If I have, for instance,


it evaluates to

(2 DiracDelta[-1 + Sqrt[13] - 2 x])/Sqrt[13] + (2 DiracDelta[1 + Sqrt[13] + 2 x])/Sqrt[13]

so that I am able to get the delta of the argument, even if not in the form

$\qquad \delta(x-x_0),\ x_0: f(x_0)=0.$

Now, if I have a much more complicated function inside the DiracDelta:

2 DiracDelta[-l m Δ + t Sqrt[4 k^2 vF^2 + Δ^2] - l Sqrt[4 vF^2 x^2 + Δ^2] + t Δ χ]

the same trick of expanding using FunctionExpand doesn't work, even if I specify the argument of the expansion: FunctionExpand[expr, x], where for expr I have my complicated delta function. Is there any way to get the simplified form?

  • $\begingroup$ First, it looks like a multivariate function (esp. to Mathematica). Second, whether or not the argument has real zeros may depend on the values of the parameters; FunctionExpand might balk in that case, even if one figures out a way to distinguish the variable from the parameters. $\endgroup$ – Michael E2 Oct 29 '15 at 11:25
  • $\begingroup$ For example, consider FunctionExpand[DiracDelta[x^2 - a^2], Assumptions -> a > 0] $\endgroup$ – Michael E2 Oct 29 '15 at 16:13

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