Contour Deformation and Jordan's Lemma in Complex Analysis

Click For Summary
SUMMARY

This discussion focuses on the application of Jordan's lemma in complex analysis, particularly in the context of contour integration. It establishes that when closing a contour with a large loop, the integral's value is determined solely by the contributions along the real axis, equating to the sum of the residues within the contour. The conversation raises a critical question about the ability to deform the contour without affecting the integral's value, especially when considering Sommerfeld integration paths and the presence of poles. The conclusion emphasizes that as long as the contour encloses all poles, Jordan's lemma asserts that the integral evaluates to zero when traversing from one real infinity to the other.

PREREQUISITES
  • Understanding of contour integration techniques in complex analysis
  • Familiarity with Jordan's lemma and its implications
  • Knowledge of holomorphic functions and residue theory
  • Basic concepts of Sommerfeld integration paths in electromagnetics
NEXT STEPS
  • Study the detailed proofs of Jordan's lemma and its applications in complex integrals
  • Explore the concept of residues and their role in evaluating complex integrals
  • Investigate Sommerfeld integration paths and their significance in electromagnetics
  • Learn about the implications of contour deformation in the presence of poles in complex analysis
USEFUL FOR

Mathematicians, physicists, and engineering students focusing on complex analysis, particularly those interested in contour integration and its applications in theoretical physics.

HasuChObe
Messages
31
Reaction score
0
Lets say you're doing one of those integrals from -\infty to \infty on the real axis and you chose to do it by contour integration. Let's say your integral is one of those integrals that's resolved by using Jordan's lemma. If you close the contour by making a giant loop such that Jordan's lemma holds, then you've established that the entire contribution to the integral is due solely to the integration over the real axis, which is equal to the sum of the residues enclosed.

Since you now know the entirety of the integral comes from integration over the real axis, can you arbitrarily deform the contour that's supposed to close over infinity to something arbitrary and still get zero contribution from the part of the contour that's not the real axis? This would be better with pictures, but hopefully you can understand what I'm asking.

I'm under the assumption that the integral over the real axis does not change its value depending on how you modify the part of the contour that's not on the real axis. This would imply that no matter how you deformed that part of the contour, its contribution remains zero. However, I'm looking at these things called Sommerfeld integration paths (electromagnetics) that would seem to suggest otherwise.
 
Physics news on Phys.org
Assuming you don't pass over any poles, if you deform a path in the complex plane you don't change the integral over that path.

You can check this in total generality - if you have a holomorphic function f in some region, and a path gamma, pick a new path beta whose endpoints are the same as gamma. Since the integral going up gamma and down beta is zero, the integral going up gamma and the integral going up beta must be equal.

However if there are poles in between your old and new contour this obviously isn't true anymore.
 
So, assuming that the contour in the complex plane encloses all the poles, does Jordan's lemma basically say that the integral in the complex plane is always zero (provided that you start at one of the real infinities and end at the other)?
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K