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?

  • $\begingroup$ I'm not sure if it helps but $\log(x) \sum\limits_{m=1}^\infty c_m x^m=\log(x) \sum\limits_{m=1}^{\infty} \frac{H_{m-1}}{m} x^m=$ $\log (x) \left(\text{Li}_2\left(\frac{1}{1-x}\right)-\text{Li}_2(x)+\log^2(1-x)-\log(-x) \log(1-x)-\frac{\pi^2}{6}\right)$. $\endgroup$ Apr 13, 2022 at 14:39
  • $\begingroup$ @StevenClark Yeah I noticed that. Actually resumming the series expansion is my ultimate goal with this expression. The other term with the $\pi^2$ can also easily be resummed. $\endgroup$
    – Pxx
    Apr 13, 2022 at 14:41
  • $\begingroup$ I'm not sure what you mean as the sum $\sum\limits_{n=1}^\infty -\frac{\pi^2}{6 n}=-\frac{\pi^2}{6} \zeta(1)$ doesn't converge. $\endgroup$ Apr 13, 2022 at 14:47
  • $\begingroup$ @StevenClark But you are missing $x^n$ in your sum. $\endgroup$
    – Pxx
    Apr 13, 2022 at 14:49
  • $\begingroup$ Are you sure about $c_m$ ? Or is there a shift of index ? Interesting problem. $(\to +1)$ $\endgroup$ Apr 13, 2022 at 14:53

2 Answers 2


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 $$\operatorname{Li}_{3}(1-x) = \zeta(3) - \frac{\pi^{2}}{6}x + \frac{x^{2}}{12} \left(-6 \ln(x) - \pi^{2}+9 \right) + \mathcal{O}(x^{3}).$$

  • $\begingroup$ Thanks! But is it really more efficient than the Series of Mathematica, with which I reach at present $\mathcal{O}(x^{19})$? I would like to try it, but I don't understand what are your limits of integration. There is such a recursive formula at the top of the Wikipedia article, but it does not have a constant. $\endgroup$
    – Pxx
    Apr 13, 2022 at 17:33
  • $\begingroup$ Aah sorry in Wikipedia it has $x$ in the denominator and not $1-x$. I guess your limits of integration are also just $0$ and $x$. Gonna test the efficiency now! $\endgroup$
    – Pxx
    Apr 13, 2022 at 17:34
  • $\begingroup$ @Pxx It's the same series. I was just showing how to get the first few terms. $\endgroup$ Apr 13, 2022 at 17:44
  • $\begingroup$ This method is very efficient by the way! I could get easily 30 coefficients and obtain the same result as @Maxim gave in the comments. $\endgroup$
    – Pxx
    Apr 13, 2022 at 19:47

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.


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