复杂的对数积分（五）
\[\Large\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x
\]
\(\Large\mathbf{Solution:}\)
Using the following known results:
\[\begin{align*}
\sum_{n=1}^{\infty }\frac{H_n}{n2^{n}}&=\frac{\pi ^{2}}{12}\\
\sum^\infty_{n=1}\frac{H_n}{n^22^n}&=\zeta(3)-\frac{\pi^2}{12}\ln{2}\\
\sum^\infty_{n=1}\frac{H_n}{n^32^n}&={\rm Li}_4\left(\frac{1}{2}\right)+\frac{\pi^4}{720}-\frac{1}{8}\zeta(3)\ln{2}+\frac{1}{24}\ln^4{2}\\
\sum^\infty_{n=1}\frac{H_n}{n^42^n}
&=2{\rm Li}_5\left(\frac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}\\&~~~-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}\\
\sum^\infty_{n=1}\frac{H_n}{n^52^n}
&=3{\rm Li}_6\left(\frac{1}{2}\right)-\frac{1}{2}\zeta(\bar{5},1)-\frac{17\pi^6}{10080}+{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}-\frac{1}{32}\zeta(5)\ln{2}+\frac{\pi^4}{1440}\ln^2{2}\\
&~~~-\frac{1}{4}\zeta^2(3)-\frac{1}{6}\zeta(3)\ln^3{2}+\frac{\pi^2}{12}\zeta(3)\ln{2}+\frac{\pi^2}{144}\ln^4{2}-\frac{1}{240}\ln^6{2}
\end{align*}\]
For the simplicity, let
\[\color{blue}{\mathbf{H}_{m}^{(k)}(x)}=\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^m}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=\sum_{n=1}^\infty H_{n}x^n
\]
Introduce a generating function for the generalized harmonic numbers for \(|x|<1\)
\[\color{blue}{\mathbf{H}^{(k)}(x)}=\sum_{n=1}^\infty H_{n}^{(k)}x^n=\frac{\operatorname{Li}_k(x)}{1-x}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=-\frac{\ln(1-x)}{1-x}
\]
and the following identity
\[H_{n+1}^{(k)}-H_{n}^{(k)}=\frac1{(n+1)^k}\qquad\Rightarrow\qquad H_{n+1}-H_{n}=\frac1{n+1}
\]
Using \(\mathcal{I}\) to denote the integral in question.Start with integration by parts (IBP),we have
\[\begin{align*}
\mathcal{I}&=-\frac{3}{4} \int_{0}^{1}\frac{\ln^4x\ln^2\left ( 1+x \right )}{1+x}\, \mathrm{d}x\\&=-\frac{3}{4}\int_{1}^{2}\frac{\ln^2x\ln^4\left ( 1-x \right )}{x}\, \mathrm{d}x\\
&=-\frac{3}{4}{\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln^4\left ( 1-x \right )\ln^2x}{x}\, \mathrm{d}x}_{\mathcal{I}_1}+3{\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln^3\left ( 1-x \right )\ln^3x}{x}\, \mathrm{d}x}_{\mathcal{I}_2}-\frac{9}{2}{\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln^2\left ( 1-x \right )\ln^4x}{x}\, \mathrm{d}x}_{\mathcal{I}_3}}}}\\
&~~~+3\underbrace{\int_{\frac{1}{2}}^{1}\frac{\ln\left ( 1-x \right )\ln^5x}{x}\, \mathrm{d}x}_{\mathcal{I}_4}-\frac{1}{7}\ln^72
\end{align*}\]
Let us integrating the indefinite form of \(\mathcal{I}_4\),
\[\begin{align*}
\int \frac{\ln\left ( 1-x \right )\ln^5x}{x}&=-\int \sum_{n=1}^{\infty }\frac{x^{n-1}}{n}\ln^5x\, \mathrm{d}x\\
&=-\sum_{n=1}^{\infty }\frac{1}{n}\frac{\partial ^{5}}{\partial n^{5}}\left [ \int x^{n-1}\, \mathrm{d}x \right ]\\
