5
$\begingroup$

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 $N$.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ @JerzyKowalski-Glikman fixed the imaginary part. $\endgroup$
    – Andrew
    Jun 25 '13 at 16:00
3
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.