A sum over all permutations I want to consider the sum 
$$\sum_{\sigma\in\mathfrak{S}_n}\frac{x^{s(\sigma)}}{|\sigma|}$$
where $x$ is a real number, $s(\sigma)$ is the number of cycles of $\sigma$ and $|\sigma|$ is the product of lengths for cycles of $\sigma$. Will it possible to have an exact formula or a generating series for this number when $n$ ranges? 
 A: The  exponential formula  tells us  that the  OGF of  the  cycle index
$Z(S_n)$ of the symmetric group is given by
$$Z(S_n) = [w^n] 
\exp\left(\sum_{l\ge 1} a_l \frac{w^l}{l}\right).$$
In the present case we are evaluating at $a_l = x/l$ so we obtain
for $P_n(x)$ the polynomial in question
$$P_n(x) =  n! [w^n] 
\exp\left(\sum_{l\ge 1} x \frac{w^l}{l^2}\right)
= n! [w^n] \exp(x\mathrm{Li}_2(w)).$$
This is an EGF for $P_n(x).$ 
For computational purposes we may  use the recurrence by Lovasz of the
cycle index  $Z(S_n)$ which  yields (complexity is  partition function
rather than factorial)
$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n  a_l Z(S_{n-l})
\quad\text{where}\quad
Z(S_0) = 1.$$
We get for $P_n(x)$ that
$$P_n(x) =
(n-1)! \sum_{l=1}^n \frac{x}{l} \frac{P_{n-l}(x)}{(n-l)!}
\quad\text{where}\quad
P_0(x) = 1.$$
This yields e.g.
$$P_5(x) = {x}^{5}+5\,{x}^{4}+{\frac {125\,{x}^{3}}{12}}
+{\frac {65\,{x}^{2}}{6}}+{\frac {24\,x}{5}}$$
and
$$P_6(x) = {x}^{6}+15/2\,{x}^{5}+{\frac {295\,{x}^{4}}{12}}
+{\frac {355\,{x}^{3}}{8}}+{\frac {8009\,{x}^{2}}{180}}+20\,x.$$
We also have that the coefficients of $P_n(x)$ are given by
$$[x^k] P_n(x) = 
\frac{n!}{k!} [w^n] 
\left(\sum_{l=1}^n \frac{w^l}{l^2}\right)^k.$$
