Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$ How would you solve the following 
$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$
I might be able to relate the integral to Euler sums .
 A: A related problem. You can have the following closed form

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx=  -\sum _{k=1}^{\infty }{\frac {\psi' \left( k+1 \right) }{{k}^{3}}}\sim -0.7115661976, $$

where $\psi(x)$ is the digamma function.
Another possible solution:

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx = \zeta(5) -\sum _{k=1}^{\infty }{\frac {\psi' \left( k \right) }{{k}^{3}}}\sim -0.7115661976. $$

Added: For the first one just use the power series expansions of $\operatorname{Li}_3(x)$ and $\frac{1}{1-x}$ and you end up with
$$\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty} \int_{0}^{1} x^{k+n}\ln(x)dx=-\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty}\frac{1}{(n+k+1)^2} $$
$$ = -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}. $$
If you manipulate the last sum, you will be able to relate it to the Euler sums as 

$$ -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}= \zeta(5)-\sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n^2}.$$ 

Note: Notice that, we are getting identities for $\zeta(5)$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x:
     \ {\large ?}}$.

In this post
  it's shown, in general grounds, that:
  $$
\int_{0}^{1}{{\rm Li}_{q}\pars{x}\ln^{r - 1}\pars{x} \over 1 - x}\,\dd x
=\pars{-1}^{r - 1}\pars{r - 1}!\bracks{\zeta\pars{r}\zeta\pars{q}%
-\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{r}} \over n^{q}}}
$$
  where $\ds{\zeta\pars{z}}$ is the Riemann Zeta Function and
  $\ds{H_{n}^{\rm \pars{r}} \equiv \sum_{k = 1}^{n}{1 \over k^{\rm r}}}$ is a
  Generalized Harmonic Number

such that $\ds{\pars{~\mbox{with}\ q = 3\ \mbox{and}\ r = 2~}}$:
$$
\int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x
=\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} - \zeta\pars{2}\zeta\pars{3}\,,\qquad
H_{n}^{\rm\pars{2}} = \sum_{k = 1}^{n}{1 \over k^{2}}
$$

In a comment of this answer, the OP $\pars{\tt @Zaid\ Alfayeai}$ pointed out that
  $\ds{\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} = 3\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}}$ such that

$$\color{#66f}{\large%
\int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x
=2\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}} \approx {\tt -0.7115}
$$
A: \begin{align}
\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}{\rm d}x
&=\sum^\infty_{n=1}H_n^{(3)}\int^1_0x^n\log{x} \ {\rm d}x\\
&=-\sum^\infty_{n=1}\frac{H_n^{(3)}}{(n+1)^2}\\
&=\sum^\infty_{n=1}\frac{1}{(n+1)^5}-\sum^\infty_{n=1}\frac{H_{n+1}^{(3)}}{(n+1)^2}\\
&=\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{H_{n}^{(3)}}{n^2}}_{S}
\end{align}
Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(2)}(-z)}{z^2}$. We know that
\begin{align}\pi\cot{\pi z}&=\frac{1}{z-n}-2\sum^\infty_{k=1}\zeta(2k)(z-n)^{2k-1}\\&\approx\frac{1}{z-n}-2\zeta(2)(z-n)\end{align}
(see here for a proof) and
\begin{align}\Psi^{(2)}(-z)&=\frac{2}{(z-n)^3}+\sum^\infty_{k=2}(-1)^{k}k(k-1)\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-2}\\&\approx\frac{2}{(z-n)^3}+2\left(H_n^{(3)}-\zeta(3)\right)\end{align}
At the positive integers,
\begin{align}
{\rm Res}(f,n)
&=\operatorname*{Res}_{z=n}\left[\frac{2}{z^2(z-n)^4}-\frac{4\zeta(2)}{z^2(z-n)^2}+\frac{2\left(H_n^{(3)}-\zeta(3)\right)}{z^2(z-n)}\right]\\
&=-\frac{8}{n^5}+\frac{8\zeta(2)}{n^3}+\frac{2H_n^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}
\end{align}
At the negative integers,
\begin{align}
{\rm Res}(f,-n)
&=\frac{\Psi^{(2)}(n)}{n^2}\\
&=\frac{2H_{n}^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}-\frac{2}{n^5}
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)
&=[z^1]\left(\frac{1}{z}-2\zeta(2)z\right)\left(-2\zeta(3)-12\zeta(5)z^2\right)\\
&=-12\zeta(5)+4\zeta(2)\zeta(3)
\end{align}
Hence
\begin{align}4S&=8\zeta(5)-8\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(5)+12\zeta(5)-4\zeta(2)\zeta(3)\\&=22\zeta(5)-8\zeta(2)\zeta(3)\end{align}
which implies
$$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$
A: \begin{align}
I&=\int_0^1\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx=\sum_{n=1}^\infty\left(H_n^{(3)}-\frac1{n^3}\right)\int_0^1x^{n-1}\ln x\ dx\\
&=\sum_{n=1}^\infty \left(H_n^{(3)}-\frac1{n^3}\right)\left(-\frac{1}{n^2}\right)=\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\tag{1}
\end{align}
lets start with the following sum
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\left(\zeta(2)-\sum_{k=1}^\infty\frac{1}{(n+k)^2}\right)\\
&=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac1{n^3(n+k)^2}\right)\\
&=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\left(-\frac3{k^4}\left(\frac1n-\frac1{n+k}\right)+\frac2{k^3n^2}+\frac1{k^3(n+k)^2}-\frac1{k^2n^3}\right)\right)\\
&=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(-\frac{3H_k}{k^4}+\frac{2\zeta(2)}{k^3}+\frac1{k^3}\left(\zeta(2)-H_k^{(2)}\right)-\frac{\zeta(3)}{k^2}\right)\\
&=\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}+2\zeta(2)\zeta(3)+\zeta(2)\zeta(3)-S-\zeta(2)\zeta(3)\\
2S&=3\zeta(2)\zeta(2)-3\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\
S&=3\zeta(2)\zeta(3)-\frac92\zeta(5)
\end{align}
using the well known formula $$\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}+\sum_{n=1}^\infty\frac{H_n^{(b)}}{n^a}=\zeta(a)\zeta(b)+\zeta(a+b)$$
therefore $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\frac{11}{2}\zeta(5)-2\zeta(2)\zeta(3)\tag{2}$$
plugging $(2)$ in $(1)$, we get
$$I=2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)$$
A: A nice way to calculate the main sum with a bonus
Using the well-known identity 
$$\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}+\sum_{n=1}^\infty\frac{H_n^{(b)}}{n^a}=\zeta(a)\zeta(b)+\zeta(a+b)$$
gives
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)\tag1$$
On the other hand, by Cauchy product we have 
$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty x^n\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)$$
set $x=1$ and use $\sum_{n=1}^\infty 
\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ we get 
$$3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=7\zeta(2)\zeta(3)-8\zeta(5)\tag2$$
By $(1)$ and $(2)$ we get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$
and
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$
A: $$\int_0^1\ln(x)\frac{Li_1(x)}{1-x}dx=-\zeta(3)$$
$$\int_0^1\ln(x)\frac{Li_2(x)}{1-x}dx=-\frac{3}{10}\zeta^2(2)$$
The same method may work for the present case. (To be continued.)
