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Does anybody know where I could find the expression for $$\frac{\partial}{\partial s}\mathrm{Li}_s(z)\bigg|_{s=0}$$ or something similar?

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    $\begingroup$ Please be more clear. $\endgroup$ – ABC Mar 21 '13 at 14:16
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    $\begingroup$ A good question. There was a question yesterday where the answer was the derivative of $\mathrm{Li}_s(1/2)$ with respect to $s$. $\endgroup$ – GEdgar Mar 21 '13 at 14:40
  • $\begingroup$ +1. I never found anything about it. However, we can use integral representations. The problem is that we usually evaluate those integrals in terms of the above derivative. So, it's a closed circle as the dog trying to eat its tail. $\endgroup$ – Felix Marin Aug 13 '14 at 3:27
  • $\begingroup$ Related to: Find the derivative of a polylogarithm function, URL (version: 2015-03-18): math.stackexchange.com/q/1194499 $\endgroup$ – Erich Sep 17 '16 at 8:41
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For every $|z|<1$, we have $$ \frac{\partial}{\partial s}\mathrm{Li}_s(z)\Big|_{s=0} = -\sum_{n=1}^\infty \log (n)\frac{z^n}{n^s} \Big|_{s=0} = -\sum_{n=1}^\infty \log(n)\,z^n $$

If it is not what you are looking for, could you be more specific?

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Use partial integration, then you can relate the differentiation of $\text{Li}(j,x)$ to $\text{Li}(j,x)$ and $\text{Li}(j-1,x)$.

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  • $\begingroup$ Using Mathematica, Li[n,x] is the same as PolyLog[n,x]. Mathematica gives D[PolyLog[n. -Exp[x],x]=PolyLog[n-1,Exp[x]]. I do not know if this helps. The Fermi Dirac integral Integrate[x^j/(1+Exp[x-u]),{x,0,Infinity}]=PolyLog[1+j,-Exp[u]]. $\endgroup$ – Hong-Yee Chiu Dec 7 '15 at 23:11

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