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Multidimensional DiracDelta: reference point AFTER integration??

Posted 10 years ago
Hi dear community,

I am beginning to work with the multidimensional DiracDelta. I dont understand, why it is necessary to replace a reference point AFTER the integration in order to obtain the correct result (see code below)
Integrate[DiracDelta[x - a, y - b], {x, 0, 1}, {y, 0, 1}] /. {a -> 1, b -> 1/2}
Integrate[DiracDelta[x - 1, y - 1/2], {x, 0, 1}, {y, 0, 1}]

1

0
Or may be you can just tell what trivial error I am making since I really dont see it. Thank you very much for your time!
POSTED BY: Mauricio Lobos
3 Replies
Posted 10 years ago
This is a somewhat interesting point. replacement after integration results in 1 because the result of the integration is a UnitStep, which has a defined value at the point where the DiracDelta has an undefined value, as a generalized function when discussed outside the integral. Replacement before integration results in 0,apparently because the integration is formally over the open interval, which well might agree with integration theory.

It is interesting to note that Mathematica uses the UnitStep as the antiderivative for DiracDelta, but uses the DiracDelta as the derivative of the HeavisideTheta. If the integration of the DiracDelta resulted in the HeavisideTheta, then in the case where replacement occured after integration, the integral would be undefined at the endpoint of the interation limit. I suspect this interesting inconsistency is needed so that in general the integral of functions containing the DiracDelta are defined everywhere, which they should be. This is probably related to questions like "How many angels can dance on the head of a pin?"

 In[11]:= Integrate[
   DiracDelta[x - a, y - b], {x, 0, 1}, {y, 0, 1}] /. {a -> 1, b -> 1/2}
 
 Out[11]= 1
 
 In[12]:= out =
  Integrate[DiracDelta[x - a, y - b], {x, 0, 1}, {y, 0, 1}]
 
 Out[12]= ConditionalExpression[
UnitStep[1 - a] UnitStep[a] UnitStep[1 - b, b], a \[Element] Reals]

In[13]:= out /. {a -> 1, b -> 1/2}

Out[13]= 1

In[14]:= Integrate[DiracDelta[x - 1, y - 1/2], {x, 0, 1}, {y, 0, 1}]

Out[14]= 0
POSTED BY: David Keith
Hi,

I might be wrong, but looks like integration is performed to 1, to get the correct answer it should be done to 1+.
Limit[Integrate[ DiracDelta[x - 1, y - 1/2], {x, 0, 1 + \[Epsilon]}, {y, 0,1}], \[Epsilon] -> 0]
For DiracDelta you can always take integration limits to be infinite, that will give the right answer too.
Integrate[ DiracDelta[x - 1, y - 1/2], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
I.M.
POSTED BY: Ivan Morozov
Hi Ivan,

thank you very much. That was completely bananas! Thank you again for your time!
POSTED BY: Mauricio Lobos
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