复杂的对数积分（四）
\[\Large\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^2x}x\mathrm{d}x
\]
\(\Large\mathbf{Solution:}\)
I will be using the following results:
\[2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)
\]
\[\sum^\infty_{n=1}\frac{H_n}{n^22^n}=\zeta(3)-\frac{\pi^2}{12}\ln{2}
\]
\[\sum^\infty_{n=1}\frac{H_n}{n^32^n}={\rm Li}_4\left(\frac{1}{2}\right)+\frac{\pi^4}{720}-\frac{1}{8}\zeta(3)\ln{2}+\frac{1}{24}\ln^4{2}
\]
\[\begin{align*}
\sum^\infty_{n=1}\frac{H_n}{n^42^n}
=&2{\rm Li}_5\left(\frac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}\\&-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}
\end{align*}\]
Using \(I\) to denote the integral in question,
\[\begin{align*}
\mathcal{I}
&=-\int^1_0\frac{\ln^3{x}\ln^2(1+x)}{1+x}{\rm d}x\\&=-\int^2_1\frac{\ln^2{x}\ln^3(x-1)}{x}{\rm d}x\\
&=\underbrace{-\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x}_{\mathcal{I}_1}\underbrace{+3\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln^2(1-x)}{x}}_{\mathcal{I}_2}\underbrace{-3\int^1_{\frac{1}{2}}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x}_{\mathcal{I}_3}-\frac{1}{6}\ln^6{2}
\end{align*}\]
For \(\mathcal{I}_1\), integration by parts gives
\[\mathcal{I}_1=\frac{1}{3}\ln^6{2}-\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x
\]
On the other hand, \(x\mapsto1-x\) yields
\[\mathcal{I}_1=-\int^\frac{1}{2}_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x
\]
Combining these two equalities, we have
\[\begin{align*}
\mathcal{I}_1
&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\int^1_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x\\&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\frac{\partial^5\beta}{\partial a^3\partial b^2}(1,0^+)\\
&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\left[\frac{1}{b}+\mathcal{O}(1)\right]\left[\left(12\zeta^2(3)-\frac{23\pi^6}{1260}\right)b+\mathcal{O}(b^2)\right]_{b=0}\\
&=\frac{23\pi^6}{2520}-6\zeta^2(3)+\frac{1}{6}\ln^6{2}
\end{align*}\]
Even with the help of Wolfram Alpha, evaluating that fifth derivative was horribly unpleasant to say the least. As for \(\mathcal{I}_2\),
\[\begin{align*}
\mathcal{I}_2&=6\sum^\infty_{n=1}\frac{H_n}{n+1}\int^1_\frac{1}{2}x^n\ln^3{x}\ {\rm d}x\\&=6\sum^\infty_{n=1}\frac{H_n}{n+1}\frac{\partial^3}{\partial n^3}\left(\frac{1}{n+1}-\frac{1}{(n+1)2^{n+1}}\right)\\
&=\color{Magenta}{-\sum^\infty_{n=1}\frac{36H_n}{(n+1)^5}}\, \color{Black}{+}\, \color{Orange}{\sum^\infty_{n=1}\frac{36H_n}{(n+1)^52^{n+1}}}\, \color{Black}{+}\, \color{Green}{\sum^\infty_{n=1}\frac{36\ln{2}H_n}{(n+1)^42^{n+1}}}\, \color{Black}{+}\, \color{Cyan}{\sum^\infty_{n=1}\frac{18\ln^2{2}H_n}{(n+1)^32^{n+1}}}\\&~~~+\color{Purple}{\sum^\infty_{n=1}\frac{6\ln^3{2}H_n}{(n+1)^22^{n+1}}}\\
&=\color{Magenta}{-\frac{\pi^6}{35}+18\zeta^2(3)}+\color{Orange}{\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\frac{1}{2}\right)}+\color{Green}{36{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}}\\
&~~~+\color{Green}{36{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{20}\ln^2{2}+18\zeta(3)\ln^3{2}-3\pi^2\zeta(3)\ln{2}-\pi^2\ln^4{2}+\frac{9}{10}\ln^6{2}}\\
&~~~+\color{Cyan}{\frac{\pi^4}{40}\ln^2{2}-\frac{9}{4}\zeta(3)\ln^3{2}+\frac{3}{4}\ln^6{2}}+\color{Purple}{\frac{3}{4}\zeta(3)\ln^3{2}-\ln^6{2}}\\
&=\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\frac{1}{2}\right)-\frac{\pi^6}{35}+36{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}+36{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}\\
&~~~-\frac{\pi^4}{40}\ln^2{2}+18\zeta^2(3)-3\pi^2\zeta(3)\ln{2}+\frac{33}{2}\zeta(3)\ln^3{2}-\pi^2\ln^4{2}+\frac{13}{20}\ln^6{2}
\end{align*}\]
For \(\mathcal{I}_3\),
\[\begin{align*}
\mathcal{I}_3
=&3\sum^\infty_{n=1}\frac{1}{n}\int^1_\frac{1}{2}x^{n-1}\ln^4{x}\ {\rm d}x\\
=&3\sum^\infty_{n=1}\frac{1}{n}\frac{\partial^4}{\partial n^4}\left(\frac{1}{n}-\frac{1}{n2^n}\right)\\
=&\sum^\infty_{n=1}\left(\frac{72}{n^6}-\frac{72}{n^62^n}-\frac{72\ln{2}}{n^52^n}-\frac{36\ln^2{2}}{n^42^n}-\frac{12\ln^3{2}}{n^32^n}-\frac{3\ln^4{2}}{n^22^n}\right)\\
