Re: DiracDelta Function question
- To: mathgroup at smc.vnet.net
- Subject: [mg21106] Re: [mg21082] DiracDelta Function question
- From: Hartmut Wolf <hwolf at debis.com>
- Date: Fri, 17 Dec 1999 01:21:06 -0500 (EST)
- Organization: debis Systemhaus
- References: <199912130451.XAA16337@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Julian Francis schrieb:
>
> Folks,
>
> Why is mathematica unable to evaluate:
> Integrate[DiracDelta[2 x - 1], {x, -Infinity, +Infinity}]
>
> Also interestingly:
> Integrate[DiracDelta[2 x - 2], {x, -Infinity, +Infinity}]
>
> returns (1/2)
>
> I should have thought that the answer in both cases would be 1.
>
> I am using Mathematica v4.0 on MS Windows 95.
>
> Thanks for any light shed on this.
>
> Julian Francis.
> acz43 at dial.pipex.com
Hello Julian,
In[158]:= Integrate[DiracDelta[2( x - a)] f[a], {x, -Infinity,
+Infinity}]
Out[158]= f[a]/2
In[159]:= a
Out[159]= a
which is correct (for any real constant a, and test function f).
If you integrate DiracDelta[2 x - 2] it will simplify to 1/2
DiracDelta[x-1] (as an intermediate), but not so with 2 x - 1, 2 x -3,
yet again with 2 x - 4, 2 x - 6 etc.
Besides Mathematica won't evaluate the Integral DiracDelta[2. x]. But again
In[173]:=
Integrate[DiracDelta[b x - a] f[x], {x, -Infinity, +Infinity}]
Out[173]=
f[a/b]/Abs[b]
So use of DiracDelta is somewhat inconvenient, a rule to explore would
be: use only symbolic (algebraic) expressions as argument to DiracDelta
and substitute for numerical values later. Before Version 4.0 DiracDelta
had been defined in a standard package, if you have access to it, look
into it (to see how complicated things are).
Kind regards, Hartmut
- References:
- DiracDelta Function question
- From: "Julian Francis" <acz43@dial.pipex.com>
- DiracDelta Function question