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 $N$.
2 Answers
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 $$ \int_{\varkappa a}^{\varkappa N}\log(e^x1)\,dx= \text{Li}_2(e^{\varkappa a}) \text{Li}_2(e^{\varkappa N})+ i \pi \varkappa( aN), $$ where $\text{Li}_2(x)$ is the polylogarithm. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N}) \text{Li}_2(e^{\varkappa a}))+i \pi (Na)}. $$

$\begingroup$ @JerzyKowalskiGlikman fixed the imaginary part. $\endgroup$– AndrewJun 25, 2013 at 16:00
I think I would go for a EulerMaclaurin expansion of the second term of
$$ \sum_{k=a}^N \log(e^{k\kappa}1) = \kappa\frac{(N+a)(Na+1)}{2}\sum_{k=a}^N\log(1e^{k\kappa}). $$