Strange representation of Heaviside and Delta function

Click For Summary
SUMMARY

The discussion centers on the representation of the Dirac delta function and Heaviside function as presented in the article found at http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6027/1/jfs080104.pdf. The delta function is expressed as \(\delta(x) = -\frac{1}{2i \pi} [z^{-1}]_{z=x}\), while the Heaviside function is represented similarly using \(\log(-z)\). The discussion explores the implications of these representations in the context of contour integration, particularly focusing on the conditions under which the integrals yield valid results, especially regarding the behavior of the function \(f(x)\) in the upper half-plane.

PREREQUISITES
  • Understanding of complex analysis, specifically contour integration.
  • Familiarity with the Dirac delta function and Heaviside function.
  • Knowledge of residue theorem and poles in complex functions.
  • Basic principles of integration in the context of complex variables.
NEXT STEPS
  • Research the properties of the Dirac delta function in complex analysis.
  • Study the residue theorem and its applications in contour integration.
  • Explore the implications of poles in the upper half-plane on integral evaluations.
  • Learn about the conditions required for the convergence of integrals involving complex functions.
USEFUL FOR

Mathematicians, physicists, and engineers interested in advanced calculus, particularly those working with distributions and complex analysis in theoretical contexts.

mhill
Messages
180
Reaction score
1
in the .pdf article http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6027/1/jfs080104.pdf

i have found the strange representation

\delta (x) = -\frac{1}{2i \pi} [z^{-1}]_{z=x}

and a similar formula for Heaviside function replacing 1/z by log(-z) , what is the meaning ? of this formula
 
Last edited by a moderator:
Physics news on Phys.org
Probably the idea is, that we make it a contour integration. Then the only pole is at z = 0 and the result of an integration would be e.g.
\int \delta(x) f(x) \, dx = \int \frac{-1}{2i\pi} \frac{f(z)}{z} \, dz = (2i\pi) \operatorname{Res}_{z = 0} \frac{f(z)}{2 \pi i} = f(0)
if f(x) does not have any poles in the upper half plane.
but I actually doubt how valid this is (even if it works, one would need requirements on f(x) for x \to \pm i\infty to close the countour; and I wonder what happens if f(x) itself has poles).

So I think that is the idea, but I wonder if it works and under what conditions.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K