Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$

Question

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 .

Answer 1

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 $\zeta(5)$.

Answer 2

$$\int_0^1\ln(x)\frac{Li_1(x)}{1-x}dx=-\zeta(3)$$ $$\int_0^1\ln(x)\frac{Li_2(x)}{1-x}dx=-\frac{3}{10}\zeta^2(2)$$ The same method may work for the present case. (To be continued.)