Integrals over Dirac delta function

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}]


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


Another example:

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


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


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

• sin (x) is not the same as Sin[x] -- you probably mean the latter. Mar 28 '18 at 3:16
• @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]) . Mar 28 '18 at 7:44
• @user64494: Thank you for your feedback. I read wikipedia link and feel confirmed. Unfortunately I can't use dropbox-link. Mar 28 '18 at 8:55
• @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)... Mar 28 '18 at 10:12
• @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" ? Mar 28 '18 at 13:59

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.

• Bei careful. First, one should try to remove Sin[x] by substitution. Mar 28 '18 at 7:56
• 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. Mar 28 '18 at 9:18
• @user64494 there's literally formula for this case in the page you linked. Mar 28 '18 at 16:11
• @Vsevolod A.: Are you serious? Read carefully the "Composition with a function" section there. Mar 28 '18 at 16:24
• @user64494 absolutely serious. Mar 29 '18 at 5:17

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.