# Stationary phase approximation for an integral with infinity saddle points?

I need a hand with the numerical evaluation, in Mathematica, for this integral:

$$f(t)=\int_{-\infty}^\infty Exp\{it(\omega_H-\omega_l-\omega_k) - \sum _{j\neq(l,k)} S_j [1-e^{-it\omega_j}]\}\, dt$$

This is the plot of the real and imaginary part:

And this are the values for the parameters of $f(t)$ employed in the graph:

$$\omega_H=0.408194 ~~;~~ \omega_l= \omega_k =0.102036 ~~;~~\omega_j=0.169499~~;~~S_j=1.05569$$

Where the all parameters are reals; $t$ is the integration time (s), $\omega$ are oscillation frequencies (1/s) and $S_j$ is an adimensional factor.

I find that this type of non-convergent integrals are approximated by the steepest descendent method or stationary phase approximation.

Employing this approximation, the integral can be approximated to:

$$f(t) \approx \sqrt{\left(\frac{2 \pi}{\sum _{j\neq(l,k)} S_j \omega_j^2 e^{-it^*\omega_j}} \right)} Exp\{it^*(\omega_H-\omega_l-\omega_k) - \sum _{j\neq(l,k)} S_j [1-e^{-it^*\omega_j}]\}$$

Where the saddle points value $t^*$, are determined numerically by the transcendental equation:

$$\omega_H-\omega_l-\omega_k=\sum _{j\neq(l,k)} S_j \omega_j e^{-it^*\omega_j}$$

Solving this transcendental equation in Mathematica, for the parameters above, I get this result:

$$C[1] \in Integers \, \&\& \, t ==(0.- 5.89974 i) (0.131673 + (0.+6.28319 i) \, C[1])$$

This means that the transcendental equation has number of infinity solutions, which also means that this equation has an infinity number of saddles points. Then the unity partition is used to get a convergent approximation to the integral $f(t)$. In this link make an appointment for this case (steepest descendent method)

So, can anyone give me a hand with formation of the cut off functions, using the partition unity, to get a convergent form of $f(t)$?

Thanks!!