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Integrate[DiracDelta[s (x - c)] x, {x, -Infinity, Infinity},
          Assumptions -> {s > 0, c > 0}]

gives me c/s in version 10.3.1, but 1+c+1/s in version 10.4.1. Have I missed anything here?

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5
  • $\begingroup$ I confirm the result for 10.4.1 $\endgroup$ Apr 27, 2016 at 7:22
  • $\begingroup$ do the assumptions matter? You should only need s!=0 $\endgroup$
    – george2079
    Apr 27, 2016 at 11:48
  • $\begingroup$ @george2079: The result would be slightly different (c/Abs[s]) for s < 0. But it should still work. (In fact, both integrals, with s > 0 and s < 0, work fine in 10.2.0.) $\endgroup$ Apr 27, 2016 at 13:15
  • $\begingroup$ Works fine in version 10.0 $\endgroup$
    – mattiav27
    Apr 27, 2016 at 14:31
  • 1
    $\begingroup$ New MMA soft, does not mean better.... $\endgroup$ Apr 27, 2016 at 17:02

1 Answer 1

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Confirmed for 10.4.1 as Alexei. However, WolframAlpha still gives the c/s result.

See result from Wolfram|Alpha

And using the properties of Dirac delta function, I would say that c/s is the correct one. May be a bug of the new 10.4.1?

You can test with numeric constants, and Mathematica gives the correct answer. But symbolic result is wrong,

Integrate[DiracDelta[(x - 3)*2] x, {x, -Infinity, Infinity}]

gives 3/2.

Another test: it seems it only happens with this expression. This one,

Integrate[DiracDelta[s ( x - c/s)] x, {x, -Infinity, Infinity}, 
 Assumptions -> {s > 0, c > 0}]

Gives the correct answer: c/s^2.

EDIT.- The new Mathematica 11 corrects this issue.

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