- #1

a1234

- 78

- 6

- TL;DR Summary
- Attempting to use steepest descents to develop an expansion of an exponential integral.

I am trying to develop an asymptotic expansion of the following integral using the method of steepest descents:

$$ \int_{0}^{\infty} \frac{1}{t+1}e^{ix(t^3-3)}dt $$

I rearranged it into the form ## \int e^{ix(t^3-3) - ln(t+1)} dt ##, where ## \phi(t) = -ln(t+1) ## and ## \psi(t)=x(t^3-3) ##, such that ## \rho(t)=\phi(t)+i\psi(t) ##.

When I solve for the critical points of ## \phi(t) ##, I get 3 long expressions for t. I am not sure if this is the way to proceed. What would be the correct way of going about this?

$$ \int_{0}^{\infty} \frac{1}{t+1}e^{ix(t^3-3)}dt $$

I rearranged it into the form ## \int e^{ix(t^3-3) - ln(t+1)} dt ##, where ## \phi(t) = -ln(t+1) ## and ## \psi(t)=x(t^3-3) ##, such that ## \rho(t)=\phi(t)+i\psi(t) ##.

When I solve for the critical points of ## \phi(t) ##, I get 3 long expressions for t. I am not sure if this is the way to proceed. What would be the correct way of going about this?