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Delta dirac function times zero

  1. Oct 4, 2011 #1
    Let δ(x)=∞ at x = 0, and zero elsewhere. Then

    δ(x)(1-exp(x)) = ?

    It seems the above expression is zero. But isn't it zero times infinity at x = 0?
     
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 4, 2011 #2

    CompuChip

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    As you have defined it, δ(x) is not really a function.
    In physics, we usually call it that, but if you want to treat it rigorously you should consider it as a distribution. The most important part about that, is that δ(x) only really makes sense under an integral sign, in expressions like
    [tex]\int \delta(x) f(x) \, dx = f(0)[/tex]
    (if x = 0 is inside the integration interval).
     
  4. Oct 5, 2011 #3
    OK, then let us consider the following example in a probability book:

    Let

    [itex]f(x)=(1-e^{-\lambda x})u(x)[/itex],

    where

    [itex]u(x)= 1, [/itex] if [itex] x \geq 1 [/itex], and zero elsewhere.

    Then, it is found (in the book) that

    [itex] \frac{d}{dx} f(x) = \lambda e^{-\lambda x} u(x)[/itex]

    But I think that the derivative should be:

    [itex] \frac{d}{dx} f(x) = \lambda e^{-\lambda x} u(x) + (1-e^{-\lambda x})\delta (x)[/itex]

    So, should the second term in the above expression be zero? Then how?
     
    Last edited: Oct 5, 2011
  5. Oct 5, 2011 #4

    CompuChip

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    It's easiest to split up the function first.
    For [itex]x \ge 1[/itex], [itex]f(x) = (1 - e^{-\lambda x})[/itex] and the derivative is what you got in the book.
    For [itex]x < 1[/itex], [itex]f(x) = 0[/itex] so [itex]f'(x) = 0[/itex].

    Note that in general f(x) is not differentiable - or even continuous - at x = 1. If you want to use the product rule, you have to consider f(x) as a distribution, which is a whole area of mathematics in itself. (Oh, and you'd probably get [itex]\delta(x - 1)[/itex] rather than just [itex]\delta(x)[/itex]... but that's beside the point).
     
  6. Oct 5, 2011 #5
    Thank you very much for your answer!

    Let me make a correction in the book example:

    u(x)=1, if x≥0 (not x≥1, as in my previous post), and zero elsewhere.

    In this case f(x) becomes continuous and differentiable. As far as I understand, even in this case I should consider f(x) as a distribution in order to use the product rule. Am I right?
     
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