Completing a Fourier Transform Integral

Click For Summary
Contour integration is a key technique for evaluating Fourier transform integrals, particularly when the integrand is complex analytic. To apply this method, one must select a contour that includes the desired integral as part of its path. The integral around this contour is influenced by the poles of the function located within it. Typically, the integral of interest is a segment of the contour integral, and careful contour selection can simplify the evaluation by ensuring that other segments either vanish or relate directly to the desired integral. Mastery of this technique often requires a deeper understanding of complex analysis, which is best gained through dedicated study rather than brief explanations.
ajw124
Messages
1
Reaction score
0
I was wondering if anyone could help me with this integral. I've heard of contour integration but I'm unsure of how it would be used for this integral.
fourierInt.jpg
 
Physics news on Phys.org
Well, I suggest you pick up a book on complex analysis. It's not something which lends itself to being taught via a single forum post. This one is nice enough and affordable.

The general idea is you choose a contour on which the integrand is complex analytic, which includes the desired integral as a segment of the contour. One then computes the integral around the contour. The value of the integral around the contour is related to the poles of the function that are within the contour. Often the actual integral which you want to evaluate is not the contour integral itself, but rather a segment integral of that contour. The contour is often chosen such that the integral along the other segments besides the ones desired either go to zero in some limit, or are multiples of the desired integral.
 
Last edited:
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

Replies
12
Views
11K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K