=&-72{\rm Li}_6\left(\frac{1}{2}\right)+\frac{8\pi^6}{105}-72{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}-36{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}\\&-\frac{21}{2}\zeta(3)\ln^3{2}+\frac{3\pi^2}{4}\ln^4{2}-\frac{1}{2}\ln^6{2}
\end{align*}\]
Thus
\[\boxed{\begin{align*}
\int_0^1\frac{\ln^3(1+x)\,\ln^2x}x\mathrm{d}x
=&~\color{blue}{36\sum^\infty_{n=1}\frac{H_n}{n^52^n}-108{\rm Li}_6\left(\frac{1}{2}\right)+\frac{143\pi^6}{2520}-36{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}}\\
&\color{blue}{+\, \frac{9}{8}\zeta(5)\ln{2}-\frac{\pi^4}{40}\ln^2{2}+12\zeta^2(3)-3\pi^2\zeta(3)\ln{2}}\\
&\color{blue}{+\, 6\zeta(3)\ln^3{2}-\frac{\pi^2}{4}\ln^4{2}+\frac{3}{20}\ln^6{2}}
\end{align*}}\]
We note that
\[\begin{align*}
\zeta(\bar{5},1)
=&\frac{1}{24}\int^1_0\frac{\ln^4{x}\ln(1+x)}{1+x}{\rm d}x
=\frac{1}{24}\int^2_1\frac{\ln{x}\ln^4(x-1)}{x}{\rm d}x\\
=&-\frac{1}{24}\int^1_\frac{1}{2}\frac{\ln{x}\ln^4(1-x)}{x}{\rm d}x+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x-\frac{1}{4}\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{x}{\rm d}x\\
&+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x+\frac{1}{144}\ln^6{2}\\
=&\underbrace{-\frac{1}{24}\int^\frac{1}{2}_0\frac{\ln^4{x}\ln(1-x)}{1-x}{\rm d}x}_{\mathcal{J}}-3\sum^\infty_{n=1}\frac{H_n}{n^52^n}+7{\rm Li}_6\left(\frac{1}{2}\right)-\frac{17\pi^6}{5040}+{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}\\
&-\frac{3}{32}\zeta(5)\ln{2}-{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}+\frac{\pi^4}{480}\ln^2{2}-\frac{1}{2}\zeta^2(3)+\frac{\pi^2}{4}\zeta(3)\ln{2}-\frac{19}{24}\zeta(3)\ln^3{2}\\
&+\frac{\pi^2}{24}\ln^4{2}-\frac{17}{360}\ln^6{2}
\end{align*}\]
since we have already derived the values of the last three integrals. For the remaining integral,
\[\begin{align*}
\mathcal{J}=&\frac{1}{24}\sum^\infty_{n=1}H_n\frac{\partial^4}{\partial n^4}\left(\frac{1}{(n+1)2^{n+1}}\right)=\sum^\infty_{n=1}\frac{H_n}{(n+1)^52^{n+1}}+\sum^\infty_{n=1}\frac{\ln{2}H_n}{(n+1)^42^{n+1}}\\
&+\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{2(n+1)^32^{n+1}}+\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{6(n+1)^22^{n+1}}+\sum^\infty_{n=1}\frac{\ln^4{2}H_n}{24(n+1)2^{n+1}}\\
=&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\frac{1}{2}\right)+{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{720}\ln^2{2}\\
&+\frac{1}{2}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{1}{40}\ln^6{2}+\frac{\pi^4}{1440}\ln^2{2}-\frac{1}{16}\zeta(3)\ln^3{2}\\&+\frac{1}{48}\ln^6{2}+\frac{1}{48}\zeta(3)\ln^3{2}-\frac{1}{36}\ln^6{2}+\frac{1}{48}\ln^6{2}\\
=&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\frac{1}{2}\right)+{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\frac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{1440}\ln^2{2}\\
&+\frac{11}{24}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{7}{180}\ln^6{2}
\end{align*}\]
Hence we can express \(\zeta(\bar{5},1)\) as
\[\begin{align*}
\zeta(\bar{5},1)
=&-2\sum^\infty_{n=1}\frac{H_n}{n^52^n}+6{\rm Li}_6\left(\frac{1}{2}\right)-\frac{17\pi^6}{5040}+2{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}-\frac{1}{16}\zeta(5)\ln{2}+\frac{\pi^4}{720}\ln^2{2}\\
&-\frac{1}{2}\zeta^2(3)-\frac{1}{3}\zeta(3)\ln^3{2}+\frac{\pi^2}{6}\zeta(3)\ln{2}+\frac{\pi^2}{72}\ln^4{2}-\frac{1}{120}\ln^6{2}
\end{align*}\]
This implies that
\[\begin{align*}
\sum^\infty_{n=1}\frac{H_n}{n^52^n}
=&3{\rm Li}_6\left(\frac{1}{2}\right)-\frac{1}{2}\zeta(\bar{5},1)-\frac{17\pi^6}{10080}+{\rm Li}_5\left(\frac{1}{2}\right)\ln{2}-\frac{1}{32}\zeta(5)\ln{2}+\frac{\pi^4}{1440}\ln^2{2}\\
&-\frac{1}{4}\zeta^2(3)-\frac{1}{6}\zeta(3)\ln^3{2}+\frac{\pi^2}{12}\zeta(3)\ln{2}+\frac{\pi^2}{144}\ln^4{2}-\frac{1}{240}\ln^6{2}
\end{align*}\]
Plucking this back into the original integral, we get another form in terms of \(\zeta(\bar{5},1)\)
\[\Large\boxed{\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^2x}x\mathrm{d}x=\color{blue}{-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)}}
\]
