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Re: Integral of Piecewise function involving DiracDelta
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74800] Re: Integral of Piecewise function involving DiracDelta
*From*: dh <dh at metrohm.ch>
*Date*: Thu, 5 Apr 2007 04:17:28 -0400 (EDT)
*References*: <euvmuo$epv$1@smc.vnet.net>
Hi Andrew,
Obviously there is a bug in the implementation of Pieceweise or its
integral. Also note that:
Integrate[Piecewise[{{DiracDelta[x],-1<x<1}},0],{x,-.1,.1}] evaluates to 1.
Further, concerning your second question. It is well known that
DiracDelta[a x] == DiracDelta[x] / a for a constant a>0. A handwaving
argument is, that the dirac function becomes "narrower" by a factor of
a, what makes the integral smaller by the same factor.
Daniel
Andrew Moylan wrote:
> Here is an integral that I expect Mathematica to evaluate to 1:
>
> Integrate[Piecewise[{{DiracDelta[x], -1 < x < 1}}, 0],
> {x, -Infinity, Infinity}]
>
> However, Mathematica 5.2 (Windows) gives the answer as 0. Here's a
> similar integral that I also expect to evaluate to 1:
>
> Integrate[Piecewise[{{DiracDelta[x-1/2], -1 < x < 1}}, 0],
> {x, -Infinity, Infinity}]
>
> For this integral, Mathematica doesn't return 0. It returns the
> following:
>
> Integrate[Piecewise[{{2*DiracDelta[-1 + 2*x], -1 < x < 1}},
> 0], {x, -Infinity, Infinity}]
>
> Can anyone help me understand what's happening here?
>
> Cheers,
>
> Andrew
>
>
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