During my studies of generating functions for expressions composed of harmonic numbers and its asyptotic approximations I found a simple sum which seems to have been neglected so far
$$s171219a=\sum _{k=1}^{\infty } x^k \left(H_k-(\log (k)+\gamma )\right){}^2\tag{1}$$
The tough part for me is the sum of the mixed term
$$s171219b=\sum _{k=1}^{\infty } x^k H_k \log (k)\tag{2}$$
Can you please help me with the tough part?
What I did up to now is:
Using the integral representation $H_k = \int_{0}^{1} \frac{1-u^k}{1-u} \, du$ the sum (2) under the integral becomes
$$A = \frac{1}{1-u} \sum _{k=1}^{\infty } (1-u^k) x^k \log (k)\tag{3}$$
Using the clever trick to generate $\log$-factors
$$\log(k) = \frac{\partial k^t}{\partial t}|_{t=0} \tag{4}$$
we get $A = R-S$ where
$$R =\frac{1}{1-u} \sum _{k=1}^{\infty } x^k k^t = \frac{\text{Li}_{-t}(x)}{1-u} \tag{5}$$
$$S = \frac{\text{Li}_{-t}(x\;u)}{1-u} $$
so that
$$A(t,x) = \int_{0}^1 \frac{\text{Li}_{-t}(x)-\text{Li}_{-t}(x\;u)}{1-u} \, du \tag{6}$$
and the sum (2) becomes
$$ \frac{\partial A(t,x)}{\partial t}|_{t=0}$$
I'm stuck, however, with the integral (6) over the polylog function.
