Re: DiracDelta Function question
- To: mathgroup at smc.vnet.net
- Subject: [mg21100] Re: [mg21082] DiracDelta Function question
- From: Wolfgang Schadow <schadow at netcom.ca>
- Date: Fri, 17 Dec 1999 01:21:01 -0500 (EST)
- References: <199912130451.XAA16337@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
> Folks, > > Why is mathematica unable to evaluate: > Integrate[DiracDelta[2 x - 1], {x, -Infinity, +Infinity}] > > Also interestingly: > Integrate[DiracDelta[2 x - 2], {x, -Infinity, +Infinity}] > > returns (1/2) > Mathematica gives the correct answer in both cases. You need to use the following relation: delta(g(x)) = sum_i 1/g'(x_i) * delta(x-x_i) Here x_i are the zeros of g(x), and g' is the derivative of g. Using this formula your function can be rewritten as: delta(2 x -1) = 1/2 delta(x - 1/2) delta(2 x -2) = 1/2 delta(x - 1) If you integrate over these functions the answer is 1/2. Wolfgang ======================================================================== Wolfgang Schadow Phone: +1-604-222-1047 ext. 6453 (office) TRIUMF +1-604-875-6066 (home) Theory Group FAX: +1-604-222-1074 4004 Wesbrook Mall Vancouver, B.C. V6T 2A3 email: schadow at triumf.ca Canada www : http://www.triumf.ca/people/schadow ========================================================================
- References:
- DiracDelta Function question
- From: "Julian Francis" <acz43@dial.pipex.com>
- DiracDelta Function question