How Can I Solve This Contour Integral with a Pole at Zero?

Click For Summary
SUMMARY

The discussion focuses on solving the contour integral involving the functions J(ω) and N(ω), specifically addressing the challenge of a pole at zero in N(ω). The integral in question is defined as $$\int_0^\infty J(\omega)N(\omega)$$, where J(ω) is given by $$\frac{1}{2\pi}\frac{\gamma_i\lambda^2}{(\lambda^2+(\omega_i-\Delta-\omega)^2)}$$ and N(ω) approximates to $$\frac{-T}{\omega t}$$ for small ω. The user seeks assistance in managing the pole at zero, which complicates the convergence of the integral.

PREREQUISITES
  • Complex analysis, particularly residue theory
  • Understanding of contour integration techniques
  • Familiarity with poles and singularities in functions
  • Knowledge of asymptotic expansions for small parameters
NEXT STEPS
  • Study residue theorem applications in contour integration
  • Research techniques for handling poles in integrals
  • Explore asymptotic analysis of functions near singularities
  • Learn about the convergence criteria for improper integrals
USEFUL FOR

Mathematicians, physicists, and engineers dealing with complex integrals, particularly those interested in contour integration and the behavior of functions with singularities.

Ana2015
Messages
1
Reaction score
0
New member warned about not using the homework template
I want to solve this contour integral
$$J(\omega)= \frac{1}{2\pi}\frac{\gamma_i\lambda^2}{(\lambda^2+(\omega_i-\Delta-\omega)^2)} $$
$$N(\omega)=\frac{1}{e^{\frac{-\omega t}{T}}-1}$$

$$\int_0^\infty J(\omega)N(\omega)$$
there are three poles I don't know how I get rid of pole on zero (pole in N(w))
would you please help me?
Thanks
 
Physics news on Phys.org
Doesn't look like a convergent integral to me.

For small ω,
$$N(\omega)=\frac{1}{e^{\frac{-\omega t}{T}}-1} \approx \frac{-T}{\omega t}$$

While J(ω) quickly approaches some constant.
 

Similar threads

Replies
1
Views
3K
Replies
2
Views
2K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K