum\limits_{j=0}^{+\infty }{\left( \frac{H_{j+1}}{\left( j+1 \right)^{3}}-\frac{1}{\left( j+1 \right)^{4}} \right)}=\sum\limits_{j=1}^{+\infty }{\left( \frac{H_{j}}{j^{3}}-\frac{1}{j^{4}} \right)}=-\zeta \left( 4 \right)+\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{j^{3}}}$$

$$=-\zeta \left( 4 \right)+\sum\limits_{j=1}^{+\infty }{\left( \int\limits_{0}^{1}{\frac{1-x^{j}}{1-x}}\frac{1}{j^{3}}dx \right)}=-\zeta \left( 4 \right)+\int\limits_{0}^{1}{\frac{1}{1-x}\sum\limits_{j=1}^{+\infty }{\left( \frac{1-x^{j}}{j^{3}} \right)}}dx$$

$$=-\zeta \left( 4 \right)+\int\limits_{0}^{1}{\frac{1}{1-x}\left( \zeta \left( 3 \right)-\sum\limits_{j=1}^{+\infty }{\frac{x^{j}}{j^{3}}} \right)}dx=-\zeta \left( 4 \right)+\int\limits_{0}^{1}{\frac{1}{1-x}\left( \zeta \left( 3 \right)-Li_{3}\left( x \right) \right)}dx$$

Using integration by parts Let $u=\zeta \left( 3 \right)-Li_{3}\left( x \right)$ and $dv=\frac{dx}{1-x}$ Then $du=-\frac{Li_{2}\left( x \right)}{x}$ and $v=-\log \left| 1-x \right|$

Then, the sum is equal to $$=-\zeta \left( 4 \right)+\left( Li_{3}\left( x \right)-\zeta \left( 3 \right) \right)\log \left| 1-x \right|\left| _{0}^{1} \right.-\int\limits_{0}^{1}{\log \left( 1-x \right)\frac{Li_{2}\left( x \right)}{x}}dx$$

$$=-\zeta \left( 4 \right)-\int\limits_{0}^{1}{Li_{2}\left( x \right)\frac{\log \left( 1-x \right)}{x}}dx$$

Making the following change of variable $$u=Li_{2}\left( x \right)\Rightarrow du=-\frac{\log \left( 1-x \right)}{x}dx$$

$$=-\zeta \left( 4 \right)+\int\limits_{0}^{1}{Li_{2}\left( x \right)\left( Li_{2}\left( x \right) \right)^{'}}dx=-\zeta \left( 4 \right)+\frac{1}{2}\left( Li_{2}^{2}\left( x \right) \right)\left| _{0}^{1} \right.=-\zeta \left( 4 \right)+\frac{1}{2}\left( Li_{2}^{2}\left( 1 \right)-Li_{2}^{2}\left( 0 \right) \right)$$

where $Li_{2}\left( x \right)=\sum\limits_{j=1}^{+\infty }{\frac{x^{j}}{j^{2}}}$ Then $$=-\zeta \left( 4 \right)+\frac{1}{2}\zeta ^{2}\left( 2 \right)=\frac{\pi ^{4}}{72}-\frac{\pi ^{4}}{90}=\frac{\pi ^{4}}{18}\left( \frac{1}{4}-\frac{1}{5} \right)=\frac{\pi ^{4}}{18}\left( \frac{1}{20} \right)=\frac{\pi ^{4}}{360}$$

Finally we conclude $$I=-2\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}=-2\left( -\zeta \left( 4 \right)+\frac{1}{2}\zeta ^{2}\left( 2 \right) \right)=-2\cdot \frac{\pi ^{4}}{360}=-\frac{\pi ^{4}}{180}$$