一个很火的积分题

\[S=\sum\limits_{j=1}^{+\infty }{H_{j}z^{j}}=-\frac{\ln \left( 1-z \right)}{1-z}.\]

Expanding the logarithm and the geometric series

\begin{align*}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}\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}}.\end{align*}

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 $\displaystyle H_0=0$ by convention. Consider the polylogarithm function,We know that

\[\frac{d}{dx}Li_{n}\left( x \right)=\frac{Li_{n-1}\left( x \right)}{x}.\]

Then

\begin{align*}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}}}\end{align*}

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$\displaystyle S=\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}.$

\begin{align*}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.\end{align*}

Using integration by parts

Let $\displaystyle u=\zeta \left( 3 \right)-Li_{3}\left( x \right)$ and $\displaystyle dv=\frac{dx}{1?x}$.Then$\displaystyle du=-\frac{Li_{2}\left( x \right)}{x}$and$\displaystyle v=-\log \left| 1-x \right|$.So, the sum is equal to

\begin{align*}S&=-\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.\end{align*}

Making the following change of variable

\begin{align*}u&=Li_{2}\left( x \right)\Rightarrow du=-\frac{\log \left( 1-x \right)}{x}dx\\S&=-\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).\end{align*}

where $\displaystyle Li_{2}\left( x \right)=\sum\limits_{j=1}^{+\infty }{\frac{x^{j}}{j^{2}}}.$Then

\[S=-\zeta \left( 4 \right)+\frac{1}{2}\zeta ^{2}\left( 2 \right)=\frac{\pi ^{4}}{72}-\frac{\pi ^{4}}{90}=\frac{\pi ^{4}}{18}\left( \frac{1}{4}-\frac{1}{5} \right)=\frac{\pi ^{4}}{18}\left( \frac{1}{20} \right)=\frac{\pi ^{4}}{360}.\]

Finally we conclude

\[I=-2\sum\limits_{j=1}^{+\infty }{\frac{H_{j}}{\left( j+1 \right)^{3}}}=-2\left( -\zeta \left( 4 \right)+\frac{1}{2}\zeta ^{2}\left( 2 \right) \right)=-2\cdot \frac{\pi ^{4}}{360}=-\frac{\pi ^{4}}{180}.\]