&=-\sum_{n=1}^{\infty }\frac{1}{n}\frac{\partial ^{5}}{\partial n^{5}}\left [ \frac{x^{n}}{n} \right ]\\
&=-\sum_{n=1}^{\infty }\frac{1}{n}\Bigg [ \frac{x^{n}\ln^5x}{n}-\frac{5x^{n}\ln^4x}{n^{2}}+\frac{20x^{n}\ln^3x}{n^{3}}-\frac{60x^{n}\ln^2x}{n^{4}}\\
&~~~~~~~~~~~~~~~+\frac{120x^{n}\ln x}{n^{5}}-\frac{120x^{n}}{n^{6}}\Bigg ]\\
&=\sum_{n=1}^{\infty }\Bigg [ -\frac{x^{n}\ln^5x}{n^{2}}+\frac{5x^{n}\ln^4x}{n^{3}}-\frac{20x^{n}\ln^3x}{n^{4}}+\frac{60x^{n}\ln^2x}{n^{5}}\\
&~~~~~~~~~~~~~~~-\frac{120x^{n}\ln x}{n^{6}}+\frac{120x^{n}}{n^{7}} \Bigg ]\\
&=-\ln^5x\mathrm{Li}_2\left ( x \right )+5\ln^4x\mathrm{Li}_3\left ( x \right )-20\ln^3x\mathrm{Li}_4\left ( x \right )+60\ln^2x\mathrm{Li}_5\left ( x \right )\\
&~~~-120\ln x\mathrm{Li}_6\left ( x \right )+120\mathrm{Li}_7\left ( x \right )
\end{align*}\]
Therefore,
\[\begin{align*}
\color{Teal}{\mathcal{I}_4}&=-\ln^52\mathrm{Li}_2\left ( \frac{1}{2} \right )+5\ln^42\mathrm{Li}_3\left ( \frac{1}{2} \right )-20\ln^32\mathrm{Li}_4\left ( \frac{1}{2} \right )-60\ln^22\mathrm{Li}_5\left ( \frac{1}{2} \right )\\
&~~~-120\ln 2\mathrm{Li}_6\left ( \frac{1}{2} \right )-120\mathrm{Li}_7\left ( \frac{1}{2} \right )\\
&=\color{Teal}{-\frac{1}{3}\ln^72+\frac{1}{3}\pi ^{2}\ln^52-\frac{35}{8}\zeta \left ( 3 \right )\ln^42-20\ln^32\mathrm{Li}_{4}\left ( \frac{1}{2} \right )-60\ln^22\mathrm{Li}_5\left ( \frac{1}{2} \right )}\\
&~~~\color{Teal}{-120\ln 2\mathrm{Li}_6\left ( \frac{1}{2} \right )-120\mathrm{Li}_7\left ( \frac{1}{2} \right )}
\end{align*}\]
Applying IBP again to evaluate \(\mathcal{I}_3\),
\[\mathcal{I}_3=-\frac{1}{5}\ln^72+\frac{2}{5}\underbrace{\color{Red}{\int_{\frac{1}{2}}^{1}\frac{\ln^5x\ln\left ( 1-x \right )}{1-x}\, \mathrm{d}x}}_\mathcal{J}
\]
Using the similar approach as calculating the red integral, then
\[{\small\begin{align*}
\int\frac{\ln^5x\ln\left ( 1-x \right )}{1-x}\, \mathrm{d}x &= -\int \sum_{n=1}^{\infty }H_nx^{n}\ln^5x\, \mathrm{d}x\\
&=-\sum_{n=1}^{\infty }H_n\int x^{n}\ln^5x\, \mathrm{d}x\\
&=-\sum_{n=1}^{\infty }H_n\frac{\partial ^{5}}{\partial n^{5}}\left [ \int x^{n}\, \mathrm{d}x \right ]\\
&=-\sum_{n=1}^{\infty }H_n\frac{\partial ^{5}}{\partial n^{5}}\left [ \frac{x^{n+1}}{n+1} \right ]\\
&=-\sum_{n=1}^{\infty }H_n\Bigg [ \frac{x^{n+1}\ln^5x}{n+1}-\frac{5x^{n+1}\ln^4x}{\left (n+1 \right )^{2}}+\frac{20x^{n+1}\ln^3x}{\left (n+1 \right )^{3}}-\frac{60x^{n+1}\ln^2x}{\left (n+1 \right )^{4}} \\
&~~~~~~~~~~~~~~~+\frac{120x^{n+1}\ln x}{\left (n+1 \right )^{5}}-\frac{120x^{n+1}}{\left (n+1 \right )^{6}}\Bigg ]\\
&=-\ln^5x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{n+1}+\ln^5x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{2}}+5\ln^4x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{\left ( n+1 \right )^{2}}\\
