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Dirac Delta Function (x+a)

  1. Aug 25, 2016 #1
    1. The problem statement, all variables and given/known data
    hi
    i have to find the result of
    ##\int_{0}^{\pi}[\delta(cos\theta-1)+ \delta(cos\theta+1)]sin\theta d\theta##

    2. Relevant equations
    i know from Dirac Delta Function that
    ##\int \delta(x-a)dx=1##
    if the region includes x=a and zero otherwise
    3. The attempt at a solution
    first i make the substitution
    let ##\cos\theta=x##
    then ##\sin\theta d\theta=-dx##
    at ##\theta=0 \rightarrow x=1##
    at ##\theta=\pi \rightarrow x=-1##

    ##\int_{1}^{-1}[\delta(x-1)+ \delta(x+1)](-dx)##

    ##\int_{-1}^{1}[\delta(x-1)+ \delta(x+1)](dx)##

    ##\int_{-1}^{1}\delta(x-1)dx+ \int_{-1}^{1}\delta(x+1)(dx)##

    so the first term is equal one but what about the second term. I know that the result must be 2
     
  2. jcsd
  3. Aug 25, 2016 #2

    DrClaude

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    Staff: Mentor

    In ##\delta(x-a) = \delta(x+1)##, what is the value of ##a##?
     
  4. Aug 25, 2016 #3
    In L.H.S
    X must be equal a. Else dirac delta will be zero.
    R.H.S
    X=-1
    Or a=-1
    Maybe I didn't understand your question.
     
  5. Aug 25, 2016 #4

    DrClaude

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    Staff: Mentor

    Correct. So what is the value of the corresponding integral?
     
  6. Aug 25, 2016 #5

    George Jones

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    Why??? I think that the first term is undefined. Something strange is happening at ##x = 1##, so write it as the limit,
    $$\int_{-1}^{1} \delta \left( x -1 \right) dx = \lim_{ a \to 1} \int_{-1}^{a} \delta \left( x -1 \right) dx.$$
    This two-sided limit, however, does not exist, because
    $$\lim_{ a \to 1^-} \int_{-1}^{a} \delta \left( x -1 \right) dx = 0,$$
    and
    $$\lim_{ a \to 1^+} \int_{-1}^{a} \delta \left( x -1 \right) dx = 1.$$
    In the first one-sided limit, ##x = 1## is not in the region of integration, while in the second one-sided limit, ##x = 1## is in the region of integration. Since the two one-sided limits are not equal, the two-sided limit does not exist.

    I suppose that you mean the second one-sided limit, i.e., that
    $$\int_{-1}^{1} \delta \left( x -1 \right) dx = \lim_{ a \to 1^+} \int_{-1}^{a} \delta \left( x -1 \right) dx.$$
     
  7. Aug 25, 2016 #6
    Since a=-1 is in the range of integral. The integral is equal one?
     
  8. Aug 25, 2016 #7

    Ray Vickson

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    How do you know the result must be 2? You are integrating through only half of a ##\delta## function. Some people would say that the result is not defined at all, others would say the final result should be 1 (not the 2 that you give). We could also argue that for ##a > -1## we have ##\int_{-1}^a \delta(x+1) \, dx = 0.15865## or ##= 0.84134## (for both of which there exist perfectly good arguments/justifications). However, I would personally take ##\int_{-1}^a \delta(x+1) \, dx = 1/2##, for more-or-less good reasons.
     
    Last edited: Aug 25, 2016
  9. Aug 25, 2016 #8
    The reason why i think the result is 2 because of a problem in Jackson Electrodynamic.
    I have now uploaded a pdf file and at page 1 you can see that the integral of dirac delta should be 2.
     

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