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

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SUMMARY

The integral of the Dirac delta function from -infinity to 0 is defined as ∫_{-∞}^{0}δ(t) dt = 1/2 based on the Heaviside step function Θ(x), which is defined as Θ(x) = ∫_{-∞}^{x}δ(t) dt. This definition introduces ambiguity at the singularity point x = 0, as different interpretations of the limit can yield different results. Specifically, the integral can be interpreted as either 0 or 1 depending on whether the limit approaches from the left (0-) or right (0+). The discussion highlights the need for a rigorous definition of the delta function and its properties in mathematical contexts.

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  • #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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