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Integrating over a discontinuity

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    I would like to prove that [tex]{d \over {dx}} \sigma(x)=2\delta(x)[/tex],
    where [tex]\sigma(x>0)=1[/tex]
    [tex]\sigma(x=0)=0[/tex]
    [tex]\sigma(x<0)=-1[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I think that I need to show that
    [tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=2=\int_{-\infty}^{\infty}2 \delta(x)dx[/tex].
    The integral looks rather harmless, and I would like to write
    [tex]\int_{-\infty}^{\infty}{d \over {dx}} \sigma(x)dx=\sigma(x) |^{\infty}_{-\infty}=1-(-1)=2[/tex].
    This looks like it works, but it seems to me that there should be some complication when integrating close to zero where we have discontiuous behaviour.
     
  2. jcsd
  3. Mar 2, 2008 #2
    Consider this: is there a difference between integrating (some integrable function) over the closed interval [0,1] versus integrating it over the half-open interval (0, 1] ? If you're not sure, look up the concept of "measure zero".
     
    Last edited: Mar 2, 2008
  4. Mar 2, 2008 #3
    I think the answer to your question slider is no. But how do I integrate "over" the discontinuity with a clear conscience.
     
  5. Mar 2, 2008 #4

    Dick

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    What you are integrating is d(sigma(x))/dx. That's worse than being discontinuous, it doesn't even exist at x=0 in the usual sense. That is a complication. But the whole point to inventing delta functions is to make sense of problems like this. There is nothing wrong with what you are doing.
     
  6. Mar 2, 2008 #5
    What is bothering me is that it looks like I am applying the fundamental theorem of calculus. The the FTC is only applicable if the integrand is continuous over the interval of integration. Here, it clearly isn't so how can my calculation be valid (even if it is giving the right answer)??.
     
  7. Mar 2, 2008 #6

    Dick

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    When you extend the family of functions you are dealing with by including distributions like delta functions, the FTC remains valid provided the antiderivative is well defined at the endpoints. You do have to formally justify this stuff, but I don't think you are being asked to do this.
     
  8. Mar 2, 2008 #7
    Ok. I'm satisfied with that. I guess I would have to study the theory of generalized functions to see such justifications??
     
  9. Mar 2, 2008 #8

    Dick

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    Yep. Exactly. You are right, the usual FTC doesn't apply. The generalized one does.
     
  10. Mar 2, 2008 #9
    Great. Thank you very much.
     
  11. Mar 2, 2008 #10

    Hurkyl

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    It turns out that what you really need to show is that, for any test function (i.e. infinitely differentiable, and 'rapidly' goes to zero at infinity) [itex]\varphi[/itex]:

    [tex]
    \int_{-\infty}^{+\infty} \left( \frac{d}{dx} \sigma(x) \right) \cdot \varphi(x) \, dx
    =
    \int_{-\infty}^{+\infty} \left( 2 \delta(x) \right) \cdot \varphi(x) \, dx
    [/tex]

    but you need to know more; the definition of the derivative of a generalized function is that it satisfies:

    [tex]
    \int_{-\infty}^{+\infty} \left( \frac{d}{dx} \sigma(x) \right) \cdot \varphi(x) \, dx
    :=
    -\int_{-\infty}^{+\infty} \sigma(x) \cdot \left( \frac{d}{dx} \varphi(x) \right) \, dx
    [/tex]
     
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