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A point about dirac delta

  1. Mar 6, 2013 #1

    ShayanJ

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    In texts about dirac delta,you often can find sentences like "The delta function is sometimes thought of as an infinitely high, infinitely thin spike at the origin".
    If we take into account the important property of dirac delta:
    [itex] \int_\mathbb{R} \delta(x) dx=1 [/itex]
    and the fact that it is zero everywhere except at the origin,it seems that we can have an explanation for the quoted sentence.It can be said that the area under the curve of dirac delta should be one but it's width is zero so its height should be infinite.
    Everything seems to work out well until it is said that,rigorously,dirac delta isn't a function because no function which is zero everywhere except at one point,can have non-zero definite integral .

    But the explanation I gave for the dirac delta being infinite at origin,can be applied to a function like f(x) defined as [itex] f(0)=\infty \ \mbox{and} \ f(x \neq 0)=0 [/itex]

    The definite integral of f(x) can be thought of as the area of a rectangle having zero width
    and infinite height so having a finite and possibly non-zero area.

    So it seems that's not a good reason for telling that dirac delta isn't a function.
    I'll appreciate any idea.
     
  2. jcsd
  3. Mar 6, 2013 #2

    Fredrik

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    Yes, you can define a function that way. This function would be "extended real-valued" instead of "real-valued", but that's not a problem. The problem is that the standard definition of the integral assigns the value 0 to this function, not 1.
     
  4. Mar 6, 2013 #3

    ShayanJ

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    What you mean by the standard definition of the integral?
    Is it the limit of the Riemann sum?
    You mean by doing that sum and taking the limit,the integral of f(x) comes out to be zero?
     
  5. Mar 6, 2013 #4

    Fredrik

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    I'm referring to the Lebesgue definition of the integral. It assigns the same value to any two functions that differ only on a set of Lebesgue measure 0. Your f differs from the constant function 0 only on the set {0}, which has Lebesgue measure 0. So the value of f at 0 is actually irrelevant to the value of the integral.
     
  6. Mar 6, 2013 #5

    mathman

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