Integral of a delta function from -infinity to 0 or 0 to +infinity

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Discussion Overview

The discussion centers on the interpretation of the integral of the Dirac delta function, particularly the expression \(\int_{-\infty}^{0}\delta(t) dt\) and its relationship to the Heaviside step function \(\Theta(x)\). Participants explore the implications of defining \(\Theta(0) = 1/2\) and the resulting ambiguities when evaluating the integral at the singular point \(x = 0\). The conversation touches on theoretical aspects, definitions, and practical applications in quantum mechanics and signal processing.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether \(\int_{-\infty}^{0}\delta(t) dt = \frac{1}{2}\) is correct, citing the ambiguity at the singularity point \(x = 0\).
  • There is a discussion about the definition of the Heaviside step function and its value at zero, with some sources suggesting it could take any value between 0 and 1.
  • Participants propose that the interpretation of the upper limit of the integral as \(0-\) or \(0+\) leads to different results: 0 for \(0-\) and 1 for \(0+\).
  • Some participants suggest that the result of the integral may depend on the sequence of functions used to approximate the delta function, indicating that different sequences could yield different integrals.
  • One participant emphasizes the distinction between the mathematical treatment of the delta function as a distribution versus its practical use in engineering contexts.
  • Another participant notes the utility of symmetric definitions of the Dirac delta function for simplifying calculations in signal processing.

Areas of Agreement / Disagreement

Participants express differing views on the value of the integral \(\int_{-\infty}^{0}\delta(t) dt\) and the implications of defining \(\Theta(0)\). There is no consensus on the correct interpretation, and multiple competing views remain regarding the treatment of the delta function and its integration.

Contextual Notes

The discussion highlights limitations in the definitions and assumptions surrounding the delta function and the Heaviside step function, particularly regarding the treatment of singular points and the implications of different sequences of approximating functions.

  • #31
i think i have a missing point in the delta function

if
\delta_{}n (x) = (n/\Pi) . (1/(1+n^{}2 x^{}2 ))

so how can we show that
\int\delta_{}n (x) d(x) = 1
 
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  • #32
Assuming you tried to say
\delta_n(x) = \frac{n}{\pi} \frac{1}{1 + n^2 x^2}
you can easily work out that
\int_{-a}^{a} \delta_n(x) \, \mathrm{d}x = \frac{2}{\pi} \operatorname{arctan}(a n).
This is no coincidence of course, it is the reason the ugly factor of 1/pi was added in the first place.
Of course, for a \to \infty this converges to 1. So if we define
\int_{-\infty}^\infty \delta_n(x) \, \mathrm dx = \lim_{a \to \infty} \int_{-a}^{a} \delta_n(x) \, \mathrm{d}x = \frac{2}{\pi}
it follows.

One has to be careful in applying limits on integrals though, in particular
0 = \int_{-\infty}^{\infty} \lim_{n \to \infty} \delta_n(x) \, \mathrm dx \neq \lim_{n \to \infty}\left( \int_{-\infty}^\infty \delta_n(x) \, \mathrm dx \right) = 1
 

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