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I Lebesgue Integral of Dirac Delta "function"

  1. Nov 18, 2017 #26

    WWGD

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    Yes, it does exist ( while you need to correct the codomain of your function) and it equals 0 . I think the Lebesgue sum will show you this, as , in Lebesgue theory we define ## a \cdot \inft=0 ## for finite ##a##. The convergence criterion for Lebesgue integrals show it exists, and it equals ##0##. Only one of your partitions will contain a non-zero value. The problem is this contradicts the fact that the integral is supposed to equal ##1##.
     
  2. Nov 19, 2017 #27

    Svein

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    Here goes: Start by defining [itex] F(x)=\int_{-\infty}^{x}\delta (t)dt[/itex]. Then F(x)=0 for x<0 and F(x)=1 for x≥0. It is therefore tempting to say that [itex]\delta (x)=dF(x) [/itex], except for the fact that neither side of the expression has any meaning at x=0. But for any interval [a, b] which does not contain {0}, the integral [itex] \int_{a}^{b}f(x)dF(x)=0[/itex] for any integrable function f(x). But going back to the definition of F(x), the integral [itex] \int_{-\varepsilon}^{\varepsilon}dF(x)[/itex] should be finite and equal to 1 for any ε>0.

    The Stieltjes solution was to define dF(x) as a measure and it extends the concept of a measure as we know it from Lebesgue integration theory. In this case the measure of the point {0} is 1 and the measure of any interval that does not contain {0} is 0. Observe that the Stieltjes measure concept is much wider than the Dirac delta "function", as it contains all the standard interval measures. In addition, more than one singular point can have a measure different from 0.
     
  3. Nov 19, 2017 #28
    Thanks for the link, I'll certainly will watch it. My initial question has been motivated from the fact a physics teacher stated that the Dirac Delta distribution is the only correct way of doing the integral, from - oo to +oo, of:

    f(x) = 0, if x is different fro 0
    f(x) = +oo if x =0

    (and up to here I certainly agree with him)

    /because the lebesgue integral of f(x) is zero/.

    I replied to him that f(0) = +oo is meaningless (even in the codomain - oo +oo) and it's this the /first/ reason why the lebesgue integral of that f(x) cannot give the right answer.
    If I "approximate" f(x) with the functions

    f_n(x) = k*n for -1/n < x < 1/n
    f_n(x) = 0 else

    I get k as integral.
    If I approximate f(x) with

    g_n(x) = n^2 for -1/n < x < 1/n
    g_n(x) = 0 else
    I get +oo.

    If I use:
    h_n(x) = sqrt(n) for -1/n < x < 1/n
    h_n(x) = 0 else
    I get 0.

    So, to me, it's impossibile to do the lebesgue integral of f(x) without precisly specifing what "f(0) = +oo" means. Or, the Lebesgue integral simply cannot be computed.

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  4. Nov 19, 2017 #29

    FactChecker

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    Oh, I agree with you there. I was careless when I said that the value on a set of measure zero didn't matter. It is not proper to include points where the function is not properly defined.

    Regarding the video series, it is a long series (over 30), but I enjoyed it so much that I practically binge-watched it. I may watch it again.
     
  5. Nov 19, 2017 #30

    WWGD

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    Ok, maybe I did sweep under the rug the whole issue of ##f## taking values ##\pm \infty ##. Will try to define more carefully.
     
  6. Nov 19, 2017 #31

    WWGD

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    Still, this is better and more rigorous than what I did , but you still do not get an actual value of 0 unless you extend continuously. You can also redefine , e.g., ##f(x)=1/x## so that ##f(0)= +\infty##
     
  7. Nov 19, 2017 #32
    Ok, I can also take ##f_n(x) = (n/\sqrt(\pi)) exp[-(nx)^2]##

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  8. Nov 20, 2017 #33

    WWGD

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    Sorry, don't see it here, ##f_n(0)=\frac {n}{\sqrt{\pi}} ##. My point is that , you can use sequential continuity in order to define ##f_n(0)##, and do not define it explicitly; just a (relatively minor) point.
     
  9. Nov 21, 2017 #34
    Thanks to every one for the answers.

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