I am trying to solve using the saddle point method (large a>0): $$I(\alpha)= \int_{-i\pi/2}^{\pi/2}dz\, (1+z^2)e^{-a\cos(z)}$$ So I find that the point I want to expand about is z=0, because $\partial_z\cos(z)=0\implies z=0,n\pi$ So at $z_0=0$, I get $$I(\alpha)=1\int_{-\epsilon}^\epsilon e^{-a(1-z^2/2+...)}\approx e^{-a}\int_{-\infty}^\infty e^{az^2/2}\, dz\approx i\frac{\sqrt{2\pi}}{\sqrt{a}}e^{-a}$$

My question is if this is a valid approach. Mostly, did I correctly choose to expand about z=0. I get confused on which saddle point to select, because I can deform the integral in many ways.

And then if I want $a<0$, would I approach it the same way?

• u have to make sure that your path of integration is a path of steepest descent. have u checked that? furthermore your integral as it is diverges.... Oct 18, 2015 at 20:17
• Not sure how to figure out if it is or isn't. As for the integral, wolfram told me it integrates to sqrt(pi/-a) Oct 18, 2015 at 21:29
• read something about saddle point method :), yes the integral might be given in closed form. is the $i$ in the integration limits really there? Oct 18, 2015 at 21:38
• Yes it is in the limit Oct 18, 2015 at 22:34

I) Let us define $$f(z)~:=~1+z^2 \tag{1}$$ for later convenience. OP's integrand $f(z) \exp\left(-a\cos z\right)$ is an entire function, so the integral is independent on the integration contour (as long as the endpoints are the same). Let us choose an $L$-shaped contour:

$$I~:=~ \int_{-\frac{i\pi}{2}}^{\frac{\pi}{2}} \!\mathrm{d}z~f(z)\exp\left(-a\cos z\right)~=~J+iK, \tag{2}$$

where

$$J~:=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f(x)\exp\left(-a\cos x\right) \tag{3}$$ $$~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f\left(\frac{\pi}{2}-x\right)\exp\left(-a\sin x\right) \tag{4}$$ $$~=~e^{-a} \int_0^{\frac{\pi}{2}}\!\mathrm{d}x~f(x)\exp\left(2a\sin^2\frac{x}{2}\right), \tag{5}$$

and where

$$K~:=~\int_{-\frac{\pi}{2}}^0\!\mathrm{d}y~f(iy)\exp\left(-a\cosh y\right) ~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}y~f(-iy)\exp\left(-a\cosh y\right). \tag{6}$$ $$~=~\int_0^{\frac{\pi}{2}}\!\mathrm{d}y~f\left(iy-\frac{i\pi}{2}\right)\exp\left(-a\cosh \left(\frac{\pi}{2}-y\right)\right). \tag{7}$$

This $L$-shaped contour is aligned with the steepest descents of the endpoints.

II) Case $a\gg 0$:

$$J~\stackrel{(4)}{\approx}~\int_0^{\infty}\!\mathrm{d}x~f\left(\frac{\pi}{2}-x\right)\exp\left(-|a|x \right) ~=~\left(1+\frac{\pi^2}{4}\right)|a|^{-1}+O(|a|^{-2}) , \tag{8}$$

$$K~\stackrel{(6)}{\approx}~\frac{1}{2}f(0)\int_{-\infty}^{\infty} \!\mathrm{d}y~\exp\left(-|a|\left(1+\frac{y^2}{2}\right)\right) ~=~e^{-|a|}\sqrt{\frac{\pi}{2|a|}}. \tag{9}$$

$$I~\sim~\left(1+\frac{\pi^2}{4}\right)|a|^{-1} \qquad\text{for}\qquad a~\to ~\infty.\tag{10}$$
III) Case $a \ll 0$:
$$J~\stackrel{(5)}{\approx}~\frac{e^{|a|}}{2}f(0)\int_{-\infty}^{\infty} \!\mathrm{d}x~\exp\left(-2|a|\left(\frac{x}{2}\right)^2\right)~=~e^{|a|}\sqrt{\frac{\pi}{2|a|}},\tag{11}$$
$$K~\stackrel{(7)}{\approx}~\int_0^{\infty}\!\mathrm{d}y~f\left(iy-\frac{i\pi}{2}\right)\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)-|a|y\sinh \left(\frac{\pi}{2}\right) \right)$$ $$~=~\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)\right)\left(\frac{1-\frac{\pi^2}{4}}{|a|\sinh \left(\frac{\pi}{2}\right)}+O(|a|^{-2})\right).\tag{12}$$
$$I~\sim~i\exp\left(|a|\cosh \left(\frac{\pi}{2}\right)\right) \frac{1-\frac{\pi^2}{4}}{|a|\sinh \left(\frac{\pi}{2}\right)} \qquad\text{for}\qquad a~\to ~-\infty.\tag{13}$$