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I am trying to evaluate the following integral using the saddle point method:

\begin{equation} \int_0^\infty \exp\left\{-N\left(\frac{\epsilon}{\sigma}+\ln (\sigma+1)+\frac{y^{2}}{\sigma+1-\tau}+\frac{\left(a^2+x^2\right)}{\sigma +\tau +1}\right)+\ln\left[2\cosh \left(\frac{2 a N x}{\sigma +\tau +1}\right)\right]\right\}\mathrm{d}\sigma \end{equation}

The problem I have is that I do not have the form: \begin{equation} \int_0^\infty e^{-Nf(\sigma)}\mathrm{d}\sigma \end{equation}

where $f(\sigma)$ is independent of $N$ due to this term: $$\ln\left[2\cosh \left(\frac{2 a N x}{\sigma +\tau +1}\right)\right]$$ $\implies$ Is there a way to extract the $N$? Or modify the term to some approximation?

$\implies$ Is there any other way to evaluate this integral?

If I missed to provide some important details or context, please let me know and I will edit the question.

Matt
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  • Often the method works anyway, but it's difficult to assert when and why https://math.stackexchange.com/questions/2234358/asymptotics-justification-for-abusing-laplaces-method – leonbloy Nov 12 '20 at 17:23
  • in the region where $2 a Nx-\tau-1\gg \sigma$ we have indeed $\log(2\cosh(...))$ is $\sim N $... – asgeige Nov 12 '20 at 18:45
  • I think the leading order asymptotics for the special case $a=x,y=0$ and $O(1)\sim a,\epsilon,\tau >0$ of your integral ,for large $N$, are given by $$ I_N\sim\sqrt{\frac{2 \pi}{N F''(\sigma^)}}e^{-N \mathcal F(\sigma^)} $$

    where $\sigma^*$ is the unique positive solution to the equation $-4a^2 \sigma^2(\sigma+1)-\epsilon (\sigma+1)(\sigma+\tau+1)^2+(\sigma+\tau+1)^2\sigma^2=0 $

    and $$ \mathcal{F}(\sigma)=\frac{4a^2}{(\sigma+\tau+1)^2}+\log(\sigma+1)+\frac{\epsilon}{\sigma} $$

     – asgeige Nov 13 '20 at 22:04
  • maybe i write this as an answer some day, key idea is that $\log(2 \cosh(...))\sim (...)$ if the expression the brackets is large enough.

    I think my reasoning caries over to your more general case as long as the parameters stay postive (maybe $1>\tau>0$ is additionaly required since the term $\sim y^2$ could change its charcteristic radically otherwise, but this is up to you to figure out) .

     – asgeige Nov 13 '20 at 22:08

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