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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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