Calculating Residues and using Cauchy Integral Formula

Click For Summary
SUMMARY

This discussion focuses on evaluating the integral \(\int_{-\infty}^{\infty}dE\frac{1}{E^{2}-\mathbf{p}^{2}-m^{2}+i\epsilon}\) using residue calculus and the Cauchy Integral Formula. The poles identified are at \(-(l-\frac{i\epsilon}{2l})\) and \((l-\frac{i\epsilon}{2l})\), with corresponding residues of \(-\frac{1}{2(l-\frac{i\epsilon}{2l})}\) and \(\frac{1}{2(l-\frac{i\epsilon}{2l})}\). The discussion suggests that simplifying the integrand using partial fraction decomposition is essential and that evaluating the integral can be done by considering contributions from residues in the complex plane.

PREREQUISITES
  • Complex analysis fundamentals, specifically residue theory.
  • Understanding of the Cauchy Integral Formula.
  • Familiarity with integral calculus and contour integration techniques.
  • Knowledge of partial fraction decomposition methods.
NEXT STEPS
  • Study the application of the Cauchy Integral Formula in complex analysis.
  • Learn about residue calculus and its role in evaluating complex integrals.
  • Explore partial fraction decomposition techniques in detail.
  • Investigate the evaluation of integrals over semicircular contours in the complex plane.
USEFUL FOR

This discussion is beneficial for physicists, mathematicians, and students engaged in advanced calculus, particularly those working with quantum field theory and complex analysis.

Sekonda
Messages
201
Reaction score
0
Hey,

I have a problem with this integral:

\int_{-\infty}^{\infty}dE\frac{1}{E^{2}-\mathbf{p}^{2}-m^{2}+i\epsilon}\: ,\: l^{2}=\mathbf{p}^{2}+m^{2}

The integration over all energies (arising in the loop function for calculating the scattering), I understand we write the above in this form:

\int_{-\infty}^{\infty}dE\frac{1}{(E+(l-\frac{i\epsilon}{2l}))(E-(l-\frac{i\epsilon}{2l}))}

Where ε is small and so the factor arising from it multiplying by itself can be neglected. It seems to evaluate this we can either calculate the residues of the two poles and sum them up and multiply by 2pi*i or we can use Cauchy Integral's formula - though I think it's the same thing... not really sure.

Our poles are at

-(l-\frac{i\epsilon}{2l})\: ,(l-\frac{i\epsilon}{2l})

and we find the residues to be

-\frac{1}{2(l-\frac{i\epsilon}{2l})},\frac{1}{2(l-\frac{i\epsilon}{2l})}

But I'm not sure how we see this or do this exactly...

Any help is appreciated,
Thanks.
 
Last edited:
Physics news on Phys.org
you can take a large semicircle in either half planes.then go for it's evaluation.Only one residue will give contribution.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
0
Views
2K