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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 $\theta$?

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

1 Answer 1

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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 $w$.

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