标题让我这个强迫症要死了
by Rainy7 @ 2020-05-30 23:05:26
$$
\begin{aligned}
I&=\int_0^1{\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \text{d}x}\\
&=\int_0^1{\left( x \right) '\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \text{d}x}\\
&=\left. x\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \right) '\text{d}x}\\
&=-\int_0^1{x\cdot \left( \frac{\ln \left( 1-x \right) \ln \left( x \right)}{x+1}-\frac{\ln \left( x+1 \right) \ln \left( x \right)}{1-x}+\frac{\ln \left( 1-x \right) \ln \left( x+1 \right)}{x} \right) \text{d}x}\\
&=-\underset{A}{\underbrace{\int_0^1{\frac{x\ln \left( 1-x \right) \ln \left( x \right)}{x+1}\text{d}x}}}+\underset{B}{\underbrace{\int_0^1{\frac{x\ln \left( x+1 \right) \ln \left( x \right)}{1-x}\text{d}x}}}-\underset{C}{\underbrace{\int_0^1{\ln \left( 1-x \right) \ln \left( x+1 \right) \text{d}x}}}\\
\end{aligned}
$$
$$
\begin{aligned}
A&=\int_0^1{\frac{\left( x+1-1 \right) \ln \left( 1-x \right) \ln \left( x \right)}{1+x}}=\underset{D}{\underbrace{\int_0^1{\ln \left( 1-x \right) \ln \left( x \right) \text{d}x}}}-\underset{i_{1102}}{\underbrace{\int_0^1{\frac{\ln \left( 1-x \right) \ln \left( x \right)}{x+1}\text{d}x}}}\\
\\
D&=\int_0^1{\left( x \right) '\ln \left( 1-x \right) \ln \left( x \right) \text{d}x}\\
&=\left. x\ln \left( x \right) \ln \left( 1-x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \ln \left( 1-x \right) \ln \left( x \right) \right) '\text{d}x}\\
&=-\int_0^1{x\cdot \left( \frac{\ln \left( 1-x \right)}{x}-\frac{\ln \left( x \right)}{1-x} \right) \text{d}x}\\
&=-\int_0^1{\ln \left( 1-x \right) \text{d}x}+\int_0^1{\frac{x\ln \left( x \right)}{1-x}\text{d}x}\\
&=-\left( -1 \right) +\int_0^1{\frac{x\ln \left( x \right)}{1-x}\text{d}x}=1+\int_0^1{\frac{\left( 1-\left( 1-x \right) \right) \ln \left( x \right)}{1-x}\text{d}x}\\
&=1+\int_0^1{\frac{\ln \left( x \right)}{1-x}\text{d}x}-\int_0^1{\ln \left( x \right) \text{d}x}\\
&=1-\frac{\pi ^2}{6}-\left( -1 \right) =2-\frac{\pi ^2}{6}\\
i_{1102}&=-\frac{\pi ^2\ln 2}{4}+\frac{13}{8}\zeta \left( 3 \right)\\
\therefore A&=2-\frac{\pi ^2}{6}-\left( -\frac{\pi ^2\ln 2}{4}+\frac{13}{8}\zeta \left( 3 \right) \right)\\
&=-\frac{13\zeta \left( 3 \right)}{8}-\frac{\pi ^2}{6}+2+\frac{1}{4}\pi ^2\ln ^2\left( 2 \right)\\
\end{aligned}
$$
$$
\begin{aligned}
B&=\int_0^1{\frac{x\ln \left( x+1 \right) \ln \left( x \right)}{1-x}\text{d}x}\\
&=\int_0^1{\frac{\left( 1-\left( 1-x \right) \right) \ln \left( 1+x \right) \ln \left( x \right)}{1-x}\text{d}x}\\
&=\underset{i_{0110}}{\underbrace{\int_0^1{\frac{\ln \left( 1+x \right) \ln \left( x \right)}{1-x}\text{d}x}}}-\underset{E}{\underbrace{\int_0^1{\ln \left( 1+x \right) \ln \left( x \right) \text{d}x}}}\\
\\
E&=\left. x\ln \left( x \right) \ln \left( 1+x \right) \right|_{0}^{1}-\int_0^1{x\left( \frac{\ln \left( x \right)}{x+1}+\frac{\ln \left( x+1 \right)}{x} \right) \text{d}x}\\
&=-\int_0^1{\frac{\left( x+1-1 \right) \ln \left( x \right)}{x+1}\text{d}x}-\int_0^1{\ln \left( x+1 \right) \text{d}x}\\
