Asymptotic Expansion of I

This figure shows the original path of integration (C1) from -∞ to ∞ on the real axis. If we try to split the ∫ into 2 parts ( and 1) we will get ∞-∞ which is useless.

The first idea is to deform the contour of the interation from C1 (Above) to C3. In our case we will take Since we don't have any poles enclosed we have:

Now Let's look at the integral on C2 and C4:

Which will ->0 as R->∞.

This now can be safely split into 2 parts the second of which is:

Ie. this integral has a finite value and infact is equal to ZERO (According to Mathematica; even if it is not zero it is a finite value and can easily be approximated).

Now we are left with the other part of the integral which is:

and

Now the Saddle Points are:

Lets Take the substitution z=τ

Now take

And now if we solve for the stationary points we get

These points are on the unit circule and are:

Let's evaluate the function at these 3 points

This means that there is no contribution from the bottom point (-I) and the contribution comes from . The next plot shows the path of the original integration as well as the path that we will have to take to go through in order to go though the Saddel point.

We have to ensure that the integration on CM1 and CM2 Vanishes as we go to infinity so that we can deform our contour from the parbola to a strait line that goes to ∞. We will look at CM1 and the argument will be identical for CM2. Take the integration will become:

Now the only thing left is to expand the orignial integration around the saddle points. We will do the saddle point and the other one will be it's complex conjogate. To do this expansion we will have to make the substitution τ->R

If we plot the abs value of integrand vs R with η=1we get

Which shows that we indeed have a maximum @ R=1

A Similar plot will show that the phase of the integrand stops CHANGING at this point also.  
So, Now let us just expand f[R ] about the point R=1. We will just take the firs 3 terms.

Ofcource we don't see anything of the form since this is a stationarry point. Also if we take m=0.944941-1.63669 I then we can write the integral as:

Now Remember we have:

Which can be shown by a simple change of variables gassian one. Note that the limits of the integration were changed from 0 to ∞ to -∞ to +∞ since we are just interested in the behaiviour of the function at R=1 and it is exponentially decaying away from that point. That can be trasfered to a Γ function though an other strait forward transformation. We will eventually get the same result which is:

And in terms of λ we get

And we finally multiply by 2 to take the effect of the other significant stationary point in the second quadrant.

This would be the first term of the expantion. If we wanted to find the higher order terms we would proceed to expand the exponent to get powers of ,. We then would expand and multiply the terms of the form to obtain a polynomial multiplied by again similar substitutions would trasfer it to a sum of Γ Integrations that can be calculated to arbitrary powers of η and consequently λ.

END OF PROBLEM