\begin{align*} \frac{1}{1-z}\ln\frac{1}{1-z} &\to [z^k]A(z) = H_k\\ \frac{1}{z}\frac{1}{1-z}\ln\frac{1}{1-z} &\to [z^k]A(z) = H_{k+1} \end{align*} So, now it is enought to integrate to get the generating function. And here my problems begin.
\begin{align*} \int_0^z\frac{1}{t}\frac{1}{1-t}\ln\frac{1}{1-t}dt \end{align*}
Which i don't know how to find.
EDIT: Note that $\frac{1}{t}\frac{1}{1-t} = \frac{1}{t}+\frac{1}{1-t}$. And i think i can find the following integral:
$$\int \frac{1}{1-t}\ln\frac{1}{1-t}dt = \frac{1}{2}\left(\ln\frac{1}{1-t}\right)^2$$
Funnily, MATLAB find the second integral involved:
>> f(z) = 1/z * log(1/(1-z))
ans(z) =
-dilog(-1/(z - 1)) - log(-1/(z - 1))^2/2
But how? And is at all right way to find the generating function.
So I understand that, given $$ A(z) = {1 \over {1 - z}}\ln \left( {{1 \over {1 - z}}} \right) = \sum\limits_{0\, \le \,k} {H_{\,k} \,z^{\,k} } = \sum\limits_{1\, \le \,k} {H_{\,k} \,z^{\,k} } $$ where the $H$ do not mean the Harmonic numbers, and $H_0=0$,
then you want to find $$ \eqalign{ & \sum\limits_{1\, \le \,k} {{{H_{\,k} } \over k}\,z^{\,k} } = \sum\limits_{0\, \le \,k} {{{H_{\,k + 1} } \over {k + 1}}\,z^{\,k + 1} } = \sum\limits_{0\, \le \,k} {H_{\,k + 1} \int_0^z {t^{\,k} dt} } = \cr & = \int_0^z {\left( {\sum\limits_{0\, \le \,k} {H_{\,k + 1} t^{\,k} } } \right)d} t = \int_0^z {\left( {t^{\, - 1} \sum\limits_{0\, \le \,k} {H_{\,k + 1} t^{\,k + 1} } } \right)d} t = \int_0^z {{1 \over t}A(t)dt} \cr} $$
Making the substitution $$ 1 - u = {1 \over {1 - t}}\quad \Rightarrow \quad u = {{ - t} \over {1 - t}}\quad \Rightarrow \quad t = 1 - {1 \over {1 - u}} = {{ - u} \over {1 - u}}\quad \Rightarrow \quad dt = - {{du} \over {\left( {1 - u} \right)^{\,2} }} $$ we get $$ \int_{t = 0}^z {{1 \over t}A(t)dt} = \int_{u = 0}^{z/\left( {z - 1} \right)} {{1 \over u}\ln (1 - u)du} = - {\rm Li}_2 \left( {{z \over {z - 1}}} \right) $$
where $ {\rm Li}_2$ denotes the Dilogarithm function.