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Dirac delta function - its confusing

  1. Mar 2, 2009 #1
    Hi

    I have been trying to learn dirac delta function. but its kind of confusing. I come across 2 contrasting definitions for it. The first one states that the function delta(x-xo) is infinite at x=x0 while the other states that delta(x-x0) tends to infinite as x tends to x0. Now both of them are different and i'm not sure which one of the two is correct

    Also alternatively, the integral of the delta function from -infinity to +infinity is 1. Now in that case since the value is 0 everywhere except x0, does this means the integral of the delta function between an interval (a b) which contains the x0 is also 1?

    Also the second definition of delta function is integral between -infinity to +infinity F(x')delta(x-x')dx' is F(x). Can i derive this second definition from the above definitions? Because if its consistent, i must be able to derive this from the previous definitions right?

    I would be happy if somebody could throw light on this question of mine
     
  2. jcsd
  3. Mar 2, 2009 #2

    Ben Niehoff

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    The first definition is correct. I think you might have misread the second definition.

    Yes.

    Yes. Try to use the fact that delta(x-x0) = 0 for all x not equal to x0.
     
  4. Mar 2, 2009 #3

    Hurkyl

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    The dirac delta is not a function.

    It doesn't make sense to ask if it has a value at x0.

    (Even if it did, how would you tell [itex]\delta(x - x_0)[/itex] apart from [itex]2 \delta(x - x_0)[/itex]?)

    No, you can't.

    First of all, under many formalizations, the integral would be nonsensical if F is discontinuous at 0, so at the very least you have to make that assumption.

    Secondly, the previously defined properties do not (apparently) distinguish between [itex]\delta(x - x_0)[/itex] and [itex]\delta(x - x_0) + \delta'(x - x_0)[/itex]... however, replacing the former with the latter would change the value of the integral.
     
  5. Mar 2, 2009 #4

    alxm

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    Yeah, but 'Dirac delta distribution' is just way too alliterative.
     
  6. Mar 2, 2009 #5

    Hurkyl

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    The point is that the opening poster shouldn't be treating it exactly like a function. Distributions bear many arithmetic similarities to functions, but you get into all sorts of problems when you try to push the analogy too far. (e.g. by asking things like "what value does it have at x0?")
     
  7. Mar 3, 2009 #6
    I think the replies leave me confused. I understand Dirac delta function is not a function and its a distribution. So that means its just a tool which can simplify my integral and its transformations. But then again can I define a distribution as and how I like? or does it has to confine to any rules?
     
  8. Mar 3, 2009 #7

    arildno

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    Dearly Missed

  9. Mar 4, 2009 #8
    Hi arildno. That was a very nice post. Thank you very much for the link.
     
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