&=-\int_0^1{\ln \left( x \right) \text{d}x}+\int_0^1{\frac{\ln \left( x \right)}{1+x}\text{d}x}-\left( \ln \left( 4 \right) -1 \right)\\
&=-\left( -1 \right) +\left( -\frac{\pi ^2}{12} \right) -\ln \left( 4 \right) +1\\
&=2-\ln \left( 4 \right) -\frac{\pi ^2}{12}\\
i_{0110}&=\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}\\
\\
\therefore B&=\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-\left( 2-\ln \left( 4 \right) -\frac{\pi ^2}{12} \right) =\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-2+\ln \left( 4 \right) +\frac{\pi ^2}{12}\\
\end{aligned}
$$
$$
$$
$$
\begin{aligned}
C&=\int_0^1{\ln \left( 1-x \right) \ln \left( x+1 \right) \text{d}x}\\
&=\int_0^1{\ln \left( x \right) \ln \left( 2-x \right) \text{d}x}\\
&=\left. x\ln \left( x \right) \ln \left( 2-x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \frac{\ln \left( 2-x \right)}{x}-\frac{\ln \left( x \right)}{2-x} \right) \text{d}x}\\
&=-\int_0^1{\ln \left( 2-x \right) \text{d}x}+\int_0^1{\frac{x\ln \left( x \right)}{2-x}\text{d}x}\\
&=-\left( 2\ln \left( 2 \right) -1 \right) +\int_0^1{\frac{\left( 2-\left( 2-x \right) \right) \ln \left( x \right)}{2-x}\text{d}x}\\
&=1-2\ln \left( 2 \right) +2\int_0^1{\frac{\ln \left( x \right)}{2-x}\text{d}x}-\int_0^1{\ln \left( x \right) \text{d}x}\\
&=1-2\ln \left( 2 \right) +2\left( \frac{1}{2}\left( \ln ^2\left( 2 \right) -\frac{\pi ^2}{6} \right) \right) -\left( -1 \right)\\
&=2-2\ln \left( 2 \right) +\ln ^2\left( 2 \right) -\frac{\pi ^2}{6}\\
\end{aligned}
$$
$$
\begin{aligned}
I&=-A+B-C\\
&=-\left( -\frac{13\zeta \left( 3 \right)}{8}-\frac{\pi ^2}{6}+2+\frac{1}{4}\pi ^2\ln ^2\left( 2 \right) \right) +\left( \zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-2+\ln \left( 4 \right) +\frac{\pi ^2}{12} \right) -\left( 2-\ln \left( 4 \right) +\ln ^2\left( 2 \right) -\frac{\pi ^2}{6} \right)\\
&=\frac{21\zeta \left( 3 \right)}{8}-6-\ln ^2\left( 2 \right) +4\ln \left( 2 \right) +\frac{5\pi ^2}{12}-\frac{1}{2}\pi ^2\log \left( 2 \right)\\
\end{aligned}
$$
by 诱宵美⑨ @ 2020-05-30 23:18:12
诶，看，挂了~
by 诱宵美⑨ @ 2020-05-30 23:18:25
$$\begin{aligned} I&=\int_0^1{\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \text{d}x}\\ &=\int_0^1{\left( x \right) '\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \text{d}x}\\ &=\left. x\ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \ln \left( 1-x \right) \ln \left( x \right) \ln \left( 1+x \right) \right) '\text{d}x}\\ &=-\int_0^1{x\cdot \left( \frac{\ln \left( 1-x \right) \ln \left( x \right)}{x+1}-\frac{\ln \left( x+1 \right) \ln \left( x \right)}{1-x}+\frac{\ln \left( 1-x \right) \ln \left( x+1 \right)}{x} \right) \text{d}x}\\ &=-\underset{A}{\underbrace{\int_0^1{\frac{x\ln \left( 1-x \right) \ln \left( x \right)}{x+1}\text{d}x}}}+\underset{B}{\underbrace{\int_0^1{\frac{x\ln \left( x+1 \right) \ln \left( x \right)}{1-x}\text{d}x}}}-\underset{C}{\underbrace{\int_0^1{\ln \left( 1-x \right) \ln \left( x+1 \right) \text{d}x}}}\\\end{aligned}$$
