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I am performing lots of simple calculations with dirac delta functions. For example, this integral:

Integrate[DiracDelta[x^(2) - y^(2)] x^4, {x, -Infinity, Infinity}]

I get the answer:

(y^4 (Boole[-\[Infinity] < -y < \[Infinity]] + Boole[-\[Infinity] < y < \[Infinity]]))/(2 Abs[y])

Another example:

Integrate[DiracDelta[sin (x)] E^(-x), {x, -1, Infinity}]

Mathematica shows the answer:

ConditionalExpression[1/Abs[sin], sin \[Element] Reals]

There is a way of calculating such integrals by using Mathematica?

Thank you for your help.

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    $\begingroup$ sin (x) is not the same as Sin[x] -- you probably mean the latter. $\endgroup$ – bill s Mar 28 '18 at 3:16
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    $\begingroup$ @user64494: I believe the integral evaluates to Integrate[DiracDelta[Sin[x]] E^(-x), {x, -1, Infinity}]==Sum[Exp[-n Pi],{n,0,Infinity}]==E^\[Pi]/(-1 + E^\[Pi]) . $\endgroup$ – Ulrich Neumann Mar 28 '18 at 7:44
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    $\begingroup$ @user64494: Thank you for your feedback. I read wikipedia link and feel confirmed. Unfortunately I can't use dropbox-link. $\endgroup$ – Ulrich Neumann Mar 28 '18 at 8:55
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    $\begingroup$ @user64494: As I mentioned, I can't use dropbox! Wikipedia gives the composition I used to get my result(similar to the answer bill s)... $\endgroup$ – Ulrich Neumann Mar 28 '18 at 10:12
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    $\begingroup$ @user64494: Remarkable appearance! Why didn't you publish your dropbox-files (which I cannot open!!!) as an answer, instead of feedback like "statement is unbased" , "I still read empty words of you" ? $\endgroup$ – Ulrich Neumann Mar 28 '18 at 13:59
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You have to be explicit about your assumptions, and careful about syntax. For the first integral, you are most likely assuming that y is real, and you need to let Mathematica know that:

Integrate[DiracDelta[x^2 - y^2] x^4, {x,-Infinity,Infinity},
          Assumptions -> y \[Element] Reals]

Abs[y]^3/2

Correcting the syntax on the second integral does not allow evaluation:

Integrate[DiracDelta[Sin[x]] Exp[-x], {x, -1, Infinity}]

By the sifting property of DiracDelta, I guess this is equal to the infinite sum:

Sum[Exp[-x], {x, 0, Infinity, Pi}]

E^\[Pi]/(-1 + E^\[Pi])

which does evaluate nicely.

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  • $\begingroup$ Bei careful. First, one should try to remove Sin[x] by substitution. $\endgroup$ – Henrik Schumacher Mar 28 '18 at 7:56
  • $\begingroup$ Up to Wiki en.wikipedia.org/wiki/Dirac_delta_function , Integrate[DiracDelta[Sin[x]] E^(-x), {x, -1, Infinity}] has no sense so MMA 11.3correctly returns it unevaluated. $\endgroup$ – user64494 Mar 28 '18 at 9:18
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    $\begingroup$ @user64494 there's literally formula for this case in the page you linked. $\endgroup$ – Vsevolod A. Mar 28 '18 at 16:11
  • $\begingroup$ @Vsevolod A.: Are you serious? Read carefully the "Composition with a function" section there. $\endgroup$ – user64494 Mar 28 '18 at 16:24
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    $\begingroup$ @user64494 absolutely serious. $\endgroup$ – Vsevolod A. Mar 29 '18 at 5:17
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We can use the formula DiracDelta[g[x]] == Sum[DiracDelta[x - xi]/Abs[g'[xi] ]] where the xi are the roots of g[x] and the sum is over all the roots.

In this case:

g[x_]=Sin[x]

and the roots of Sin[x] are.

xi=n Pi

So by the formula we have for the integral in question

Sum[Integrate[DiracDelta[x-xi]/Abs[g'[x]/.x->xi]Exp[-x],{x,-1,Infinity}],{n,0,Infinity}]

(*E^Pi/(E^Pi - 1)*)

which matches bill s and Ulrich Neumann results.

It seems like MMA should be able to do this directly.

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