复杂的对数积分（二）
\[\Large\int_{0}^{1}\frac{\ln^{2}\left ( x \right )\mathrm{Li}_{2}\left ( x \right )}{1-x}\mathrm{d}x=-11\zeta \left ( 5 \right )+6\zeta \left ( 3 \right )\zeta \left ( 2 \right )
\]
\(\Large\mathbf{Proof:}\)
\[\begin{align*}
\int_{0}^{1}\frac{\ln^{2}\left ( x \right )\mathrm{Li}_{2}\left ( x \right )}{1-x}\mathrm{d}x &=\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\sum_{n=1}^{\infty }\frac{x^{n}}{n^{2}}\mathrm{d}x \\
&=\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\left [ \sum_{n=1}^{\infty }\frac{1}{n^{2}}-\sum_{n=1}^{\infty }\frac{1-x^{n}}{n^{2}} \right ]\mathrm{d}x \\
&=\zeta \left ( 2 \right )\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\mathrm{d}x-\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\sum_{n=1}^{\infty }\frac{1-x^{n}}{n^{2}}\mathrm{d}x
\end{align*}\]
However,
\[\begin{align*}
\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\mathrm{d}x&=\int_{0}^{1}\ln\left ( 1-x \right )\left [ 2\ln\left ( x \right )\frac{1}{x} \right ]\mathrm{d}x=-2\int_{0}^{1}\mathrm{Li}{}'_{2}\left ( x \right )\ln\left ( x \right )\mathrm{d}x \\
&=2\int_{0}^{1}\mathrm{Li}_{2}\left ( x \right )\frac{1}{x}\mathrm{d}x=2\int_{0}^{1}\mathrm{Li}{}'_{3}\left ( x \right )\mathrm{d}x=2\mathrm{Li}_{3}\left ( 1 \right )=2\zeta \left ( 3 \right )
\end{align*}\]
Such that
\[\int_{0}^{1}\frac{\ln^{2}\left ( x \right )\mathrm{Li}_{2}\left ( x \right )}{1-x}\mathrm{d}x=2\zeta \left ( 2 \right )\zeta \left ( 3 \right )-\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\sum_{n=1}^{\infty }\frac{1-x^{n}}{n^{2}}\mathrm{d}x
\]
Also
\[\begin{align*}
\int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\sum_{n=1}^{\infty }\frac{1-x^{n}}{n^{2}}\mathrm{d}x &=\sum_{n=1}^{\infty }\frac{1}{n^{2}}\int_{0}^{1}\ln^{2}\left ( x \right )\frac{1-x^{n}}{1-x}\mathrm{d}x \\
&=\sum_{n=1}^{\infty }\frac{1}{n^{2}}\int_{0}^{1}\ln^{2}\left ( x \right )\sum_{k=1}^{n}x^{k-1}\mathrm{d}x \\
&= \sum_{n=1}^{\infty }\frac{1}{n^{2}}\sum_{k=1}^{n}\underset{=\displaystyle \frac{2}{k^{3}}}{\underbrace{\int_{0}^{1}\ln^{2}\left ( x \right )x^{k-1}\mathrm{d}x}}=2\sum_{n=1}^{\infty }\frac{H_{n}^{\left ( 3 \right )}}{n^{2}}
\end{align*}\]
The last sum can be evaluated with the generating function\(\displaystyle \sum_{n=1}^{\infty }x^{n}H_{n}^{\left ( 3 \right )}=\frac{\mathrm{Li}_{3}\left ( x \right )}{1-x}\).Namely
\[\begin{align*}
\sum_{n=1}^{\infty }\frac{x^{n}}{n}H_{n}^{\left ( 3 \right )}&=\int_{0}^{x}\frac{\mathrm{Li}_{3}\left ( t \right )}{t}\mathrm{d}t+\int_{0}^{x}\frac{\mathrm{Li}_{3}\left ( t \right )}{1-t}\mathrm{d}t \\
&=\mathrm{Li}_{4}\left ( x \right )-\ln\left ( 1-x \right )\mathrm{Li}_{3}\left ( x \right )+\int_{0}^{x}\ln\left ( 1-t \right )\mathrm{Li}{}'_{3}\left ( t \right )\mathrm{d}t\\
&=\mathrm{Li}_{4}\left ( x \right )-\ln\left ( 1-x \right )\mathrm{Li}_{3}\left ( x \right )+\int_{0}^{x}\ln\left ( 1-t \right )\frac{\mathrm{Li}_{2}\left ( t \right )}{t}\mathrm{d}t \\
