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Let $\epsilon$ be an $N$th root of unity, and $q=\epsilon e^h$ where $h<0$. I am trying to give a derivation of the lead term of $$(z;q)_{\infty}=\prod_{n=1}^{\infty}(1-zq^n),$$ as $h\rightarrow 0^-$.

From Kashaev and Fadeev's paper on the quantum dilogarithm if $\epsilon=1$ then

$$(z;q)_{\infty}\rightarrow\frac{1}{\sqrt{1-z}}e^{\frac{L_2(z)}{h}}(1+O(h)),$$ where $$L_2(z)=\int_0^z\frac{\log{(1-t)}}{t}dt.$$

In Reshetikhin's paper on the Quasitriangularity of Quantum groups at roots of $1$, in Communications in Mathematical Physics, 170, 79-99 (1995), he gets, $$(z;q)_{\infty}\rightarrow e^{\frac{-1}{N^2h}L_2(z^n)}(1-z^N)^{1/2}\prod_{m=0}^N(1-\epsilon^mz)^{\frac{-m}{N}}(1+O(h)).$$

My sense is there should be a direct path from Kashaev and Fadeev's formula to Reshetikhin's formula via the factorization identity for Lobachevsky function.

Recall that if $\Lambda(\theta)$ is the Lobachevsky function, and $z=e^{2i\theta}$ then
$$2{\bf i}\Lambda(\theta)=L_2(e^{2i\theta})-L_2(1)+\theta(\pi-\theta).$$

The factorization identity, which can be found in chapter 7 of Thurston's notes says, $$\Lambda(N\theta)=\sum_{r\ mod N}N\Lambda(\theta+\frac{\pi r}{N}).$$

My feeling is that you can parse the product $\prod_{n=1}^{\infty}(1-zq^n)$ according to the residue class $r$ of $n$ modulo $N$, to get

$$(z;\epsilon e^h)_{\infty}=\prod_{n=1}^{\infty}(1-z(\epsilon e^h)^n)=\prod_{r=0}^{N-1}(\prod_{n=1}^{\infty}(1-z\epsilon^r e^{Nnh+rh}))= \prod_{r=0}^{N-1}(z\epsilon^r e^{hr}; e^{Nh})_{\infty}.$$ Apply the formula for the asymptotics as you got to $1$, and then use the factorization formula to combine the integrals. I get a formula that is close to Reshetikhin's, but I am missing the idea to close it up. Specifically, I can't see where the $\prod_{m=0}^N(1-\epsilon^mz)^{\frac{-m}{N}}$ comes from.

Maybe someone who is better at this sort of analysis can just see it?

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