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

Hans de Vries

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Typically you want to link the delta function and its derivatives to the differential operators,


[tex]
\begin{aligned}
&~~\delta(t)&\ast~~& f(t) &= ~~&~~f(t) \\ \\
&~\frac{\partial \delta(t)}{\partial t}&\ast~~ &f(t) &= ~~&~\frac{\partial f(t)}{\partial t} \\ \\
&\frac{\partial^2 \delta(t)}{\partial t^2}&\ast~~& f(t) &= ~~&\frac{\partial^2 f(t)}{\partial t^2} \\ \\
\end{aligned}
[/tex]


for consistent operations, where [itex]\ast[/itex] denotes convolution. This is a symmetric definition of [itex]\delta(x)[/itex],
and corresponds with 1/2 for the half-space integral.

Note that higher order derivatives of the delta function occur in physical propagators
as soon as you work in more then three dimensions. See for instance:
http://physics-quest.org/Higher_dimensional_EM_radiation.pdf



Regards, Hans
 
Last edited:

reilly

Science Advisor
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The point is, starting from a theory of distributions on R, you can't derive a theory of distributions on half-infinite intervals. While you could attempt pin down the sources of ambiguity and attempt to make choices to resolve them, the effort would be so great that it would surely not be worth the effort -- it would be better to either start with a theory that is more appropriate (e.g. to begin with some sort of theory of local distributions, or jump ship and use nonstandard analysis), or simply make ad-hoc definitions along the way to meet your needs.
Sorry that it's taken so long for me to reply.

I might be missing something in your discussion of half-infinite distributions. I say that because in the Zemanium book -- mentioned above -- there's a great deal of material on distributions defined on a half interval, some of which involves Laplace transforms.

The standard delta function of physics -- like the integral over all space of a plane wave. -- is an even function, as pointed out by Hans. The series you proposed gives a function that agrees with the physics delta function for x>0, and is zero for x<0. It is not the physics delta function.

Just a note that the Cauchy Integral Thrm provides another approach, one used very often in EE and in the dispersion relations of QM.
Regards,
Reilly
 
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Both integrals, because of the evenness or parity invariance of [tex]\delta(x)[/itex], are equal to 1/2 --[tex]\delta(x)[/itex] = [tex]\delta(-x)[/itex].
Anyway, the fact that [tex]\delta(x)\ =\ \delta(-x)[/tex] does not imply that the delta function was constructed using a sequence of even functions [tex]\delta_n(x)[/tex]; it can be proved without using that hypothesis.:

[tex]\int_{-\infty}^{+\infty}\delta(-x)f(x)dx\ =\ (-x\ =\ u)

=\ -\int_{+\infty}^{-\infty}\delta(u)f(-u)du\ =\ \int_{-\infty}^{+\infty}\delta(u)f(-u)du\ =\ f(0)[/tex]

So it would seem that we could also use non symmetric [tex]\delta_n(x)[/tex].
 
Anyway, the fact that [tex]\delta(x)\ =\ \delta(-x)[/tex] does not imply that the delta function was constructed using a sequence of even functions...


So it would seem that we could also use non symmetric [tex]\delta_n(x)[/tex].
Spot on!

That was precisely my point when I tried to demonstrate that you can use an asymmetric (neither even nor odd) aperiodic rectangular pulse to get to the delta 'function', in which case you can't use the even symmetry property of the sequence.
 

Hurkyl

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I might be missing something in your discussion of half-infinite distributions. I say that because in the Zemanium book -- mentioned above -- there's a great deal of material on distributions defined on a half interval, some of which involves Laplace transforms.
I'm not trying to say you cannot have distributions on the half-line. Indeed, the essential details of the construction of distributions work for any infinite-dimensional vector space. The point is that, unlike functions, distributions do not have a restriction map: if you're given a distribution on the real line, there is no 'good' way to turn it into a function on a half-line.

In particular, using just the property that
[tex]\int_{-\infty}^{+\infty} \delta(x) f(x) \, dx = f(0)[/tex]
for any test function f, you cannot derive the equation
[tex]\int_{-\infty}^{0} \delta(x) f(x) \, dx = \frac{1}{2}f(0)[/tex]
If you want to (rigorously) discuss such notions, you have to acknowledge that they are not a consequence of facts about distributions on the entire real line.


Incidentally, the fact that distributions are dual to test functions means that distributions do have a canonical extension map lets you turn a distribution on the half-line into a distribution on the entire real line. However, test functions do not have this feature.



The standard delta function of physics -- like the integral over all space of a plane wave. -- is an even function, as pointed out by Hans. The series you proposed gives a function that agrees with the physics delta function for x>0, and is zero for x<0. It is not the physics delta function.

Just a note that the Cauchy Integral Thrm provides another approach, one used very often in EE and in the dispersion relations of QM.
Regards,
Reilly[/QUOTE]
 
i think i have a missing point in the delta function

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

so how can we show that
[tex]\int\delta_{}n (x) d(x) = 1[/tex]
 

CompuChip

Science Advisor
Homework Helper
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Assuming you tried to say
[tex]\delta_n(x) = \frac{n}{\pi} \frac{1}{1 + n^2 x^2}[/tex]
you can easily work out that
[tex]\int_{-a}^{a} \delta_n(x) \, \mathrm{d}x = \frac{2}{\pi} \operatorname{arctan}(a n)[/tex].
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 [itex]a \to \infty[/itex] this converges to 1. So if we define
[tex]\int_{-\infty}^\infty \delta_n(x) \, \mathrm dx = \lim_{a \to \infty} \int_{-a}^{a} \delta_n(x) \, \mathrm{d}x = \frac{2}{\pi}[/tex]
it follows.

One has to be careful in applying limits on integrals though, in particular
[tex]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[/tex]
 

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