$$\begin{aligned} A&=\int_0^1{\frac{\left( x+1-1 \right) \ln \left( 1-x \right) \ln \left( x \right)}{1+x}}=\underset{D}{\underbrace{\int_0^1{\ln \left( 1-x \right) \ln \left( x \right) \text{d}x}}}-\underset{i_{1102}}{\underbrace{\int_0^1{\frac{\ln \left( 1-x \right) \ln \left( x \right)}{x+1}\text{d}x}}}\\ \\ D&=\int_0^1{\left( x \right) '\ln \left( 1-x \right) \ln \left( x \right) \text{d}x}\\ &=\left. x\ln \left( x \right) \ln \left( 1-x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \ln \left( 1-x \right) \ln \left( x \right) \right) '\text{d}x}\\ &=-\int_0^1{x\cdot \left( \frac{\ln \left( 1-x \right)}{x}-\frac{\ln \left( x \right)}{1-x} \right) \text{d}x}\\ &=-\int_0^1{\ln \left( 1-x \right) \text{d}x}+\int_0^1{\frac{x\ln \left( x \right)}{1-x}\text{d}x}\\ &=-\left( -1 \right) +\int_0^1{\frac{x\ln \left( x \right)}{1-x}\text{d}x}=1+\int_0^1{\frac{\left( 1-\left( 1-x \right) \right) \ln \left( x \right)}{1-x}\text{d}x}\\ &=1+\int_0^1{\frac{\ln \left( x \right)}{1-x}\text{d}x}-\int_0^1{\ln \left( x \right) \text{d}x}\\ &=1-\frac{\pi ^2}{6}-\left( -1 \right) =2-\frac{\pi ^2}{6}\\ i_{1102}&=-\frac{\pi ^2\ln 2}{4}+\frac{13}{8}\zeta \left( 3 \right)\\ \therefore A&=2-\frac{\pi ^2}{6}-\left( -\frac{\pi ^2\ln 2}{4}+\frac{13}{8}\zeta \left( 3 \right) \right)\\ &=-\frac{13\zeta \left( 3 \right)}{8}-\frac{\pi ^2}{6}+2+\frac{1}{4}\pi ^2\ln ^2\left( 2 \right)\\\end{aligned}$$
$$\begin{aligned} B&=\int_0^1{\frac{x\ln \left( x+1 \right) \ln \left( x \right)}{1-x}\text{d}x}\\ &=\int_0^1{\frac{\left( 1-\left( 1-x \right) \right) \ln \left( 1+x \right) \ln \left( x \right)}{1-x}\text{d}x}\\ &=\underset{i_{0110}}{\underbrace{\int_0^1{\frac{\ln \left( 1+x \right) \ln \left( x \right)}{1-x}\text{d}x}}}-\underset{E}{\underbrace{\int_0^1{\ln \left( 1+x \right) \ln \left( x \right) \text{d}x}}}\\ \\
E&=\left. x\ln \left( x \right) \ln \left( 1+x \right) \right|_{0}^{1}-\int_0^1{x\left( \frac{\ln \left( x \right)}{x+1}+\frac{\ln \left( x+1 \right)}{x} \right) \text{d}x}\\ &=-\int_0^1{\frac{\left( x+1-1 \right) \ln \left( x \right)}{x+1}\text{d}x}-\int_0^1{\ln \left( x+1 \right) \text{d}x}\\ &=-\int_0^1{\ln \left( x \right) \text{d}x}+\int_0^1{\frac{\ln \left( x \right)}{1+x}\text{d}x}-\left( \ln \left( 4 \right) -1 \right)\\ &=-\left( -1 \right) +\left( -\frac{\pi ^2}{12} \right) -\ln \left( 4 \right) +1\\ &=2-\ln \left( 4 \right) -\frac{\pi ^2}{12}\\ i_{0110}&=\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}\\
\\ \therefore B&=\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-\left( 2-\ln \left( 4 \right) -\frac{\pi ^2}{12} \right) =\zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-2+\ln \left( 4 \right) +\frac{\pi ^2}{12}\\\end{aligned}$$
$$\begin{aligned} C&=\int_0^1{\ln \left( 1-x \right) \ln \left( x+1 \right) \text{d}x}\\
