# Asymptotic expansion of q-Pochhammer symbol near q = 1

I'd like to understand the asymptotics of the q-Pochhammer symbol $$(a;q)_\infty$$ as $$q \to 1^-$$ with $$a$$ complex, where

$$(a;q)_\infty = \prod_{n = 0}^\infty (1- aq^n).$$

More specifically, I'm actually just interested in the limiting behavior as $$a$$ approaches an arbitrary point on the unit circle in the complex plane: $$(q^x e^{i \theta}; q)_\infty$$ as $$q \to 1^-$$, with $$x$$ and $$\theta$$ real. I managed to find a partial answer in this paper, which in theorem 3.2 gives the asymptotic expansion

$$(q^x; q)_\infty = \frac{\sqrt{2\pi}}{\Gamma(x)}\left(\ln\frac{1}{q}\right)^{\frac{1}{2}-x} \prod_{k = 0, k \neq 1}^\infty \exp\left[\frac{\zeta(2-k)}{k!} B_k(x) (\ln q)^{k-1}\right]$$

for $$x > 0$$, where $$\zeta(2-k)$$ is the Riemann zeta function and $$B_k(x)$$ are Bernoulli polynomials. This is just the $$\theta = 0$$, $$x > 0$$ version of what I'm looking for. Doing some numerical checks, it seems that this expansion is still valid when extended to complex $$x$$ (at least in some neighborhood of $$x = 0$$, which is all I've checked), but this generalization isn't sufficient to get the expansion I want: in the expansion above, the argument $$a = q^x$$ always approaches 1 as $$q \to 1$$ even for complex $$x$$, whereas I want $$a = q^x e^{i\theta}$$ to approach an arbitrary point on the unit circle.

Is there some other expansion that applies in the regime I'm interested in? Or perhaps is there a way to modify the above expansion to work for general $$\theta$$?

• Have you checked whether the proof in the linked paper can be modified to handle the more general case of yours?
– Gary
Apr 4, 2021 at 12:29

After doing some more digging around, I finally found an answer. Let $$m$$ be a positive integer, $$\zeta$$ be a primitive $$m^\mathrm{th}$$ root of unity, $$q = \zeta e^{-\epsilon/m}$$, $$w$$ a complex number with $$|w| < 1$$, and $$\nu$$ a complex number with $$\epsilon \nu = o(1)$$. Then Lemma 2.1 of this paper gives an asymptotic expansion for $$(q w e^{-\nu \epsilon/m}; q)_\infty$$ at small $$\epsilon$$. Taking the case $$m = 1$$ (hence $$\zeta = 1$$) of their result, I get
$$(e^{-\epsilon x} w; e^{-\epsilon})_\infty = \exp\left(\sum_{r = 0}^\infty B_r(x)\mathrm{Li}_{2-r}(w) \frac{(-\epsilon)^{r-1}}{r!} \right),$$
where $$\mathrm{Li}_{2-r}(w)$$ are polylogarithms. Technically this result is restricted to $$|w| < 1$$, whereas I wanted $$w = e^{i\theta}$$, but doing some numerical checking it looks like this expansion is perfectly well-behaved for general complex $$w \neq 1$$ (at least for those $$w$$ that I've checked). I suspect the reason for the restriction to $$|w| < 1$$ in the Garoufalidis and Zagier paper is that for the general-$$m$$ case this restriction is necessary to ensure that you stay away from the singularities of the polylogarithms. A far as I can tell this restriction doesn't seem necessary for $$m = 1$$, though.
So assuming the extension to general $$w \neq 1$$ holds, supplementing this new expansion with the $$w = 1$$ case I gave in my original post gives a set of expansions for $$(q^x w; q)_\infty$$ as $$q \to 1^-$$ for any complex $$w$$.