&~~~-5\ln^4x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{3}}-20\ln^3x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{\left ( n+1 \right )^{3}}+20\ln^3x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{4}}\\
&~~~+60\ln^2x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{\left ( n+1 \right )^{4}}-60\ln^2x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{5}}-120\ln x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{\left ( n+1 \right )^{5}}\\
&~~~+120\ln x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{6}}+120\ln^2x\sum_{n=1}^{\infty }\frac{H_{n+1}x^{n+1}}{\left ( n+1 \right )^{6}}-120\ln x\sum_{n=1}^{\infty }\frac{x^{n+1}}{\left ( n+1 \right )^{7}}\\
&=-\sum_{n=1}^{\infty }\Bigg [ \frac{H_nx^{n}\ln^5x}{n}-\frac{x^{n}\ln^5x}{n^{2}}-\frac{5H_nx^{n}\ln^4x}{n^{2}}+\frac{5x^{n}\ln^4x}{n^{3}}\\
&~~~~~~~~~+\frac{20H_nx^{n}\ln^3x}{n^{3}}-\frac{20x^{n}\ln^3x}{n^{4}}-\frac{60H_nx^{n}\ln^2x}{n^{4}}+\frac{60x^{n}\ln^2x}{n^{5}}\\
&~~~~~~~~~+\frac{120H_nx^{n}\ln x}{n^5}-\frac{120x^{n}\ln x}{n^{6}}-\frac{120H_nx^{n}}{n^{6}}-\frac{120x^{n}}{n^{7}} \Bigg ]\\
&=-\color{blue}{\mathbf{H}_{1}\left ( x \right )}\ln^5x+\mathrm{Li}_{2}\left ( x \right )\ln^5x+5\color{Blue}{\mathbf{H}_{2}\left ( x \right )}\ln^4x-5\mathrm{Li}_{3}\left ( x \right )\ln^4x-20\color{Blue}{\mathbf{H}_{3}\left ( x \right )}\ln^3x\\
&~~~+20\mathrm{Li}_{4}\left ( x \right )\ln^3x+60\color{Blue}{\mathbf{H}_4\left ( x \right )}\ln^2x-60\mathrm{Li}_{5}\left ( x \right )\ln^2x-120\color{Blue}{\mathbf{H}_5\left ( x \right )}\ln x\\
&~~~+120\mathrm{Li}_6\left ( x \right )\ln x+120\color{blue}{\mathbf{H}_6\left ( x \right )}-120\mathrm{Li}_7\left ( x \right )
\end{align*}}\]
Hence
\[\begin{align*}
\mathcal{J}&=-\color{blue}{\mathbf{H}_1\left ( \frac{1}{2} \right )}\ln^52+\mathrm{Li}_2\left ( \frac{1}{2} \right )\ln^52-5\color{blue}{\mathbf{H}_2\left ( \frac{1}{2} \right )}\ln^42+5\mathrm{Li}_3\left ( \frac{1}{2} \right )\ln^42\\
&~~~-20\color{blue}{\mathbf{H}_3\left ( \frac{1}{2} \right )}\ln^32+20\mathrm{Li}_4\left ( \frac{1}{2} \right )\ln^32-60\color{blue}{\mathbf{H}_4\left ( \frac{1}{2} \right )}\ln^22+60\mathrm{Li}_5\left ( \frac{1}{2} \right )\ln^22\\
&~~~-120\color{blue}{\mathbf{H}_5\left ( \frac{1}{2} \right )}\ln2+120\mathrm{Li}_6\left ( \frac{1}{2} \right )\ln2-120\color{blue}{\mathbf{H}_6\left ( \frac{1}{2} \right )}+120\mathrm{Li}_7\left ( \frac{1}{2} \right )\\
&=-2\ln^72-\frac{225}{8}\zeta \left ( 3 \right )\ln^42+\frac{1}{18}\pi ^{4}\ln^22-60\ln^22\mathrm{Li}_5\left ( \frac{1}{2} \right )\\
&~~~-60\ln^32\mathrm{Li}_4\left ( \frac{1}{2} \right )-\frac{15}{8}\ln^22\zeta \left ( 5 \right )+5\pi ^{3}\ln^22\zeta \left ( 3 \right )+\frac{5}{3}\pi ^{2}\ln^52\\
