Solving an indefinite integral

Click For Summary
SUMMARY

The discussion focuses on evaluating the contour integral INTEGRAL(-INF TO +INF) [S exp(iwt)/(t2 + s2)2dt, which is related to the probability of excitation of an atom from the ground state to an excited state. Participants suggest using the calculus of residues to solve the integral by closing the integration path in the complex plane. The integral is identified as improper rather than indefinite, highlighting the need for precise terminology in mathematical discussions.

PREREQUISITES
  • Understanding of contour integrals
  • Familiarity with the calculus of residues
  • Knowledge of Laplace transforms, specifically LAPLACE TRANSFORM OF (COS at)
  • Basic concepts of Fourier transforms
NEXT STEPS
  • Study the application of the calculus of residues in complex analysis
  • Learn about improper integrals and their evaluation techniques
  • Explore the relationship between Laplace transforms and Fourier transforms
  • Investigate advanced topics in quantum mechanics related to atomic excitation probabilities
USEFUL FOR

Students and professionals in mathematics, physics, or engineering who are dealing with complex integrals, particularly in quantum mechanics and signal processing.

GAGS
Messages
53
Reaction score
0

Homework Statement



1. The calculation of the probability of excitation of an atom originally in the
ground state to an excited state, involves the contour integral
INTEGRAL(-INF TO +INF) [S exp(iwt)/(t2 + s2)2dt
Evaluate the above integral.

Homework Equations





The Attempt at a Solution


Attempt: what i know that LAPLACE TRANSFORM OF (COS at) = s/(s2+ a2)
Moreover this integral is looking inverse FOURIER TRANSFORM, But how to solve it. I am not getting it.
 
Physics news on Phys.org
Well, you mentioned the term "contour integral" and then wrote an integral over the real axis. So I guess the point is that you make it into a contour integral (close the integration by an infinitely large semi-arc in the complex plane) and use calculus of residues to evaluate it. Have you seen / done this before?

<nitpick>By the way, it is an improper integral, but not an indefinite one.</nitpick>
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
12
Views
3K