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

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The discussion centers on the integral of the delta function, specifically whether the expression ∫_{-∞}^{0} δ(t) dt = 1/2 is correct. Participants highlight the ambiguity at the singular point x = 0, where different definitions of the Heaviside step function lead to varying interpretations of the integral. The conversation emphasizes that while some sequences of functions converging to δ(t) yield a result of 1/2, this is not universally applicable due to the nature of the delta function as a distribution rather than a conventional function. The need for rigorous definitions and the implications of integration limits are also underscored, particularly in the context of engineering versus mathematical perspectives. Ultimately, the integral's value remains contentious and context-dependent.
  • #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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