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Let $\operatorname{Li}_s(z)$ denote the polylogarithm function $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s}.$$

Does there exists a closed form or a known function which generates the polylogarithm on powers of the order $s$? I mean How to formulate $G$ such that

$$G(x,z)=\sum_{n=0}^\infty \operatorname{Li}_n(z) x^n$$ or the exponential generating function for the polylogarithm $$G(x,z)=\sum_{n=0}^\infty \frac{\operatorname{Li}_n(z)}{n!}x^n$$ instead?

For example we know for $z=1$ and starting from a higher order, its generating function is Digamma, as shown in Generating functions and the Riemann Zeta Function

Dr Potato
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    Your polylogarithm does not depend on $n$. – Gary Jun 29 '21 at 05:27
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    If you substitute $$ {\rm Li}n (z) = \frac{z}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{{t^{n - 1} }}{{{\rm e}^t - z}}{\rm d}t} $$ for $n\geq 1$ and change the order of summation and integration, you obtain a representation in terms of the modified Bessel function: $$ z\sqrt x \int_0^{ + \infty } {\frac{{t^{ - 1/2} }}{{{\rm e}^t - z}}I_1 (2\sqrt {xt} ){\rm d}t} = \sum\limits{n = 1}^\infty {\frac{{{\rm Li}_n (z)}}{{n!}}x^n } . $$ – Gary Jun 29 '21 at 06:54
  • You should have posted it as an answer! although such form is not so closed and it would need a bit more of simplification! – Dr Potato Jun 29 '21 at 14:43

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Supposing $|z|<1$ and that we have the right to reverse the order of the series we get : \begin{align} G(x,z)&:=\sum_{n=0}^\infty \sum_{k=1}^\infty \frac{z^k}{k^n} x^n\\ &= \sum_{k=1}^\infty z^k \sum_{n=0}^\infty \left(\frac xk\right)^n\\ &= \sum_{k=1}^\infty \frac{z^k}{1-\frac xk},\quad\text{for $x\not\in \mathbb{N}$}\\ &= \sum_{k=1}^\infty \frac{k\;z^k}{k-x},\quad\text{for $x\not\in \mathbb{N}$}\\ &= z\frac d{dz}\sum_{k=1}^\infty \frac{z^k}{k-x},\quad\text{for $x\not\in \mathbb{N}$}\\ &= z\frac d{dz}\left(z\,\Phi(z, 1, 1-x)\right),\quad\text{for $x\not\in \mathbb{N}$}\\ &= z\left(\frac1{1-z}+x\,\Phi(z, 1, 1-x)\right),\quad\text{for $x\not\in \mathbb{N}$}\\ \end{align} using the Lerch zeta function $\displaystyle\;\Phi(z, s, \alpha) := \sum_{n=0}^\infty \frac { z^n} {(n+\alpha)^s}$

Your second generating function appears not so easy : \begin{align} H(x,z)&:=\sum_{n=0}^\infty \frac{\operatorname{Li}_n(z)}{n!}x^n\\ &= \sum_{k=1}^\infty z^k \sum_{n=0}^\infty \frac 1{n!}\left(\frac xk\right)^n\\ &= \sum_{k=1}^\infty z^k \exp{\frac xk}\\ \end{align} but it seems that Gary already provided an answer!

Raymond Manzoni
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  • Yes indeed also I obtained this formula for the factorial version, but I have not been able to simplify it. – Dr Potato Jun 29 '21 at 14:44
  • Well, the formula provided by Gary is not really simple (even if there is a long list of nearly similar integrals producing Bessel functions). Another idea could be rewrite my last formula as a PDE. Anyway the first $G$ function is simpler and you may omit the $\frac z{1-z}$ term if you start at $n=1$. Excellent continuation, – Raymond Manzoni Jun 29 '21 at 15:16
  • Another integral expression would be taking @gary 's approach to the first version, and $e^z$ as the generating function of the reciprocal of the factorial, and then apply the termwise Hadamard product of two generating functions. – Dr Potato Jun 29 '21 at 17:29
  • An integral expression for the factorial version is obtained from substituting the Laurent series of the exponential function centered and evaluated at zero, $$\frac{1}{n!}=\frac{1}{2\pi i} \oint \frac{e^u}{u^{1+n}}du,$$ in the second version: $$H(x,z)=\sum_{n=0}^\infty \frac{\operatorname{Li}n(z)}{n!}x^n=\sum{n=0}^\infty \frac{1}{2\pi i} \oint \frac{e^u}{u^{1+n}}du \operatorname{Li}n(z)x^n=\frac{1}{2\pi i} \oint \sum{n=0}^\infty \operatorname{Li}_n(z) (\frac{x}{u})^n e^u \frac{du}{u}$$ thus $$H(x,z)=\frac{1}{2\pi i} \oint G(\frac{x}{u},z) e^u \frac{du}{u}.$$ – Dr Potato Jul 01 '21 at 17:06
  • @DrPotato: well you may put all this in an answer and add all the details you'll find relevant (self-answers are recommended at SE!). Excellent continuation, – Raymond Manzoni Jul 01 '21 at 18:46
  • For now we have that one. If someone else comes up with a closed form of it I will really appreciate it and change the right answer to that one! – Dr Potato Jul 02 '21 at 03:31
  • Nevertheless, it seems to equal zero: For simplicity, following the same procedure but changing $\operatorname{Li}n$ by $\operatorname{Li}{1+n}$ leads to $$\sum_{n=0}^\infty \frac{Li_{1+n}(z)}{n!} x^n =\frac{1}{2\pi i} \oint e^u \sum_{k=0}^\infty \frac{z^k}{k+1-\frac{x}{u}} \frac{du}{u}$$ By the Residue Theorem this would equal the same as approximating to $u=0$ in the other integrand factor than $\frac{du}{u}$, which tends to have an $-\infty$ denominator! Where am I wrong in this reasoning? Is it not possible to change the order of integration and summation in this case ..or what? – Dr Potato Jul 02 '21 at 04:32