What partial sum formulae exist for this basic hypergeometric series?

Question

I've run into:

$$\sum_{x=1}^{\infty} {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$

I am interested mostly in the cases where $a = 1$ or $ a = 2$

Things I've done so far: reference a few places on basic hypergeometric series, not limited to something that looks remotely like what I'm interested in: Take a look at the "Simple Series" section, the first example

I'm looking for formulae that are "short and simple", ideally. Of course, if that cannot be done, I'll settle for computationally efficient with a static number of terms regardless of input.

EDIT: I've made some progress, please see: JJacquelin's answer to this question, can someone help get their attention (I cannot comment yet). If we cannot, should not, or do not wish to get their attention through contact, perhaps explaining some of the manipulations in their answer might help us here. For example, pulling out the $1 \over m$ from the sum in the second to last line to obtain the integral shown, why was that done? What technique would apply if it were $1 \over m!$ instead?

Secondly, I've found that theta functions may be involved somehow:See GEdgar's answer here as well as Paramanand Singh's answer here.

Answer 1

Not an answer, but a research idea: first, replace $x$ by $k$ then write your sum as follows:

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{k=1}^\infty \left(c_k \frac{k^a}{q^k}\right) \frac{q^k}{1-q^k}$

where $c_k=0$ for $k>n$ and 1 otherwise. This is a Lambert series; see wikipedia. That is, let

$a_k = \left(c_k \frac{k^a}{q^k}\right)$

and then apply a Dirichlet convolution (see wikipedia) to get

$b_m = (a*1)(m) = \sum_{k|m} a_k$

and thus

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{m=1}^\infty b_m q^m$

So now you have an infinite sum, in place of a finite one. A different, less fancy way to say what I just said is to just write

$\frac{1}{1-q^k} = \sum_{m=0}^\infty (q^k)^m$

and just plug that in, and then exchange order of summation. That way, you don't have a pesky denominator; you traded it for an infinite sum, which is pesky in a different way.

I suspect my suggestion above is completely useless; I cant quite figure out what you want. You might also find more willing help on math exchange instead of mathoverflow?

Answer 2

Please note that, if I am not mistaken, for the case of $q \in (0,1)$ the sum diverges, since

$$ \sum_{x=1}^{\infty} {x^a\over 1-q^{x}} = \sum_{x=1}^{\infty} \sum_{l=0}^{\infty}{x^a q^{l x}} > \sum_{x=1}^{\infty} x^a = \infty $$

About your question concerning JJacquelin's manipulations, the reason he integrates is because he wants to get rid of the term $1/m$ in the denominator that prevents him from using his previous result. Doing this integration yields

$$ \int \frac{1}{x} \sum_{m=1}^\infty \frac{(x y)^m}{1-y^m} dx = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int x^{m-1}dx= \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m} = \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m} $$

If it had $m!$ in the denominator, as you ask, the same manipulation would have led to multiple integrations

$$ \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \sum_{m=1}^\infty \frac{y^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\sum_{m=1}^\infty \frac{(x_1 y)^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\left [ \psi_y \left( 1 + \frac{\ln(x_1)}{\ln(y)} + \ln(1-y) \right) \right]dx_1 dx_2 \dots dx_m $$