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Re: Re: Folding Deltas
Zhengji Li wrote:
> 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)
>>
Actually the product of delta functions is not defined. This is because
it is not possible to make it well defined. For example I could give
limiting approximations to the integrand that make that integral give
any result, not just zero. There are also other ways to show it is not
well defined.
The original note was a bit unclear but I the correct form of the
identity in question uses convolution of deltas, not product. Also there
is a minor typo at the web page for Rota's talk. The idenity boils down
simply to
delta(a_i) * delta(b_j) = delta(a_i+b_j)
where '*' denotes convolution, not ordinary product. This follows from
basic rules of convolution, translation, and the fact that delta(0) is
the identity element for convolution.
Daniel Lichtblau
Wolfram Research
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