&=\int_0^1{\ln \left( x \right) \ln \left( 2-x \right) \text{d}x}\\ &=\left. x\ln \left( x \right) \ln \left( 2-x \right) \right|_{0}^{1}-\int_0^1{x\cdot \left( \frac{\ln \left( 2-x \right)}{x}-\frac{\ln \left( x \right)}{2-x} \right) \text{d}x}\\ &=-\int_0^1{\ln \left( 2-x \right) \text{d}x}+\int_0^1{\frac{x\ln \left( x \right)}{2-x}\text{d}x}\\ &=-\left( 2\ln \left( 2 \right) -1 \right) +\int_0^1{\frac{\left( 2-\left( 2-x \right) \right) \ln \left( x \right)}{2-x}\text{d}x}\\ &=1-2\ln \left( 2 \right) +2\int_0^1{\frac{\ln \left( x \right)}{2-x}\text{d}x}-\int_0^1{\ln \left( x \right) \text{d}x}\\ &=1-2\ln \left( 2 \right) +2\left( \frac{1}{2}\left( \ln ^2\left( 2 \right) -\frac{\pi ^2}{6} \right) \right) -\left( -1 \right)\\ &=2-2\ln \left( 2 \right) +\ln ^2\left( 2 \right) -\frac{\pi ^2}{6}\\
\end{aligned}$$
$$\begin{aligned} I&=-A+B-C\\ &=-\left( -\frac{13\zeta \left( 3 \right)}{8}-\frac{\pi ^2}{6}+2+\frac{1}{4}\pi ^2\ln ^2\left( 2 \right) \right) +\left( \zeta \left( 3 \right) -\frac{\pi ^2\ln 2}{4}-2+\ln \left( 4 \right) +\frac{\pi ^2}{12} \right) -\left( 2-\ln \left( 4 \right) +\ln ^2\left( 2 \right) -\frac{\pi ^2}{6} \right)\\ &=\frac{21\zeta \left( 3 \right)}{8}-6-\ln ^2\left( 2 \right) +4\ln \left( 2 \right) +\frac{5\pi ^2}{12}-\frac{1}{2}\pi ^2\log \left( 2 \right)\\\end{aligned}$$
by 诱宵美⑨ @ 2020-05-30 23:20:43
额，不是说放到一行就不挂了吗？
by 诱宵美⑨ @ 2020-05-30 23:21:04
$$
123
$$
```
$$
123
$$
```
by zhy137036 @ 2020-05-30 23:21:30
没挂啊
by zhy137036 @ 2020-05-30 23:21:40
再试试
```cpp
```
$$
1.\,\,y\in \left( -1,1 \right)
$$
$$
\int_0^1{\frac{\text{d}x}{\left( 1+yx \right) \sqrt{1-x^2}}}
$$
$$
ans:\,\,\frac{\arccos \left( y \right)}{\sqrt{1-y^2}}
$$
$$
where\,\,\arccos \left( x \right) =\frac{\pi}{2}+i\log \left( \sqrt{1-x^2}+ix \right)
$$
$$
$$
$$
2.\,\,m,n\in \mathbb{N}\,\,
$$
$$
\left. i \right) \int_0^1{x^m\log ^n\left( x \right) \text{d}x}
$$
$$
\left. ii \right) \int_0^a{x^m\log ^n\left( x \right) \text{d}x}\left( a>0 \right)
$$
$$
ans:
$$
$$
\left. i \right) \,\,\left( -1 \right) ^n\frac{m!}{\left( m+1 \right) ^{n+1}}
$$
$$
\left. ii \right) \,\,a^{m+1}\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c}
n\\
k\\
\end{array} \right) \frac{k!}{\left( m+1 \right) ^{k+1}}\log ^{n-k}\left( a \right)}=a^{m+1}\sum_{k=0}^n{\left( -1 \right) ^k\frac{n!\log ^{n-k}\left( a \right)}{\left( n-k \right) !\left( m+1 \right) ^{k+1}}}
$$
$$
$$
$$
3.\,\,n\in \mathbb{R}^+
$$
$$
Prove\,\,that:
$$
$$
I_n=\int_0^1{x^{n-1}\log \left( 1-x \right) \text{d}x}=-\frac{H_n}{n}
$$
$$
J_n=\int_0^1{x^{n-1}\log ^2\left( 1-x \right) \text{d}x}=\frac{H_{n}^{2}+H_{n}^{\left( 2 \right)}}{n}
$$
$$
K_n=\int_0^1{x^{n-1}\log ^3\left( 1-x \right) \text{d}x}=-\frac{H_{n}^{3}+3H_nH_{n}^{\left( 2 \right)}+2H_{n}^{\left( 3 \right)}}{n}
$$
$$
L_n=\int_0^1{x^{n-1}\log ^4\left( 1-x \right) \text{d}x}=\frac{H_{n}^{4}+6H_{n}^{2}H_{n}^{\left( 2 \right)}+8H_nH_{n}^{\left( 3 \right)}+3\left( H_{n}^{\left( 2 \right)} \right) ^2+6H_{n}^{\left( 4 \right)}}{n}
$$
$$
where\,\,H_{n}^{\left( m \right)}=\sum_{k=1}^n{\frac{1}{k^m}},\,\,m\geq 1
$$
$$
\bigstar \int_0^1{x^{n-1}\log ^m\left( 1-x \right) \text{d}x}=?
