Re: DiracDelta Function question

• To: mathgroup at smc.vnet.net
• Subject: [mg21100] Re: [mg21082] DiracDelta Function question
• 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

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