- #1

- 31

- 0

## Main Question or Discussion Point

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.

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.