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Principle Value of 1/x

  1. Jan 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that, in a distributional sense,

    x* Pv(1/x) = 1

    2. Relevant equations

    The function 1/x cannot be integrated locally in the origin. Nevertheless,

    int(1/x, x=-1..1) =0, and thus convergent.

    Therefore, one defines the (non-regular) distribution Pv(1/x), as follows:

    <Pv(1/x), phi(x)> = Pv*int(phi(x)/x, x=-infinity..infinity)

    which is defined to be equal to:

    limit( eps->0) ( int( phi(x)/x,x=-infinity..-eps) + int( phi(x)/x, x=eps..infinity) )

    3. The attempt at a solution

    My problem is that I don't really understand how I can write x*Pv(1/x) as a distribution. My attempt at a solution is as follows::

    <x*Pv(1/x), phi(x)> = limit(eps->0) (int( phi(x)/x*x,x=-infinity..-eps) = int( phi(x)/x*x, x=eps..infinity)

    If this step is allowed, then of course, this integration is equal to :

    int(phi(x),x=-infinity..infinity) = 1 , because this is a distribution. But surely, it can't be as simple as that!
  2. jcsd
  3. Jan 10, 2007 #2
    What you need to show is that the distribution
    [tex] x \mathcal{P} \frac{1}{x} [/tex]
    is the unity distribution. That is, you can multiply this distribution with another distribution without changing it. Alternatively you can show that the effect of using this distribution on a test function [tex] \psi [/tex] is the same as using the distribuition 1 on a test function.

    Since [tex] \left< 1, \psi \right> = \int_{-\infty}^{\infty} dx \psi(x) [/tex]
    and this is exactly what you have shown.

    I think your last sentence is a little off though. Your "phi" is not a distribution, but a test function.

    Remember that distributions are defined by the effect they have on test functions. So they take test functions (from some pre-defined function space) and produces a number.
  4. Jan 10, 2007 #3


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    Homework Helper

    There is not a general way of multiplying distributions. The only reason you can do it here is because x is not only a distribution, but is also a function. So what you need to do is apply P(1/x) to the function [itex]x\phi(x)[/itex].
  5. Jan 10, 2007 #4
    Hey Status,

    of course! That's great, I didn't realize you were allowed to treat simple functions as distributions/test functions interchangably. That is, I'd seen the property, but not what it was useful for.


    indeed, when aforementioned trick is allowed, it comes down to something as simple as that.
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