I am trying to find the asymptotic expansion of the following integral as $t \rightarrow \infty$ (I will not be careful with overall constants):
$G(t) = \int_{-\pi}^{\pi} dp_1 \int_{-\pi}^{\pi} dp_2 \, \frac{e^{\frac{i}{\sqrt{2}} (p_1+p_2)t}}{m^2 + 4 - 2 \left( \cos(p_1) + \cos(p_2)\right)}$.
Here $m > 0$. If we make the transformation $u = \frac{1}{2}(p_1 + p_2)$, $v = \frac{1}{2}(p_1-p_2)$, we obtain:
$G(t) \propto \int_{-\pi}^{\pi} dv\ \int_{-\pi + v}^{\pi+v} du\, \frac{e^{i\sqrt{2}ut}}{m^2+4-4\cos(u)\cos(v)}$.
I want to use residue theorem:
$G(t) \propto \int_{-\pi}^{\pi} dv\ \left( \int_{-\pi+v+i\infty}^{-\pi+v}+ \int_{-\pi + v}^{\pi+v} + \int_{\pi+v}^{\pi + v + i\infty} - \int_{-\pi+v+i\infty}^{-\pi+v} - \int_{\pi+v+i\infty}^{\pi+v}\right) du\, \frac{e^{i\sqrt{2}ut}}{m^2+4-4\cos(u)\cos(v)}$
Of the $5$ integrals in the parenthesis, the first three forms a closed contour in the upper-half semi plane. Then by residue theorem, if we define $\omega = \ln \left( \frac{m^2+4}{4\cos(v)}+ \sqrt{\left(\frac{m^2+4}{4\cos(v)}\right)^2 +1 } \right)$:
$G(t) \propto \int_{-\pi}^{\pi} dv \, \frac{e^{-\sqrt{2}\omega t}}{4\cos(v)\sinh(\omega)} - \int_{-\pi}^{\pi} dv \left(\int_{-\pi+v+i\infty}^{-\pi+v} + \int_{\pi+v+i\infty}^{\pi+v}\right)du\,\frac{e^{i\sqrt{2}ut}}{m^2+4-4\cos(u)\cos(v)}$
The first term can be approximated by finding minimum value of $\omega(v)$. I don't know how to proceed with the second term, let me assume that it is negligible compared to the first term. Then one finds:
$G(t) \propto e^{-\sqrt{2}t\, \text{arccosh} (1+\frac{m^2}{4})}$.
This problem is taken from page 7 of: https://www2.ph.ed.ac.uk/~s0948358/mysite/Intro-LQCD-basics.pdf. Here the author doesn't show the calculation explicitly but the result is identical to what I have found by calculating residue and ignoring the second term above.