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The following is defined, where $\epsilon \to 0^+$ is a cutoff: $$ \mathcal{F}(Z)=\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \frac{1}{\sinh^2 s/2} e^{-sx}. $$

Question: how do we see that $\mathcal{F}$ has a perturbative + non-perturbative expansion (in $x \gg 0$) given by

$$ \mathcal{F}(Z) = \sum_g \mathcal{F}_g x^{2-2g} + O(e^{-x}),$$ for some coefficients $\mathcal{F}_g$ ?

Reference: the issue is discussed in arXiv:hep-th/9809187 eq. (3.2) and below. The book Quantum Field Theory by Itzykson and Zuber eq. (4-117) and below should help clarify.

It should be possible to see this kind of expansion by rescaling the variable $s$ in the integral and making a Taylor expansion of the $\sinh^2$ in the denominator. The non-perturbative terms should arise from picking up poles at imaginary values of $s= 2\pi i n.$

Note that the $\epsilon$ parameter is supposed to ensure convergence of the integral.


I think I found a partial answer in the paper http://arxiv.org/abs/0907.4082 paragraphs 2.2 and 2.4, although I'm not 100% sure how one relates formula (2.20) here with (2.1) in http://arxiv.org/abs/hep-th/9812127

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First of all, you can simplify this by rewriting it as an indefinite integral of $x = Z/F$ like so: \begin{equation} \int_\epsilon^\infty \frac{ds}{s^3} \left ( \frac{s/2}{\sinh s/2} \right )^2 e^{-s x} = - \frac{1}{4} \int_\epsilon^\infty ds \int_0^x dx \frac{1}{\sinh^2 s/2} e^{-s x} \end{equation}

From here, you have roughly \begin{equation} \int_\epsilon^\infty ds \int_0^x dx \frac{e^{-s}}{(1 - e^{-s} )^2} e^{-s x} \end{equation} and the fraction can be Taylor expanded and the integrals evaluated order by order. I believe that's how that would work.

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  • $\begingroup$ I'm sorry, that limit of integration should be from $x$ to $\infty$. $\endgroup$ – webb Nov 6 '13 at 21:29
  • $\begingroup$ what you get is $-\int_{\epsilon}^\infty \mathrm{d}s \int_x^\infty \mathrm{d}m \frac{e^{-s}}{(1-e^{-s})^2}e^{-sm}$: are you suggesting to expand near $s=\infty$ the fraction $\frac{e^{-s}}{(1-e^{-s})^2}$ ? $\endgroup$ – jj_p Nov 8 '13 at 13:47
  • $\begingroup$ Because then, I'm not sure how to proceed to obtain the two contributions above $\endgroup$ – jj_p Nov 8 '13 at 14:46
  • $\begingroup$ $1/(1 - x)^2 = \frac{d}{d x} \sum_n x^n$ which is more or less the form you have for $x = e^{-s}$ and since $s > 0$ the series will converge. Then you have an infinite series of exponentials and you can integrate them that way. $\endgroup$ – webb Nov 8 '13 at 18:49
  • $\begingroup$ OK, but have you been able to reproduce the above result in this way? $\endgroup$ – jj_p Nov 12 '13 at 9:00

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