TITLE: Asymptotics of q-Pochhammer symbol
QUESTION [6 upvotes]: In one of the papers I'm reading there is an asymptotic formula for q-Pochhammer symbol it is written as: $ (x,q)_n \propto e^{\frac{1}{\hbar}(Li_2(x)-Li_2(xq^n))} $ where $q=e^{\hbar}$. How can I derive this formula and how do I generalize it to for example $(q^2,q^2)_n$ ? I'm interested in the limit $\hbar \rightarrow 0$, $n \rightarrow \infty$ where $q^n$ is fixed if it isn't clear.

REPLY [5 votes]: The $q$-Pochhammer symbol $(x;q)_n=\prod_{k=0}^{n-1}(1-xq^k)$ can more generally be written as 
$$ (x;q)_n = \frac{(x;q)_\infty}{(q^n x;q)_\infty}$$
as you can check for yourself. The "infinite" $q$-Pochhammer symbol can be expanded as
$$ (x;q)_\infty= \exp \bigg[ \frac 1 \hbar \sum_{k=0}^\infty \frac{B_k \hbar^k}{k!} \text{Li}_{2-k}(x)\bigg],  \quad |x|<1$$
where we have set $q=e^\hbar$. This expansion is a small exercise which you can complete yourself: start by expanding $\log (x;q)_\infty$ in powers of $x$ using Taylor series and then some results about the Lambert series should finish the proof. 
These two results prove your expression to leading order in $\hbar$. Applying it for $(q^2;q^2)_n$ is then straight-forward, but the limit $\hbar \to 0$ with $\hbar n$ constant is trickier. I hope this helps.
PS. May I ask where you read this?