Steepest descent contour includes singularity (asymptotic expansions)

[Jerbearrrrrr]

Homework Statement

We require an asymptotic expansion of (t in general complex):
$$\int _{-1} ^\infty \frac{e^{i \lambda t^2} }{\sqrt{1+t}}$$ dt
in the limit (lambda) tends to infinity.
Hint given is to sketch the path of Im(it^2)=const through t=0 and t=-1 in the complex t-plane.

The Attempt at a Solution

I have a candidate steepest descent path (it's kind of a standard one - a bit of a hyperbola and a bit of y=-x) but the integral 'starts' at a singularity. What do we do about that?
Could perhaps try starting the integral from $$-1+\epsilon$$ but how do we go about doing that in practice? (Since we have to choose a convenient direction for epsilon to tend to -1 from)

thanks

 

[mighty2000]

for the hint, let me know if my approach is on the right track.

Hello,

Thank you for your question. The approach you suggested is definitely on the right track. As you mentioned, the integral starts at a singularity when t = -1, so we need to choose a convenient direction for epsilon to tend to -1 from in order to avoid this singularity.

One approach could be to take the path of Im(it^2) = const to approach -1 from above, so that epsilon is always positive. This would avoid the singularity at t = -1 and allow us to calculate the integral without any issues.

Another approach could be to use a change of variables, such as u = 1 + t, to move the singularity to infinity. This would then allow us to use standard asymptotic expansion techniques for large values of u to calculate the integral.

I hope this helps. Let me know if you have any further questions.