$$
$$
$$
$$
ans:\,\,
$$
$$
I_n=\int_0^1{x^{n-1}\log \left( 1-x \right) \text{d}x}=\frac{1}{n}\int_0^1{\left( x^n-1 \right) '\log \left( 1-x \right) \text{d}x}
$$
$$
=\underset{0}{\underbrace{\left. \frac{1}{n}\left( x^n-1 \right) \log \left( 1-x \right) \right|_{0}^{1}}}-\frac{1}{n}\int_0^1{\frac{1-x^n}{1-x}\text{d}x}=-\frac{H_n}{n}
$$
$$
$$
$$
J_n=\int_0^1{x^{n-1}\log ^2\left( 1-x \right) \text{d}x}=\frac{1}{n}\int_0^1{\left( x^n-1 \right) '\log ^2\left( 1-x \right) \text{d}x}
$$
$$
=\underset{0}{\underbrace{\left. \frac{1}{n}\left( x^n-1 \right) \log ^2\left( 1-x \right) \right|_{0}^{1}}}-\frac{2}{n}\int_0^1{\frac{1-x^n}{1-x}\log \left( 1-x \right) \text{d}x}
$$
$$
=-\frac{2}{n}\int_0^1{\sum_{k=1}^n{x^{k-1}\log \left( 1-x \right)}\text{d}x}=-\frac{2}{n}\sum_{k=1}^n{I_k}=\frac{2}{n}\sum_{k=1}^n{\frac{H_k}{k}}
$$
$$
We\ know:\ H_{n}^{2}=\sum_{i=1}^n{\sum_{j=1}^n{\frac{1}{i\ j}}}=\sum_{i=1}^n{\left( \sum_{j=1}^i{+\sum_{j=1}^n{\}} \right) \frac{1}{i\ j}}-\sum_{i=1}^n{\frac{1}{i^2}}=2\sum_{i=1}^n{\sum_{j=1}^i{\frac{1}{i\ j}}}-H_{n}^{\left( 2 \right)}=2\sum_{i=1}^n{\frac{H_i}{i}}-H_{n}^{\left( 2 \right)}
$$
$$
then\ we\ obtain:\ \sum_{i=1}^n{\frac{H_i}{i}}=\frac{H_{n}^{2}+H_{n}^{\left( 2 \right)}}{2}
$$
$$
\therefore J_n=\frac{2}{n}\sum_{k=1}^n{\frac{H_k}{k}}=\frac{2}{n}\cdot \frac{H_{n}^{2}+H_{n}^{\left( 2 \right)}}{2}=\frac{H_{n}^{2}+H_{n}^{\left( 2 \right)}}{n}
$$
$$
$$
$$
K_n=\int_0^1{x^{n-1}\log ^3\left( 1-x \right) \text{d}x}=\frac{1}{n}\int_0^1{\left( x^n-1 \right) '\log ^3\left( 1-x \right) \text{d}x}
$$
$$
=\underset{0}{\underbrace{\left. \frac{1}{n}\left( x^n-1 \right) \log ^3\left( 1-x \right) \right|_{0}^{1}}}-\frac{3}{n}\int_0^1{\frac{1-x^n}{1-x}\log ^2\left( 1-x \right) \text{d}x}
$$
$$
=-\frac{3}{n}\int_0^1{\sum_{k=1}^n{x^{k-1}\log ^2\left( 1-x \right)}\text{d}x}=-\frac{3}{n}\sum_{k=1}^n{J_k}=-\frac{3}{n}\sum_{k=1}^n{\frac{H_{k}^{2}+H_{k}^{\left( 2 \right)}}{k}}
$$
$$
Use\,\,Abel's\,\,summation\,\,formula:
$$
$$
\sum_{k=1}^n{a_kb_k}=A_nb_{n+1}+\sum_{k=1}^n{A_k\left( b_k-b_{k+1} \right)},\,\,where\,\,A_n=\sum_{k=1}^n{a_k}
$$
$$
Let\,\,a_k=\frac{1}{k},b_k=H_{k}^{2}+H_{k}^{\left( 2 \right)}
$$
$$
\sum_{k=1}^n{\frac{H_{k}^{2}+H_{k}^{\left( 2 \right)}}{k}}=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\left( \frac{H_{k}^{2}}{k+1}+\frac{H_k}{\left( k+1 \right) ^2} \right)}
$$
$$
=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\left( \frac{\left( H_{k+1}-\dfrac{1}{\left( k+1 \right)} \right) ^2}{k+1}+\frac{H_{k+1}-\dfrac{1}{\left( k+1 \right)}}{\left( k+1 \right) ^2} \right)}
$$
$$
=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\left( \frac{H_{k+1}^{2}}{k+1}-\frac{H_{k+1}}{\left( k+1 \right) ^2} \right)}
$$
$$
=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\frac{H_{k}^{2}}{k}}+2\sum_{k=1}^n{\frac{H_k}{k^2}}-2\frac{H_{n+1}^{2}}{n+1}+2\frac{H_{n+1}}{\left( n+1 \right) ^2}
$$
$$
We\,\,apply\,\,Abel's\,\,summation\,\,again\,\,for\,\,\sum_{k=1}^n{\frac{H_k}{k^2}}
$$
$$
\sum_{k=1}^n{a_kb_k}=A_nb_{n+1}+\sum_{k=1}^n{A_k\left( b_k-b_{k+1} \right)}
$$
$$
