Solving Integral of e^wix for Particle Location

  • Context: Graduate 
  • Thread starter Thread starter skateboarding
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The integral of the function \( \int_0^t e^{i\omega t'} dt' \) is evaluated as \( \frac{e^{i\omega t}}{i\omega} - \frac{1}{i\omega} \), which simplifies to \( \frac{1}{\omega} \sin(\omega t) - \frac{1}{\omega} i \cos(\omega t) + \frac{i}{\omega} \). It is established that this integral approaches zero when the frequency \( \omega \) is large, as the oscillations between 0 and \( t \) cancel out. Conversely, if \( t \) is close to \( \omega \), the integral yields a non-zero value. The discussion emphasizes the dependence of the integral's value on the relationship between \( t \) and \( \omega \).

PREREQUISITES
  • Understanding of complex exponentials and Euler's formula
  • Familiarity with integral calculus
  • Knowledge of oscillatory functions and their properties
  • Basic concepts of perturbation theory in physics
NEXT STEPS
  • Study the properties of oscillatory integrals in complex analysis
  • Learn about perturbation theory applications in quantum mechanics
  • Explore the implications of frequency domain analysis in signal processing
  • Investigate the relationship between oscillation frequency and integral convergence
USEFUL FOR

Physicists, mathematicians, and engineers involved in particle location analysis, perturbation theory, and those interested in the behavior of oscillatory integrals.

skateboarding
Messages
11
Reaction score
0
For particle location, perturbation theory, etc, I see the following integral.

\LARGE \int_0^t { e^{i\omega t^'}}dt^'

Where \omega is some constant, or frequency. It says in my text that this is equal to 0 if \omega is not close to 0. My logic leads me to think that when \omega is large, the function oscillates many times between 0 and t, so it's integral is 0. However, when I carry out the integral explicitly, it is less clear.

\LARGE {\int_0^t {e^{i\omega t^'}}dt^'} = \frac{e^{i\omega t}}{i\omega} = \cos{\omega t} + i\sin{\omega t} - 1

My question is, from the expression above, how to I show that this integral is 0 or close to zero? I think it depends on the values chosen for \omega and t. If t is greater than \omega, the integral should give a small number, and if t is close to \omega, the integral should give some non zero value. Any help would be appreciated.
 
Physics news on Phys.org
skateboarding said:
For particle location, perturbation theory, etc, I see the following integral.

\LARGE \int_0^t { e^{i\omega t^'}}dt^'

Where \omega is some constant, or frequency. It says in my text that this is equal to 0 if \omega is not close to 0. My logic leads me to think that when \omega is large, the function oscillates many times between 0 and t, so it's integral is 0. However, when I carry out the integral explicitly, it is less clear.

\LARGE {\int_0^t {e^{i\omega t^'}}dt^'} = \frac{e^{i\omega t}}{i\omega} = \cos{\omega t} + i\sin{\omega t} - 1
How did you get that last part?
\LARGE {\int_0^t {e^{i\omega t^&#039;}}dt^&#039;} = \frac{e^{i\omega t}}{i\omega}- \frac{1}{i\omega{ = \frac{1}{i\omega}(cos(\omega t)+ i sin(\omega t)+ 1)= \frac{1}{\omega}sin(\omega t)- \frac{1}{\omega}i cos(\omega t)+ \frac{i}{\omega}[/itex]<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> My question is, from the expression above, how to I show that this integral is 0 or close to zero? I think it depends on the values chosen for \omega and t. If t is greater than \omega, the integral should give a small number, and if t is close to \omega, the integral should give some non zero value. Any help would be appreciated. </div> </div> </blockquote>
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
2K