&=\mathrm{Li}_{4}\left ( x \right )-\ln\left ( 1-x \right )\mathrm{Li}_{3}\left ( x \right )+\int_{0}^{x}\mathrm{Li}_{2}\left ( t \right )\mathrm{Li}{}'_{2}\left ( t \right )\mathrm{d}t \\
&=\mathrm{Li}_{4}\left ( x \right )-\ln\left ( 1-x \right )\mathrm{Li}_{3}\left ( x \right )-\frac{1}{2}\mathrm{Li}_{2}^{2}\left ( x \right )
\end{align*}\]
\[\begin{align*}
\sum_{n=1}^{\infty }\frac{H_{n}^{\left ( 3 \right )}}{n^{2}}&=\int_{0}^{1}\frac{\mathrm{Li}_{4}\left ( t \right )}{t}\mathrm{d}t-\int_{0}^{1}\frac{\ln\left ( 1-t \right )\mathrm{Li}_{3}\left ( t \right )}{t}\mathrm{d}t-\frac{1}{2}\int_{0}^{1}\frac{\mathrm{Li}_{2}^{2}\left ( t \right )}{t}\mathrm{d}t \\
&= \zeta \left ( 5 \right )+\mathrm{Li}_{2}\left ( 1 \right )\mathrm{Li}_{3}\left ( 1 \right )-\int_{0}^{1}\mathrm{Li}_{2}\left ( t \right )\frac{\mathrm{Li}_{2}\left ( t \right )}{t}\mathrm{d}t-\frac{1}{2}\int_{0}^{1}\frac{\mathrm{Li}_{2}^{2}\left ( t \right )}{t}\mathrm{d}t\\
&=\zeta \left ( 5 \right )+\zeta \left ( 2 \right )\zeta \left ( 3 \right ) -\frac{3}{2}\int_{0}^{1}\frac{\mathrm{Li}_{2}^{2}\left ( t \right )}{t}\mathrm{d}t
\end{align*}\]
The last integral integrate once by parts,we find
\[\begin{align*}
I &=\int_{0}^{1}\frac{\mathrm{Li}_{2}^{2}\left ( x \right )}{x}\mathrm{d}x=\int_{0}^{1}\mathrm{Li}_{2}\left ( x \right )\mathrm{d}\left ( \mathrm{Li}_{3}\left ( x \right ) \right ) \\
&=\mathrm{Li}_{2}\left ( 1 \right )\mathrm{Li}_{3}\left ( 1 \right )+\int_{0}^{1}\frac{\mathrm{Li}_{3}\left ( x \right )\ln\left ( 1-x \right )\mathrm{d}x}{x} \\
&=\zeta \left ( 2 \right ) \zeta \left ( 3 \right )-\sum_{n=1}^{\infty }\sum_{k=1}^{\infty }\frac{1}{n^{3}k\left ( n+k \right )}
\end{align*}\]
where the last line is obtained by expanding \(\displaystyle \mathrm{Li}_{3}\left ( x \right )\) and \(\ln\left ( 1-x \right )\) into series and integrating. Now if we donote\(\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}\) the \(n\)th harmonic number, the sum with respect to \(k\) can be written as$$\sum_{k=1}^{\infty }\frac{1}{k\left ( n+k \right )}=\frac{1}{n}\sum_{k=1}^{\infty }\left ( \frac{1}{k}-\frac{1}{n+k} \right )=\frac{H_{n}}{n}$$
so that$$I=\zeta \left ( 2 \right )\zeta \left ( 3 \right )-\sum_{n=1}^{\infty }\frac{H_{n}}{n^{4}}=2\zeta \left ( 2 \right )\zeta \left ( 3 \right )-3\zeta \left ( 5 \right )$$
so we have
\[\sum_{n=1}^{\infty }\frac{H_{n}^{\left ( 3 \right )}}{n^{2}}=\frac{11}{2}\zeta \left ( 5 \right )-2\left ( 2 \right )\zeta \left ( 3 \right )
\]
\[\Rightarrow \int_{0}^{1}\frac{\ln^{2}\left ( x \right )}{1-x}\sum_{n=1}^{\infty }\frac{1-x^{n}}{n^{2}}\mathrm{d}x=11\zeta \left ( 5 \right )-4\zeta \left ( 2 \right )\zeta \left ( 3 \right )
\]
Now we have the final result:
\[\boxed{\displaystyle \color{blue}{\int_{0}^{1}\frac{\ln^{2}\left ( x \right )\mathrm{Li}_{2}\left ( x \right )}{1-x}\mathrm{d}x=-11\zeta \left ( 5 \right )+6\zeta \left ( 3 \right )\zeta \left ( 2 \right )}}
\]