&~~~-120\ln 2\color{blue}{\mathbf{H}_5\left ( \frac{1}{2} \right )}+120\ln2\mathrm{Li}_6\left ( \frac{1}{2} \right )--120\color{blue}{\mathbf{H}_6\left ( \frac{1}{2} \right )}+120\mathrm{Li}_7\left ( \frac{1}{2} \right )
\end{align*}\]
Combine these all together,we get
\[{\small\boxed{\displaystyle \begin{align*}
\int_{0}^{1}\frac{\ln^{3}\left ( 1+x \right )\ln^{3}x}{x}\mathrm{d}x =& \color{Teal} {-\int_{\frac{1}{2}}^{1}\frac{\ln^{3}x\ln^{3}\left ( 1-x \right )}{1-x}\mathrm{d}x+3\int_{\frac{1}{2}}^{1}\frac{\ln^{3}\left ( 1-x \right )\ln^{3}x}{x}\mathrm{d}x}\\
&\color{Teal} {+\frac{15}{4}\ln^{7}2+\frac{75}{2}\zeta \left ( 3 \right )\ln^{4}2-\frac{1}{10}\pi ^{4}\ln^{3}2-2\pi ^{2}\ln^{5}2+\frac{27}{8}\zeta \left ( 5 \right )\ln^22}\\
&\color{Teal} {-9\pi ^{3}\zeta \left ( 3 \right )\ln^22-12\ln^22\mathrm{Li}_{5}\left ( \frac{1}{2} \right )+48\ln^32\mathrm{Li}_{3}\left ( \frac{1}{2} \right )-576\ln2\mathrm{Li}_{6}\left ( \frac{1}{2} \right )}\\
&\color{Teal} {+216\ln2\sum_{n=1}^{\infty }\frac{H_n}{n^{5}2^{n}}+216\sum_{n=1}^{\infty }\frac{H_n}{n^{6}2^{n}}-576\mathrm{Li}_{7}\left ( \frac{1}{2} \right )}
\end{align*}}}\]
In order to use the Beta function,we make the following transformations
\[\begin{align*}
&-\int_{\frac{1}{2}}^{1}\frac{\ln^{3}x\ln^{3}\left ( 1-x \right )}{1-x}\mathrm{d}x+3\int_{\frac{1}{2}}^{1}\frac{\ln^{3}\left ( 1-x \right )\ln^{3}x}{x}\mathrm{d}x\\
=&-\int_{0}^{1}\frac{\ln^3x\ln^{3}\left ( 1-x \right )}{1-x}\mathrm{d}x+4\int_{\frac{1}{2}}^{1}\frac{\ln^{3}\left ( 1-x \right )\ln^{3}x}{x}\mathrm{d}x
\end{align*}\]
And using the defination of Nielsen Generalized Polylogarithm Function
\[S_{n,p}\left ( z \right )=\frac{\left ( -1 \right )^{n+p-1}}{\left ( n-1 \right )!p!}\int_{0}^{1}\frac{\ln^{n-1}t\ln^p\left ( 1-zt \right )}{t}\mathrm{d}t
\]
and
\[\sum_{m=1}^{\infty }\frac{z^{m}H_m}{m^{n+1}}=S_{n,2}\left ( z \right )+\mathrm{Li}_{n+2}\left ( z \right )
\]
Hence we get
\[{\small\boxed{\displaystyle \begin{align*}
\int_{0}^{1}\frac{\ln^{3}\left ( 1+x \right )\ln^{3}x}{x}\mathrm{d}x =& \color{Teal} {-\frac{\partial ^{6}}{\partial a^{3}\partial b^{3}}\mathrm{B}\left ( 1,0^+ \right )+4\int_{\frac{1}{2}}^{1}\frac{\ln^{2}\left ( 1-x \right )\ln^{4}x}{1-x}\mathrm{d}x}\\
&\color{Teal} {+\frac{77}{20}\ln^{7}2+\frac{75}{2}\zeta \left ( 3 \right )\ln^{4}2-\frac{51}{140}\pi ^{6}\ln2+\frac{1}{20}\pi ^{4}\ln^{3}2-\frac{1}{2}\pi ^{2}\ln^{5}2}\\
&\color{Teal} {-\frac{27}{8}\zeta \left ( 5 \right )\ln^22-9\pi ^{3}\zeta \left ( 3 \right )\ln^22+18\pi ^{2}\zeta \left ( 3 \right )\ln^22-36\zeta \left ( 3 \right )\ln^42}\\