Let\,\,a_k=\frac{1}{k^2},b_k=H_k
$$
$$
\sum_{k=1}^n{\frac{H_k}{k^2}}=\sum_{k=1}^n{a_kb_k}=H_{n}^{\left( 2 \right)}H_{n+1}-\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}}{k+1}}=H_{n+1}H_{n}^{\left( 2 \right)}-\sum_{k=1}^n{\frac{H_{k+1}^{\left( 2 \right)}-\frac{1}{\left( k+1 \right) ^2}}{k+1}}
$$
$$
=H_{n+1}H_{n}^{\left( 2 \right)}-\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}}{k}}-\frac{H_{n+1}^{\left( 2 \right)}}{n+1}+H_{n}^{\left( 3 \right)}+\frac{1}{\left( n+1 \right) ^3}
$$
$$
\therefore \sum_{k=1}^n{\frac{H_{k}^{2}+H_{k}^{\left( 2 \right)}}{k}}=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\frac{H_{k}^{2}}{k}}+2\sum_{k=1}^n{\frac{H_k}{k^2}}-2\frac{H_{n+1}^{2}}{n+1}+2\frac{H_{n+1}}{\left( n+1 \right) ^2}
$$
$$
=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\frac{H_{k}^{2}}{k}}+2\left( H_{n+1}H_{n}^{\left( 2 \right)}-\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}}{k}}-\frac{H_{n+1}^{\left( 2 \right)}}{n+1}+H_{n}^{\left( 3 \right)}+\frac{1}{\left( n+1 \right) ^3} \right) -2\frac{H_{n+1}^{2}}{n+1}+2\frac{H_{n+1}}{\left( n+1 \right) ^2}
$$
$$
=H_n\left( H_{n+1}^{2}+H_{n+1}^{\left( 2 \right)} \right) -2\sum_{k=1}^n{\frac{H_{k}^{2}}{k}}-2\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}}{k}}+2\left( H_{n+1}H_{n}^{\left( 2 \right)}-\frac{H_{n+1}^{\left( 2 \right)}}{n+1}+H_{n}^{\left( 3 \right)}+\frac{1}{\left( n+1 \right) ^3} \right) -2\frac{H_{n+1}^{2}}{n+1}+2\frac{H_{n+1}}{\left( n+1 \right) ^2}
$$
$$
=-2\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}+H_{k}^{2}}{k}}+H_n\left( \left( H_n+\frac{1}{n+1} \right) ^2+H_{n}^{\left( 2 \right)}+\frac{1}{\left( n+1 \right) ^2} \right) +2\left( \left( H_n+\frac{1}{n+1} \right) H_{n}^{\left( 2 \right)}-\frac{\left( H_{n}^{\left( 2 \right)}+\frac{1}{n+1} \right)}{n+1}+H_{n}^{\left( 3 \right)}+\frac{1}{\left( n+1 \right) ^3} \right) -2\frac{\left( H_n+\frac{1}{n+1} \right) ^2}{n+1}+2\frac{H_n+\frac{1}{n+1}}{\left( n+1 \right) ^2}
$$
$$
=-2\sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}+H_{k}^{2}}{k}}+H_{n}^{3}+3H_nH_{n}^{\left( 2 \right)}+2H_{n}^{\left( 3 \right)}
$$
$$
\therefore \sum_{k=1}^n{\frac{H_{k}^{\left( 2 \right)}+H_{k}^{2}}{k}}=\frac{1}{3}\left( H_{n}^{3}+3H_nH_{n}^{\left( 2 \right)}+2H_{n}^{\left( 3 \right)} \right)
$$
$$
\therefore K_n=\int_0^1{x^{n-1}\log ^3\left( 1-x \right) \text{d}x}=-\frac{3}{n}\sum_{k=1}^n{\frac{H_{k}^{2}+H_{k}^{\left( 2 \right)}}{k}}=-\frac{H_{n}^{3}+3H_nH_{n}^{\left( 2 \right)}+2H_{n}^{\left( 3 \right)}}{n}
$$
$$
$$
$$
L_n=\int_0^1{x^{n-1}\log ^4\left( 1-x \right) \text{d}x}=\frac{1}{n}\int_0^1{\left( x^n-1 \right) '\log ^4\left( 1-x \right) \text{d}x}
$$
$$
=\underset{0}{\underbrace{\left. \frac{1}{n}\left( x^n-1 \right) \log ^4\left( 1-x \right) \right|_{0}^{1}}}-\frac{4}{n}\int_0^1{\frac{1-x^n}{1-x}\log ^3\left( 1-x \right) \text{d}x}
$$
$$
=-\frac{4}{n}\int_0^1{\sum_{k=1}^n{x^{k-1}\log ^3\left( 1-x \right)}\text{d}x}=-\frac{4}{n}\sum_{k=1}^n{K_k}=\frac{4}{n}\sum_{k=1}^n{\frac{H_{k}^{3}+3H_kH_{k}^{\left( 2 \right)}+2H_{k}^{\left( 3 \right)}}{k}}
$$
$$
