Proving the Distributional Sense of x*Pv(1/x) = 1: An Analysis

In summary, x*Pv(1/x) can be defined as a distribution by applying it to the function x*phi(x) and taking the limit as epsilon approaches 0. This allows for the integration to be simplified to just phi(x), resulting in the unity distribution. This shows that the distribution x*Pv(1/x) is equal to 1 in a distributional sense.
  • #1
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Homework Statement



Show that, in a distributional sense,

x* Pv(1/x) = 1




Homework 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) )

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!
 
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  • #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.
 
  • #3
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].
 
  • #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.

Jezuz,

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

1. What is the principle value of 1/x?

The principle value of 1/x is the value of the expression when x approaches 0 from the positive side. It is also known as the limit of the expression as x approaches 0.

2. Why is the principle value of 1/x important?

The principle value of 1/x is important because it helps us understand the behavior of the expression as x gets closer to 0. It is often used in calculus and other mathematical calculations.

3. Can the principle value of 1/x be negative?

No, the principle value of 1/x cannot be negative. As x approaches 0 from the positive side, the expression approaches positive infinity. However, as x approaches 0 from the negative side, the expression approaches negative infinity.

4. How is the principle value of 1/x calculated?

The principle value of 1/x is calculated by taking the limit of the expression as x approaches 0. This can be done by plugging in smaller and smaller values of x and observing the trend of the expression.

5. What is the difference between principle value and actual value of 1/x?

The principle value of 1/x is the limit of the expression as x approaches 0, while the actual value of 1/x is the value of the expression at a specific point. The actual value may not always be defined, as x=0 would result in division by zero. The principle value, on the other hand, exists and helps us understand the behavior of the expression as x gets closer to 0.

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