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Why i has that result

  1. Jul 7, 2009 #1
    why
    [tex]
    \int_{0^+}^{t} (x+1)\delta (x)dx=0
    [/tex]

    delta is defined that in 0 it is infinity
    and on x differs 0 its value is 0

    there heard of some formula
    i am not sure if its the right form
    [tex]
    \int f(x)\delta (x)dx=f(0)
    [/tex]

    why i get 0
    ??
     
  2. jcsd
  3. Jul 7, 2009 #2
    I can make out what you're asking, I think, but in future please express yourself in clearer English.

    The delta function is simply defined in the way you described, yes. It has the important property that [tex]\int_{-\infty}^{\infty} \delta (x)dx = 1[/tex]
    , or rather, as long as the point x=0 is within the limits of integration, the result is 1, otherwise it is 0.

    Thus looking at your second integral, as the integration is carried out from -infinity, the delta function is 0, except when the integral crosses the point x=0, where delta(x) = 1 and f(x) = f(0), thus the integral "picks out" the value of f(x) at x=0
     
  4. Jul 7, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, the delta function is NOT defined by "in 0 it is infinity
    and on x differs 0 its value is 0" since "infinity" is not an allowable value for functions and, in fact, the delta "function" is not really a function- it is a "distribution" or "functional" and is defined by the property you give: If a< 0< b, then
    [tex]\int_a^b f(x)\delta(x) dx= f(0)[/itex]
    or, more generally, if a is in the set A then, for any function integrable on A,
    [tex]\int_A f(x)\delta(x-a)dx= f(a)[/itex]
    while, if a is not in the set,
    [tex]\int_A f(x)\delta(x-a)dx= 0[/itex]
     
  5. Jul 7, 2009 #4
    I suppose that [tex]0^+[/tex] means do the integral from [tex]a[/tex] to [tex]+\infty[/tex] with [tex]a>0[/tex], then take the limit as [tex]a[/tex] goes to [tex]0[/tex] from the right. Then your result is [tex]0[/tex], but the same thing with [tex]0^-[/tex] is [tex]1[/tex] .

    Anyone have another guess what it means?
     
  6. Jul 7, 2009 #5
    The limits of integration do not contain the x value where the delta function explodes. If this x value (exactly zero in this case) is not contained in the limits then the delta function is always zero, so the integral is always zero.
     
  7. Jul 7, 2009 #6
    i cant link your formulas with my interval

    how to explainm interval in your terms?
     
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