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(asked in MSE, but received no attention)

Suppose I need to compute a sum, $$ \sum_{n=0}^N a_n,$$ each term of which involves an integral, $$a_n=\int e^{Nf(x)+ng(x)}dx.$$

I am interested in the large-$N$ regime, so a saddle point approximation would be justified.

My problem is the following. The approximation would be different for the first terms of the sum, with $n=o(N)$, and the last terms, with $n=O(N)$ (because the stationary point of $g(x)$ must be considered in the latter case, but not in the former).

I am not sure how to go about this. Should I divide the summation range in two parts and do two different approximations? But then where do I place the boundary? Or is there a uniform approximation?

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If you perform the finite sum first, $$ \sum_{n=0}^{N} e^{ng(x)} = \frac{1-e^{(N+1)g(x)} }{1-e^{g(x)} } $$ you're left with the integral $$ \int dx\, e^{Nf(x)} \frac{1-e^{(N+1)g(x)} }{1-e^{g(x)} } = \int dx\, \exp \left( Nf(x) +\ln \frac{1-e^{(N+1)g(x)} }{1-e^{g(x)} } \right) $$ which you can still consider in saddle-point approximation - of course, attention must be paid to zeros of $g(x)$.

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  • $\begingroup$ Sure, thank you, but suppose I can't do the finite sum first. Say, because actually $a_n=b_n\int e^{Nf(x)+ng(x)}dx$ for some pesky $b_n$. $\endgroup$
    – thedude
    Commented Nov 23, 2020 at 14:40
  • $\begingroup$ Welllll ... then you should have asked that. And you'd have to say more about $b_n $ - if it depends on $N$, then we have practically nothing to go on anymore. I'd still try to do the finite sum first - after all, you only need the leading-$N$ behavior, which you might still be able to get analytically, even if the exact sum is intractable. $\endgroup$
    – Michael Engelhardt
    Commented Nov 23, 2020 at 14:46
  • $\begingroup$ Well, the title says "approximation for terms in a sum". I would like to understand how to control the approximation for the terms, if I should divide them in two parts, or not, how so, etc. Doing the finite sum first misses the point. $\endgroup$
    – thedude
    Commented Nov 23, 2020 at 14:58
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    $\begingroup$ It depends of course on how $b_n $ behaves. But I think one thing we can say is that one has to keep track of the behavior of each term with $N$ and only drop parts that fall off more quickly than $1/N$ - so one can't be too naive in disregarding the behavior of $g(x)$ in the saddle-point approximation. Note that terms of order $1/N$ end up yielding a contribution $(1/N)\sum_{n=0}^{N} \rightarrow \int dn$. $\endgroup$
    – Michael Engelhardt
    Commented Nov 23, 2020 at 15:29

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