TITLE: How to estimate of $\prod_{k=a}^N \frac{1}{e^{k\kappa}-1}$ for large $N$?
QUESTION [5 upvotes]: I have to estimate the expression $\prod_{k=a}^N \frac{1}{e^{k\kappa}-1}$ for $\kappa$ very small $\kappa \sim 10^{-19}$ and $N$ very large $N\sim 10^{26}$ and $a$ arbitrary $a=1, \ldots, N$. I do not really need an exact expression, just the leading order expression in $N$.

REPLY [4 votes]: Taking logarithm gives
$$
-\sum_{k=a}^N \log(e^{k \varkappa}-1)=
-\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa.
$$
The last sum is a Riemann sum of the integral
$$
\int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx=
\text{Li}_2(e^{\varkappa a})-
\text{Li}_2(e^{\varkappa N})+
i \pi  \varkappa( a-N),
$$
where $\text{Li}_2(x)$ is the polylogarithm.
So the product is approximately equal to
$$
e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})-
\text{Li}_2(e^{\varkappa a}))+i \pi  (N-a)}.
$$

REPLY [3 votes]: I think I would go for a Euler-Maclaurin expansion of the second term of
$$
-\sum_{k=a}^N \log(e^{k\kappa}-1) = -\kappa\frac{(N+a)(N-a+1)}{2}-\sum_{k=a}^N\log(1-e^{-k\kappa}).
$$