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Re: Apparent error integrating product of DiracDelta's

  • To: mathgroup at smc.vnet.net
  • Subject: [mg91997] Re: Apparent error integrating product of DiracDelta's
  • From: magma <maderri2 at gmail.com>
  • Date: Tue, 16 Sep 2008 19:24:46 -0400 (EDT)
  • References: <gag2lg$39k$1@smc.vnet.net> <gal3ht$dv1$1@smc.vnet.net>

On Sep 15, 9:40 am, "Nasser Abbasi" <n... at 12000.org> wrote:
> "Michael Mandelberg" <mmandelb... at comcast.net> wrote in message
>
> news:gag2lg$39k$1 at smc.vnet.net...
>
> > How do I get:
>
> > Integrate[DiracDelta[z- x]   DiracDelta[z- y], {z-Infinity, Infinity}=
]
>
> > to give DiracDelta[x-y] as the result?  Currently it gives 0.  I ha=
ve
> > all three variable assumed to be Reals.  I am using 6.0.0.
>
> > Thanks,
> > Michael Mandelberg
>
> I think you have synatx error in the limit part. I assume you mean to wri=
te
> {z, -Infinity,Infinity}
>
> Given that, I think zero is the correct answer.  When you multiply 2 de=
ltas
> at different positions, you get zero. Integral of zero is zero.
>
> Nasser

No Nasser, the correct value of the integral should be DiracDelta[x-
y], as Michael said.
This value is indeed 0 if x != y but it is not 0 if x==y.

Mathematica correctly calculates:

Integrate[f[z - x] DiracDelta[z - y], {z, -Infinity, Infinity},
 Assumptions -> y \[Element] Reals]

as

f[-x + y]

However it fails to recognize that if f[z-x] is replaced by
DiracDelta[z-x], the result should be

DiracDelta[-x + y]

or the equivalent

DiracDelta[x - y]

In the help file, under "possible issues" it is mentioned that
"Products of distributions with coinciding singular support cannot be
defined:"

So perhaps at the moment the only way to do the integral is:

Integrate[f[z - x] DiracDelta[z - y], {z, -Infinity, Infinity},
  Assumptions -> y \[Element] Reals] /. f -> DiracDelta

hth


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