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Dirac Delta fn

  1. Apr 14, 2007 #1
    1. The problem statement, all variables and given/known data
    int[d(x-a)f(x)dx]=f(a) is the dirac delta fn

    but is int[d(a-x)f(x)]=f(a) as well? If so why?


    3. The attempt at a solution
    Is it because at x=a, d(0)=infinite and integrate dirac delta over a region including x=0 when d(0) is in the value in the integral will produce 1 hence f(a).
     
  2. jcsd
  3. Apr 14, 2007 #2

    Gib Z

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    Think about it this way.

    Since
    [tex]\int f(x) d(a-x) = -\int f(x) d(x-a) = -f(a)[/tex] your asking is f(a)=-f(a).
     
  4. Apr 14, 2007 #3

    HallsofIvy

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    No, [itex]\int_{-\infty}^{\infty} \delta (a- x)f(x)dx= -f(a)[/itex]. Use the substitution u= a-x.


    That's a very rough way of putting it. More correctly the dirac delta function (actually a "generalized function" or "distribution") is DEFINED by the property that [itex]\int \delta(x) f(x)dx = f(0)[/itex] as long as the integration include x= 0.
     
    Last edited: Sep 20, 2009
  5. Apr 14, 2007 #4
    So it can be but in some cases it is not.
     
  6. Apr 14, 2007 #5

    Gib Z

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    Yup, choose your a.
     
  7. Sep 20, 2009 #6
    is it possible to explain step by step including the integration of the following:

    (1) f(x) d(x-a) = f(a) d(x-a)

    (2) \int f(a) d(a-x) = f(x)

    Thank you,..!!!
     
  8. Sep 20, 2009 #7

    HallsofIvy

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    Are you changing symbols here? does "d(x-a)" still mean the delta function? Are you asking why
    (1) [tex]\int f(x)\delta(x- a)dx= \int f(a)\delta(x-a)dx[/tex]?

    That is because the definition of [itex]\delta(x-a)[/itex] require that the
    [tex]\int f(x)\delta(x-a)dx= f(a)[/tex]
    for any function f as long as the integral of integration includes x= a.
    Of course,
    [tex]\int f(a)\delta(x-a) dx= f(a)\int \delta(x-a)dx= f(a)(1)= f(a)[/tex]

    As for
    (2)[tex]\int f(a)d(a- x) dx= f(x)[/tex]
    That's not true. In fact,the right hand side cannot be a function of x at all. The integral is just as in (1).
     
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