I have the following conceptual question (I tried briefly looking for some references but could not find any - relevant literature links would be greatly appreciated!).
Question I want to use the saddle point approximation for the following integral: $$I=\oint_{z=0, |z|\ge |w|} dz \oint_{w=0} dw \frac{1}{(z-w)^2} e^{-n f(z)}e^{-m f(w)}$$
Assume all the nice properties for $f$ (analytic etc.).
My Attempt We do the first integral without saddle point and simply get, $$I = \oint_{z=0}(-m f'(z))e^{-(n+m) f(z)} \\ = \frac{m}{(m+n)}\oint_{z=0}d \left(e^{-(m+n)f(z)}\right) = 0$$
Is there a non-zero result that can be attained? More generally I think the problem is to consider, $$I' = \oint dz ~g(z)e^{n f(z)}$$ for large $n$, with a saddle point $f'(z_0) = 0, f(z_0) \neq 0$ and also, $g(z_0)=0$. Usually we do not consider the last assumption. In such a case, how do we check if the approximation yields a non-zero/finite value? I started by considering the usual expansions $f(z) = f(z_0) + f''(z_0)(z-z_0)^2 + ..$ and $g(z) = g'(z_0) (z-z_0) + ..$. Clearly, the second possible order is a total derivative i.e. $$I' \approx g'(z_0) e^{-n f(z_0)} \oint dz ~ \exp\left(-\frac{n}{2}f''(z_0)(z-z_0)^2\right)(z-z_0) \approx \oint d \left[\exp\left(-\frac{n}{2}f''(z_0)(z-z_0)^2\right)\right] = 0 (?)$$.
Any help would be appreciated.