Asymptotics of the quantum dilogarithm Fadeev and Kashaev define the quantum dilogarithm by
$$
\Psi(x) = \prod_{n=1}^\infty (1 - x q^n)
$$
for $|q| < 1$.
For $q = \exp(\epsilon)$, $\Re \epsilon < 0$, they say the asymptotic expansion
$$
  \Psi(x) = \frac{1}{\sqrt{1 - x}} \exp( \operatorname{Li}_2(x) / \epsilon)(1 + O(\epsilon))
$$ as $\epsilon \to 0$ is "easy to see," but I'm having trouble deriving it.
Here $\operatorname{Li}_2$ is the dilogarithm
$$
  \operatorname{Li}_2(x) = - \int_0^x \frac{\log( 1-t )}{t} \, dt = \sum_{n=1}^\infty \frac{x^n}{n^2}.
$$
I've tried expanding
$$
    \log \left( \Psi(x) \sqrt{1-x} \exp(- \epsilon^{-1} \operatorname{Li}_2(x)) \right) = \log \Psi(x) + \frac{1}{2} \log(1 - x) - \frac{1}{\epsilon} \operatorname{Li}_2(x).
$$
but this doesn't seem like it can work: if you expand $\log \Psi(x)$ in $\epsilon$ there's only positive powers, so I don't see how you can get a cancellation with the $\epsilon^{-1} \operatorname{Li}_2(x)$.
 A: Not exactly an answer, but this may help.
Define the coefficients $a_n(q)$ by
$$f_q(x)=\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}a_n(q)x^n.$$
Since $f_q(0)=1$, we have $a_0(q)=1$. Then note that
$$\begin{align}
\prod_{m\ge1}(1-xq^m)&=(1-xq)\prod_{m\ge2}(1-xq^m)\\
&=(1-xq)\prod_{m\ge1}(1-xq^{m+1})\\
&=(1-xq)\prod_{m\ge1}(1-(xq)q^{m})\\
&=(1-xq)f_q(xq).
\end{align}$$
Thus
$$\begin{align}
\sum_{n\ge0}a_n(q)x^n&=(1-xq)\sum_{n\ge0}a_n(q)q^nx^n\\
&=\sum_{n\ge0}a_n(q)q^nx^n-\sum_{n\ge0}a_n(q)q^{n+1}x^{n+1}\\
&=a_0(q)(qx)^0+\sum_{n\ge1}a_n(q)q^nx^n-\sum_{n\ge1}a_{n-1}(q)q^{n}x^{n}\\
&=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\
1+\sum_{n\ge1}a_n(q)x^n&=1+\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}\\
\sum_{n\ge1}a_n(q)x^n&=\sum_{n\ge1}\left(a_n(q)-a_{n-1}(q)\right)q^{n}x^{n}.
\end{align}$$
Equating coefficients, we have
$$a_n(q)=\left(a_n(q)-a_{n-1}(q)\right)q^{n},\qquad n\ge1.$$
This is
$$a_n(q)=\frac{q^n}{q^n-1}a_{n-1}(q).$$
And since $a_0(q)=1$, we have
$$a_n(q)=\prod_{k=1}^{n}\frac{q^k}{q^k-1}=\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n},$$
that is
$$\prod_{m\ge1}(1-xq^m)=\sum_{n\ge0}\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n}x^n.$$
Hopefully this can help :)
