Find the derivative of a polylogarithm function I was trying to find to which function the next series converges.
$$
\sum_{n=1}^{\infty} \ln(n)z^n
$$
If we take the polylogarithm function $Li_s(z)$ defined as
$$
Li_s(s)=\sum_{n=1}^{\infty} \frac{z^n}{n^s}
$$
Then it is easily seen that
$$
\sum_{n=1}^{\infty} \ln(n)z^n = - \left( \frac{\partial}{\partial s}Li_s(z)\right)_{s=0}
$$

Now, my question is how to calculate $ \frac{\partial}{\partial s}Li_s(z)$, using an integral representation for $Li$, such as
  $$
Li_s(z)=\frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{zt^{s-1}}{e^t-z} dt
$$  

Is there any nice solution to this? All my attempts are unclear about it, especially because of the derivative of $\Gamma(s)$. 
 A: You might write your series as
$$ \sum_{n=1}^\infty \int_0^1 \dfrac{(n-1) z^n}{(n-1)u + 1}\; du = \int_{0}^{1}\!{\frac {{z}^{2}}{u+1}
{\mbox{$_2$F$_1$}\left(2,{\frac {u+1}{u}};\,{\frac {2\,u+1}{u}};\,z\right)}}
\, du
$$
but I don't think you'll get a closed form for the integral.
A: See the Eq.(18) of the article https://www.carma.newcastle.edu.au/jon/MTWIII.pdf
$$
Li_0^{(m)} (z) = \sum_{n\ge 0} \zeta^{(m)}(-n) \frac{\log^n (z)}{n!} - \sum_{t=0}^m (-1)^t \left(\begin{matrix}m\\t\end{matrix}\right) \Gamma^{(t)}(1) \frac{\mathcal{L}^{m-t}}{\log z}
$$
Where $\mathcal{L} = \log(-\log z)$.
The same article discuss $\zeta^{(m)}(-n)$ at sec. 3.2.1. The $\Gamma$ derivative can be rewritten using that as $\Gamma^{'}(z) = \Gamma(z) \psi(z)$, where $\psi$ is the polygamma function of zeroth order.
At the wanted situation,
$$
Li_0^{'} (z) = \sum_{n\ge 0} \zeta^{'}(-n) \frac{\log^n (z)}{n!} 
- \frac{\log(-\log z)}{\log z}
+ \frac{\psi(1)}{\log z}
$$
This is just a way to rewrite the series, but may (I don't know) converge fast.

Another way is to use the Hankel contour integral representation (valid around $s=0$) (see https://en.wikipedia.org/wiki/Polylogarithm#Integral_representations)
$$
Li_s(e^{\mu}) = - \frac{\Gamma(1-s)}{2\pi i} \oint_H \frac{(-t)^{s-1}}{e^{t-\mu}-1} dt
$$
Taking the derivative with respect to $s$ and then making $s\rightarrow 0$,
$$
Li'_0(e^{\mu}) = \frac{1}{2\pi i} \oint_H \frac{1}{e^{t-\mu}-1} \frac{dt}{t} \left[ \gamma + \log (-t) \right] = \gamma Li_0(e^{\mu}) + \frac{1}{2\pi i} \oint_H \frac{\log (-t)}{e^{t-\mu}-1} \frac{dt}{t}
$$
Well, one problem still remains.
