Integral of function involving Dirac delta

Question

I am performing lots of simple calculations with dirac delta functions. It would be awesome if Mathematica could do this routine exercise for me, eliminating any possible human errors. For example, this integral:

Integrate[DiracDelta[t - a - (r^2 + (z - a)^2)^(1/2)] /
  (4 Pi (r^2 + (z - a)^2)^(1/2)), {a, -Infinity, Infinity}]

Mathematica does not know how to evaluate it.

I tried using FunctionalExpand to expand the delta-function, but in this case it does nothing. But I can easily take the integral above manually.

Maybe there is a way of evaluating such integrals automatically?

Answer 1

Adding assumptions finds two of the cases.

Assuming[{a, t, r, z} ∈ Reals && t > z,
 Integrate[
  DiracDelta[
    t - a - (r^2 + (z - a)^2)^(1/2)]/(4 Pi (r^2 + (z - a)^2)^(1/2)),
  {a, -Infinity, Infinity}]
 ]
(*  1/(4 π t - 4 π z)  *)

Assuming[{a, t, r, z} ∈ Reals && t < z,
 Integrate[
  DiracDelta[
    t - a - (r^2 + (z - a)^2)^(1/2)]/(4 Pi (r^2 + (z - a)^2)^(1/2)),
  {a, -Infinity, Infinity}]
 ]
(*  (-t + z)/(4 π r^2)  *)

The case t == z seems to reveal a bug:

Assuming[{a, r, z} ∈ Reals,
 Integrate[
  DiracDelta[
     t - a - (r^2 + (z - a)^2)^(1/2)]/(4 Pi (r^2 + (z - a)^2)^(1/2)) /. t -> z,
  {a, -Infinity, Infinity}]
 ]

Select::normal: Nonatomic expression expected at position 1 in Select[Integrate`ImproperDump`newx,Integrate`NLtheoremDump`sel$293548]. >>

Answer 2

Try this:

    1/(4 Pi (r^2 + (z - a)^2)^(1/2)) /. (Solve[
     t - a - (r^2 + (z - a)^2)^(1/2) == 0, a][[1, 1]]) // Simplify

(*  1/(2 \[Pi] Sqrt[(r^2 + (t - z)^2)^2/(t - z)^2])  *)

Have fun!

Answer 3

Although Mathematica cannot solve

Integrate[DiracDelta[g[x]] h[x], x]

as written, it can be transformed to

Integrate[DiracDelta[x-x0] h[x] / (Abs[D[g, x]]/.x -> x0), x]

using standard formulas, where g[x0] == 0. (Note that, if g[x] has multiple real zeroes, the preceding expression must be summed over those zeroes.)

This can be applied to the specific case here as follows, with q the original integral:

arg = Cases[q, DiracDelta[z_] -> z, Infinity];
a0 = a /. First@Solve[arg == 0, a];
q /. DiracDelta[z_] :> DiracDelta[a - a0]/(Abs[D[arg, a] /. a -> a0])

(* {Boole[-Infinity < -((r^2 - t^2 + z^2)/(2*t - 2*z)) < Infinity]/
     (2*Pi*Sqrt[(r^2 + (t - z)^2)^2/(t - z)^2]*
     Abs[1 + (-r^2 + (t - z)^2)/(Sqrt[(r^2 + (t - z)^2)^2/(t - z)^2]*(t - z))])} *)

If the constants contained in the argument of Boole are real, it evaluates to 1, and the expression greatly simplifies.

Assuming[{a, t, r, z} \[Element] Reals, Simplify[%]]
(* Piecewise[{{(4*Pi*t - 4*Pi*z)^(-1), t >= z}}, (-t + z)/(4*Pi*r^2)] *)

(This is, of course, the approach used by Alexei Boulbitch in his earlier answer.)