I need to calculate an integral by the saddle point method.
$I=\int_0^\infty dt\, f(t)\, e^{i S(t)}$.
Notice that in my case $S(t)$ has not the typical form $S(t)=\lambda g(t)$, with a parameter $\lambda \rightarrow \infty$ . Instead, it has the form $S(t)=\int_0^t h(t′)dt′$, and grows continuously with the parameter $t$.
I can find analytically the point $t_s$ for which $\frac{\partial S(t_s)}{\partial t}=0$.
Nevertheless $\frac{\partial^2 S(t_s)}{\partial t^2}=0$ , as well. This is a problem because the method is only valid for $\frac{\partial^2 S(t_s)}{\partial t^2}\neq0$. If I think of the saddle point method as a simple Taylor expansion of the argument of the exponential to second order, I feel tempted to write the saddle point solution as
$I^{sp}\approx f(t_s)e^{i S(t_s)}$,
instead of
$I^{sp}\approx \frac{1}{\sqrt{\partial^2 S(t_s)/\partial t^2}} f(t_s)e^{i S(t_s)}$
i.e., just evaluate the phase at $t_s$ but do not use the second derivative with respect to $t$ in the denominator. Is it correct? Or I should go to order $>2$ when the second derivative is zero?
I think the saddle point solution should be just
$I^{sp}\approx f(t_s)e^{i S(t_s)}$
Because the original formulation can be thought of as a Taylor expansion of the exponential to second order, and the square root of the denominator would come from the gaussian integral involving the second order, which is null, and therefore that integral is not performed for getting the result. See, for instance How is the Saddle point approximation used in physics?