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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 $N$.

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2 Answers 2

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 $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))+i \pi (N-a)}. $$

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Thank you. That helps! –  Jerzy Kowalski-Glikman Jun 25 '13 at 15:36
    
@JerzyKowalski-Glikman fixed the imaginary part. –  Andrew Jun 25 '13 at 16:00

I think I would go for a Euler-Maclaurin expansion of the second term of

$$ -\sum_{k=a}^N \log(e^{k\kappa}-1) = -\kappa\frac{(N+a)(N-a+1)}{2}-\sum_{k=a}^N\log(1-e^{-k\kappa}). $$

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