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Re: Re: Integrating DiracDelta to get UnitStep
*To*: mathgroup at smc.vnet.net
*Subject*: [mg91290] Re: [mg91260] Re: Integrating DiracDelta to get UnitStep
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 13 Aug 2008 04:42:07 -0400 (EDT)
*References*: <g7p2tm$arr$1@smc.vnet.net> <200808120846.EAA18763@smc.vnet.net>
On 12 Aug 2008, at 10:46, Jean-Marc Gulliet wrote:
> Note that HeavisideTheta[x] is not defined for x == 0 and that the
> value
> of the integral is 1/2 for x form -inf to zero. This might explained
> why
> Mathematica does not returned HeavisideTheta[t].
>
>
> In[2]:= Integrate[DiracDelta[x], {x, -\[Infinity], t},
> Assumptions -> Im[t] == 0 && Re[t] < 0]
>
> Out[2]= 0
>
> In[3]:= Integrate[DiracDelta[x], {x, -\[Infinity], t},
> Assumptions -> Im[t] == 0 && Re[t] == 0]
>
> Out[3]= 1/2
>
> In[4]:= Integrate[DiracDelta[x], {x, -\[Infinity], t},
> Assumptions -> Im[t] == 0 && Re[t] > 0]
>
> Out[4]= 1
However, Mathmatica does return HeavisideTheta[t] here:
Integrate[DiracDelta[x], {x, -Infinity, t}, Assumptions -> t <= 0]
HeavisideTheta[t]
and this,appears to include to contradict
Integrate[DiracDelta[x], {x, -Infinity, 0}]
1/2
This may perhaps be "justified" by reference to the fact that
Integrate for functions with parameters returns results which, in
general, are only "generically correct" (in some informal sense of
"generically").
Andrzej Kozlowski
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