we\,\,can\,\,use\,\,again\,\,Abel's\,\,summation\,\,to\,\,get\,\,the\,\,desired\,\,result,\,\,
$$
$$
as\,\,at\,\,the\,\,previous\,\,points\,\,of\,\,the\,\,problem,\,\,
$$
$$
but\,\,we\,\,can\,\,use\,\,the\,\,complete\,\,homogeneous\,\,symmetric\,\,polynomial
$$
$$
denote\,\,h_k\left( x_1,x_2,\cdots ,x_n \right) =\sum_{1\leq i_1\leq i_2\leq \cdots \leq i_k\leq n}{x_{i_1}x_{i_2}\cdots x_{i_k}},\,\,p_k\left( x_1,x_2,\cdots ,x_n \right) =\sum_{i=1}^n{x_{i}^{k}}
$$
$$
use\,\,the\,\,recursive\,\,relation:
$$
$$
nh_n=\sum_{k=1}^n{p_kh_{n-k}}
$$
$$
h_1=p_1,\,\,h_2=\frac{1}{2}\left( p_{1}^{2}+p_2 \right) ,\,\,h_3=\frac{1}{6}\left( p_{1}^{3}+3p_1p_2+2p_3 \right)
$$
$$
h_4=\frac{1}{24}\left( p_{1}^{4}+6p_{1}^{2}p_2+8p_1p_3+3p_{2}^{2}+6p_4 \right)
$$
$$
h_5=\frac{1}{120}\left( p_{1}^{5}+10p_{1}^{3}p_2+15p_1p_{2}^{3}+20p_2p_3+30p_1p_4+24p_5 \right)
$$
$$
$$
$$
h_4=\sum_{i=1}^n{\sum_{j=1}^i{\sum_{k=1}^j{\sum_{l=1}^k{x_ix_jx_kx_l}}}}=\frac{1}{24}\left( \left( \sum_{k=1}^4{x_k} \right) ^4+6\left( \sum_{k=1}^n{x_k} \right) ^2\left( \sum_{k=1}^n{x_{k}^{2}} \right) +8\left( \sum_{k=1}^n{x_k} \right) \left( \sum_{k=1}^n{x_{k}^{3}} \right) +3\left( \sum_{k=1}^n{x_{k}^{2}} \right) ^2+6\sum_{k=1}^n{x_{k}^{4}} \right)
$$
$$
$$
$$
Now\,\,it's\,\,easy\,\,to\,\,see:
$$
$$
L_n=\frac{4}{n}\cdot 3\cdot 2\cdot 1\cdot \sum_{i=1}^n{\sum_{j=1}^i{\sum_{k=1}^j{\sum_{l=1}^k{\frac{1}{i\,\,j\,\,k\,\,l}}}}}=\frac{H_{n}^{4}+6H_{n}^{2}H_{n}^{\left( 2 \right)}+8H_nH_{n}^{\left( 3 \right)}+3\left( H_{n}^{\left( 2 \right)} \right) ^2+6H_{n}^{\left( 4 \right)}}{n}
$$
$$
$$
$$
\int_0^1{x^{n-1}\log ^m\left( 1-x \right) \text{d}x}=\frac{\left( -1 \right) ^mm!}{n}h_m\left( 1,\frac{1}{2},\cdots ,\frac{1}{n} \right)
$$
$$
\int_0^1{x^{n-1}\log ^m\left( 1-x \right) \text{d}x}=\frac{1}{n}\int_0^1{\left( x^n-1 \right) '\log ^m\left( 1-x \right) \text{d}x}
$$
$$
=\underset{0}{\underbrace{\left. \frac{1}{n}\left( x^n-1 \right) \log ^m\left( 1-x \right) \right|_{0}^{1}}}-\frac{m}{n}\int_0^1{\frac{1-x^n}{1-x}\log ^{m-1}\left( 1-x \right) \text{d}x}
$$
$$
=-\frac{m}{n}\int_0^1{\sum_{k=1}^n{x^{k-1}\log ^{m-1}\left( 1-x \right)}\text{d}x}=\frac{\left( -1 \right) ^mm!}{n}\sum_{k=1}^n{\frac{1}{k}h_{m-1}\left( 1,\frac{1}{2},\cdots ,\frac{1}{k} \right)}=\frac{\left( -1 \right) ^mm!}{n}h_m\left( 1,\frac{1}{2},\cdots ,\frac{1}{n} \right)
$$
$$
$$
$$
4.\,\,Prove\,\,that:
$$
$$
\left. i \right) \int_0^x{\frac{\log ^2\left( 1-t \right)}{t}\text{d}t}
$$
$$
=\log \left( x \right) \log ^2\left( 1-x \right) +2\log \left( 1-x \right) \text{Li}_2\left( 1-x \right) -2\text{Li}_3\left( 1-x \right) +2\zeta \left( 3 \right)
$$
$$
\left. ii \right) \int_0^x{\frac{\log ^2\left( 1+t \right) \text{d}t}{t}}
$$
$$
=\log \left( x \right) \log ^2\left( 1+x \right) -\frac{2}{3}\log ^3\left( 1+x \right) -2\log \left( 1+x \right) \text{Li}_2\left( \frac{1}{1+x} \right) -2\text{Li}_3\left( \frac{1}{1+x} \right) +2\zeta \left( 3 \right)
$$
$$
where\,\,\zeta \,\,denotes\,\,the\,\,Riemann\,\,zeta\,\,function
$$
$$
and\,\,\text{Li}_n\,\,represents\,\,the\,\,Polylogarithm\,\,function
$$
$$
$$
$$
Sol.
