How do I apply the method of steepest descents to this integral?

  • #1
a1234
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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?
 
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  • #2
You seem to be treating [itex]t[/itex] as real throughout. Your task is to find a contour in the complex [itex]t[/itex]-plane which will yield the same value as the contour along the positive real axis (ie. you can't cross any singularities of the integrand in order to get there), so you cannot assume that [itex]t[/itex] is real.

I think you can remove a constant factor of [itex]e^{-3ix}[/itex] outside the integration, so you are looking at [tex] I(x) = e^{-3ix} \int_0^\infty \frac{1}{1 + t} e^{ixt^3}\,dt.[/tex]

You are interested in the real part of the exponent [itex]ixt^3[/itex], which is [itex]-x|t|^3 \sin (3 \arg t)[/itex]. Your contour is therefore a straight line of the form [itex]\arg t = \mbox{const}[/itex], chosen such that [itex]\sin (3\arg t) = 1[/itex].
 
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