&\color{Teal} {-54\zeta ^{2}\left ( 3 \right )\ln2-108\ln2\zeta \left ( \bar{5},1 \right )+204\ln^22\mathrm{Li}_{5}\left ( \frac{1}{2} \right )+48\ln^32\mathrm{Li}_{3}\left ( \frac{1}{2} \right )}\\
&\color{Teal} {-72\ln2\mathrm{Li}_{6}\left ( \frac{1}{2} \right )+216S_{5,2}\left ( \frac{1}{2} \right )-360\mathrm{Li}_{7}\left ( \frac{1}{2} \right )}
\end{align*}}}\]
Let
\[\mathbf{O}=\frac{\partial ^{6}}{\partial a^{3}\partial b^{3}}\mathrm{B}\left ( 1,0^+ \right )~~,~~\mathbf{P}=\frac{\partial ^{6}}{\partial a^{4}\partial b^{2}}\mathrm{B}\left ( 1,0^+ \right )~~,~~\mathbf{Q}=\frac{\partial ^{6}}{\partial a^{2}\partial b^{4}}\mathrm{B}\left ( 1,0^+ \right )
\]
For \(\mathcal{I}_1\)
\[\mathcal{I}_1=-\frac{1}{4}\ln^72+\frac{4}{3}\underbrace{\color{Red} {\int_{\frac{1}{2}}^{1}\frac{\ln^3x\ln^3\left ( 1-x \right )}{1-x}\mathrm{d}x}}_\mathcal{M}
\]
For \(\mathcal{I}_2\)
\[\mathcal{I}_2=\frac{1}{4}\ln^72+\frac{3}{4}\underbrace{\color{Red} {\int_{\frac{1}{2}}^{1}\frac{\ln^4x\ln^2\left ( 1-x \right )}{1-x}\mathrm{d}x}}_\mathcal{N}
\]
Therefore,
\[\begin{align*}
&4\int_{\frac{1}{2}}^{1}\frac{\ln^4x\ln^2\left ( 1-x \right )}{1-x}\mathrm{d}x+4\mathcal{I}_1-4\mathcal{I}_1\\
&=4\mathbf{P}-4\mathcal{I}_1\\
&=4\mathbf{P}+\frac{4}{3}\ln^72-\frac{16}{3}\mathcal{M}-\frac{16}{3}\mathcal{I}_2\\
&=4\mathbf{P}+\frac{4}{3}\ln^72-\frac{16}{3}\mathbf{O}+\frac{16}{3}\mathcal{I}_2\\
&=4\mathbf{P}+\frac{8}{3}\ln^72-\frac{16}{3}\mathbf{O}-\frac{3}{4}\mathbf{Q}+\frac{3}{4}\mathcal{I}_3\\
&=4\mathbf{P}+\frac{151}{60}\ln^72-\frac{16}{3}\mathbf{O}-\frac{3}{4}\mathbf{Q}+\frac{3}{10}\mathcal{J}
\end{align*}\]
With the help of Mathematica,we can easily get the value of \(\mathbf{O},\mathbf{P},\mathbf{Q}\),and combine with the previous results,we get the Final Result:
\[{\small\boxed{\displaystyle \color{Blue}{\begin{align*}
\color{black}{\int_{0}^{1}\frac{\ln^{3}\left ( 1+x \right )\ln^{3}x}{x}\mathrm{d}x =}&~\frac{1}{5}\pi ^{4}\zeta \left ( 3 \right )+\frac{1}{4}\pi ^{2}\zeta \left ( 5 \right )-\frac{3}{20}\zeta \left ( 7 \right )+42\ln^32\zeta \left ( 3 \right )+8\ln^62\\
&-4\pi ^{2}\ln^32+\frac{71}{12}\ln^{7}2+\frac{87}{2}\zeta \left ( 3 \right )\ln^{4}2-\frac{17}{56}\pi ^{6}\ln2+\frac{1}{24}\pi ^{4}\ln^{3}2\\
&-\frac{5}{4}\pi ^{2}\ln^{5}2-\frac{45}{16}\zeta \left ( 5 \right )\ln^22-\frac{21}{2}\pi ^{3}\zeta \left ( 3 \right )\ln^22+15\pi ^{2}\zeta \left ( 3 \right )\ln^22\\
&-\frac{711}{16}\zeta \left ( 3 \right )\ln^42-45\zeta ^{2}\left ( 3 \right )\ln2-90\ln2\zeta \left ( \bar{5},1 \right )+150\ln^22\mathrm{Li}_{5}\left ( \frac{1}{2} \right )\\
&-18\ln^32\mathrm{Li}_{4}\left ( \frac{1}{2} \right )-216\ln2\mathrm{Li}_{6}\left ( \frac{1}{2} \right )+180S_{5,2}\left ( \frac{1}{2} \right )-360\mathrm{Li}_{7}\left ( \frac{1}{2} \right )
\end{align*}}}}\]
