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

**Follow-Ups**:**Re: Re: Apparent error integrating product of DiracDelta's***From:*Daniel Lichtblau <danl@wolfram.com>