Re: Integrating DircaDelta[x]
- To: mathgroup at smc.vnet.net
- Subject: [mg78543] Re: Integrating DircaDelta[x]
- From: dh <dh at metrohm.ch>
- Date: Wed, 4 Jul 2007 05:27:33 -0400 (EDT)
- References: <f6d9oo$2hm$1@smc.vnet.net>
Hi,
forgot to mention, that this is one of the new "features" of version 6.
Daniel
dh wrote:
> Hi Phillys,
>
> I think we simply have a bug here. Wolfram should take note. Consider:
>
> Integrate[DiracDelta[x],{x,-Infinity,a},Assumptions->{Element[a,Reals]}]
>
> gives 1, however if we add a<0 to the assumptions:
>
> Integrate[DiracDelta[x],{x,-Infinity,a},Assumptions->{Element[a,Reals],a<0}]
>
> we get 0.
>
> Daniel
>
>
>
> Pillsy wrote:
>
>> In Mathematica 6, integrating the DiracDelta function with specified
>
>> limits gives the expected result:
>
>
>> In[1]:= Integrate[DiracDelta[x], {x, -Infinity, -1}]
>
>
>> Out[1]:= 0
>
>
>> In[2]:= Integrate[DiracDelta[x], {x, -Infinity, 1}]
>
>
>> Out[2]:= 1
>
>
>> In[3]:= Integrate[DiracDelta[x], {x, -Infinity, 0}]
>
>
>> Out[3]:= 1/2
>
>
>> But when you replace the limit with a variable, it returns something
>
>> quite different:
>
>
>> In[4]:= Integrate[DiracDelta[x], {x, -Infinity, a}]
>
>
>> Out[4]:= If[a \[Element] Reals, 1,
>
>> Integrate[DiracDelta[x], {x, -\[Infinity], a},
>
>> Assumptions -> Im[a] < 0 || Im[a] > 0]]
>
>
>> In[5]:= Map[% /. a -> # &, {-1, 0, 1}]
>
>
>> Out[5]:= {1, 1, 1}
>
>
>> Any idea what's going on? I'm using the 32-bit x86 Mac version on OS X
>
>> 10.4.10, if it matters.
>
>
>> TIA,
>
>> Pillsy
>
>
>
>
>