Series expansion of $\text{Li}_3(1-x)$ at $x \sim 0$ My question is simple, but maybe hard to answer. I would like to have a series expansion for $\text{Li}_3 (1-x)$ at $x \sim 0$ in the following form:
$$\text{Li}_3 (1-x) = \sum_{n=0} c_n x^n + \log x \sum_{m=1} c_m x^m. \tag{1}$$
The first few terms are:
$$\text{Li}_3(1-x) = \zeta (3)-\frac{\pi ^2 x}{6}+\left(\frac{3}{4}-\frac{\pi ^2}{12}\right) x^2+\left(\frac{7}{12}-\frac{\pi
   ^2}{18}\right) x^3+\left(\frac{131}{288}-\frac{\pi ^2}{24}\right) x^4+\left(\frac{53}{144}-\frac{\pi
   ^2}{30}\right) x^5+ \left(-\frac{x^2}{2}-\frac{x^3}{2}-\frac{11 x^4}{24}-\frac{5 x^5}{12}\right) \log x+O\left(x^6\right), \tag{2}$$
however Mathematica fails to go beyond $\mathcal{O}(x^{15})$. The coefficients for the $\log$ term are easy to guess:
$$c_m = \frac{H_{m-1}}{m}. \tag{3}$$
I also managed to obtain the coefficient with the $\pi^2$:
$$- \frac{\pi^2}{6n}, \tag{4}$$
but I am struggling with the remaining coefficient, especially with so few coefficients. Any idea what this coefficient could be?
 A: I don't know if there is a closed-form expression for the coefficients, but the following is a way to get the first few terms of the series.  The approach can be used to get as many terms as you want.
Starting with the fact that $$- \int_{0}^{x} \frac{\operatorname{Li}_{n}(1-t)}{1-t} \, \mathrm dt= \int_{1}^{1-x} \frac{\operatorname{Li}_{n}(u)}{u} \, \mathrm du = \operatorname{Li}_{n+1}(1-x) - \operatorname{Li}_{n+1}(1),$$ we have
$$\begin{align} \operatorname{Li}_{2}(1-x) - \frac{\pi^{2}}{6} &= - \int_{0}^{x}\frac{\operatorname{Li}_{1}(1-t)}{1-t} \, \mathrm dt\\ & = \int_{0}^{x}\frac{\ln(t)}{1-t} \, \mathrm dt \\ &= \int_{0}^{x} \ln(t) \left(1+\mathcal{O}(t) \right) \, \mathrm dt \\ &= x \left( \ln(x)-1 \right) + \mathcal{O}(x^{2}).\end{align}$$
Then $$\begin{align} \operatorname{Li}_{3}(1-x) -\zeta(3) &= - \int_{0}^{x}\frac{\operatorname{Li}_{2}(1-t)}{1-t} \, \mathrm dt \\ &= - \int_{0}^{x} \frac{\frac{\pi^{2}}{6}+t \left(\ln(t)-1 \right) + \mathcal{O}(t^{2})}{1-t} \, \mathrm dt \\ &= - \int_{0}^{x} \left(\frac{\pi^{2}}{6}+t \left(\ln(t)-1 \right) + \mathcal{O}(t^{2}) \right)\left(1+ t+\mathcal{O}(t^{2}) \right) \, \mathrm dt \\ &= -\frac{\pi^{2}}{6}x - \frac{\pi^{2}}{6} \frac{x^{2}}{2}+ \frac{x^{2}}{4} \left(1-2 \ln(x)\right) + \frac{x^{2}}{2} + \mathcal{O}(x^{3}). \end{align} $$
Therefore, we have $$\operatorname{Li}_{3}(1-x) = \zeta(3) - \frac{\pi^{2}}{6}x + \frac{x^{2}}{12} \left(-6 \ln(x) - \pi^{2}+9 \right) + \mathcal{O}(x^{3}).$$
A: Using the method explained by @RandomVariable, it is easy to generate a lot of coefficients and to find the pattern. The full expression reads:
$$\text{Li}_3 (1-x) = \zeta_3 - \frac{\pi^2}{6} x + \sum_{n\geq 2} \left( \frac{H_n - n \psi^{(1)}_n}{n^2} - \frac{1}{n^3} \right) x^n - \log x \sum_{n \geq 1} \frac{H_{m-1}}{m} x^m.$$
However as explained in the comments this expression is not super useful, since resumming the middle term restores the $\text{Li}_3 (1-x)$ on the RHS. At least it is useful for producing an efficient expansion of the LHS.
