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MathGroup Archive 2005

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Re: Folding Deltas

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56906] Re: [mg56876] Folding Deltas
  • From: Zhengji Li <zhengji.li at gmail.com>
  • Date: Tue, 10 May 2005 03:42:09 -0400 (EDT)
  • References: <200505090545.BAA13779@smc.vnet.net>
  • Reply-to: Zhengji Li <zhengji.li at gmail.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Integrate[DiracDelta[t], {t, -a, a}] = 1, where a > 0.

Maybe Mathematica think Integrate[DiracDelta[t]  DiracDelta[t - 2] ,
{t, -3, 3}] is a little bit complicated. (As far as I know, the result
should be 0)

But, Integrate[Anything, {t, a, b}] + Integrate[Anything, {t, b, a}] =
0, so you will get the result.

On 5/9/05, baermic at yahoo.com <baermic at yahoo.com> wrote:
> Can anyone help to verify in Mathematica the expression given by Rota
> (http://xoomer.virgilio.it/maurocer/Text07.htm):
> 
> Convolution ( Sum of DiracDeltaFct ** Sum of DiracDeltaFct) == Sum
> (DiracDeltaFct + Values).
> 
> I tied
> 
> Integrate[DiracDelta[t]  DiracDelta[t - 2] , {t, -3, 3} ]
> which does not evaluate;
> but
> 
> Integrate[DiracDelta[t]  DiracDelta[t - 2] , {t, -3, 1} ] +
> Integrate[DiracDelta[t]  DiracDelta[t - 2] , {t, 1, 3} ] == 0
> True
> 
> ( I use ver 5.1 with W2k)
>


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