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Re: DiracDelta Function question

  • To: mathgroup at
  • Subject: [mg21106] Re: [mg21082] DiracDelta Function question
  • From: Hartmut Wolf <hwolf at>
  • Date: Fri, 17 Dec 1999 01:21:06 -0500 (EST)
  • Organization: debis Systemhaus
  • References: <>
  • Sender: owner-wri-mathgroup at

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

Hello Julian,

In[158]:= Integrate[DiracDelta[2( x - a)] f[a], {x, -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 

Integrate[DiracDelta[b x - a] f[x], {x, -Infinity, +Infinity}]

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

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