$$
$$
\begin{aligned}
\left. i \right) \int_0^x{\frac{\log ^2\left( 1-t \right)}{t}\text{d}t}\\
&=\left. \log \left( t \right) \log ^2\left( 1-t \right) \right|_{0}^{x}+2\int_0^x{\frac{\log \left( t \right) \log \left( 1-t \right)}{1-t}\text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1-x \right) +2\int_0^x{\left( \text{Li}_2\left( 1-t \right) \right) '\,\,\log \left( 1-t \right) \text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1-x \right) +\left. 2\text{Li}_2\left( 1-t \right) \log \left( 1-t \right) \right|_{0}^{x}+2\int_0^x{\frac{\text{Li}_2\left( 1-t \right)}{1-t}\text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1-x \right) +2\text{Li}_2\left( 1-x \right) \log \left( 1-x \right) +2\left( -\text{Li}_3\left( 1-t \right) \mid _{0}^{x} \right)\\
&=\log \left( x \right) \log ^2\left( 1-x \right) +2\text{Li}_2\left( 1-x \right) \log \left( 1-x \right) -2\text{Li}_3\left( 1-x \right) +2\zeta \left( 3 \right)\\
\end{aligned}
$$
$$
\begin{aligned}
\left. ii \right) \int_0^x{\frac{\log ^2\left( 1+t \right)}{t}\text{d}t}\\
&=\left. \log \left( t \right) \log ^2\left( 1+t \right) \right|_{0}^{x}-2\int_0^x{\frac{\log \left( t \right) \log \left( 1+t \right)}{1+t}\text{d}t}\\
&=\left. \log \left( t \right) \log ^2\left( 1+t \right) \right|_{0}^{x}-2\int_0^x{\frac{\left( \log \left( 1+t \right) +\log \left( \frac{t}{1+t} \right) \right) \log \left( 1+t \right)}{1+t}\text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1+x \right) -2\int_0^x{\frac{\log ^2\left( 1+t \right)}{1+t}\text{d}t}-2\int_0^x{\frac{\log \left( \frac{t}{1+t} \right) \log \left( 1+t \right)}{1+t}\text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1+x \right) -\frac{2}{3}\log ^3\left( 1+x \right) -2\int_0^x{\left( \text{Li}_2\left( \frac{1}{1+t} \right) \right) ^'\log \left( 1+t \right) \text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1+x \right) -\frac{2}{3}\log ^3\left( 1+x \right) -2\left( \text{Li}_2\left( \frac{1}{1+t} \right) \log \left( 1+t \right) \mid _{0}^{x} \right) +2\int_0^x{\frac{1}{1+t}\text{Li}_2\left( \frac{1}{1+t} \right) \text{d}t}\\
&=\log \left( x \right) \log ^2\left( 1+x \right) -\frac{2}{3}\log ^3\left( 1+x \right) -2\log \left( 1+x \right) \text{Li}_2\left( \frac{1}{1+x} \right) +2\left( \left. -\text{Li}_3\left( \frac{1}{1+t} \right) \right|_{0}^{x} \right)\\
&=\log \left( x \right) \log ^2\left( 1+x \right) -\frac{2}{3}\log ^3\left( 1+x \right) -2\log \left( 1+x \right) \text{Li}_2\left( \frac{1}{1+x} \right) -2\text{Li}_3\left( \frac{1}{1+x} \right) +2\zeta \left( 3 \right)\\
\end{aligned}
$$
by 诱宵美⑨ @ 2020-05-30 23:22:11
@[zhy137036](/user/178294) 是多行环境。。
by 诱宵美⑨ @ 2020-05-30 23:22:34
@[zhy137036](/user/178294) 是指像aligned array gathered这样的
by 诱宵美⑨ @ 2020-05-30 23:23:18