Proving x*delta' ~ -delta with Generalized Functions

[aphrodasic]

hi
can anyone tell me how could one go about to prove
x* (delta)' ~ -delta

where delta is the dirac delta funtion of x.
~ approximately equal

delta' = first derivative of delta

i know this can be done by using the concept of GENERALISED FUNTIONS.
WHICH INVOLVES MUTLIPLYING THE LHS OF THE EQUATION WITH A GOOD FUNCTION and taking limits to infinity to get the RHS..
I tried but there is something wrong.
can anyone help me out?

 

[Hurkyl]

What is your definition of the derivative of a generalized function?

In the one I'm most familiar with, the definition of the derivative of a generalized function $\varphi$, its that it's the generalized function satisfying
$$\int_{-\infty}^{+\infty} \varphi'(x) f(x) \, dx = -\int_{-\infty}^{+\infty} \varphi(x) f'(x) \, dx$$
for all test functions f.

 

[aphrodasic]

yes, that is the correct defination.

 

[Hurkyl]

So, your identity
$$x \delta'(x) = -\delta(x)$$
is true if and only if
$$\int_{-\infty}^{+\infty} x \delta'(x) f(x) \, dx = \int_{-\infty}^{+\infty} (-\delta(x)) f(x) \, dx = -f(0)$$
for all test functions f, right? So what did you get when you used the definition of derivative?

 

[aphrodasic]

i expanded the LHS of your equation using PArts.
where u = x(delta') and v = f
and i got stuck.
i am still not able to trace the steps which you did to land on to the expression which you posted in the previous step.

 

[matt grime]

Isn't one of the important points in doing integration by parts that you choose something to be du/dx, and the other part to be v? Setting u= somthing and v=something seems lke you're not sure how to integrate by parts.

There is clearly only one choice to make for du/dx...

 

[Hurkyl]

i am still not able to trace the steps which you did to land on to the expression which you posted in the previous step.

I was simply restating what you needed to prove, not actually providing a proof.



I'm not sure what you're trying to do with IBP; if you were setting $u = x \delta'(x)$ and $v = f(x)$, that would allow you to do something with
$$\int u \, dv = \int f'(x) \delta'(x) x \, dx$$
but it wouldn't help at all when the integrand is $\delta'(x) x f(x) \, dx$.



Did you try applying the definition of the derivative of a generalized function?
(This is equivalent to formally doing integration by parts with a certain choice, but I find it easier to think about by treating it as a rule in of itself)

 

[aphrodasic]

yes i used the defination of the GF,
taking u = delta'*f'; v = x
and solving the integral gives me zero. something is going wrong..

 

[aphrodasic]

i was wondering can i use the following defination for delta function
delta = n/(pi)^.5 *exp (-(n*x)^2)
having done this i can get my answer.
but i do not know if this is correct?

 

[Hurkyl]

Huh? There are no u's and v's in the definition I quoted...

I agree with matt's assessment, though; you seem to have forgotten integration by parts. If you want to use integration by parts here, I really think it would be worthwhile to reopen your calculus textbook and do some integration by parts exercises before proceeding with this problem.

 

[aphrodasic]

i am not sure what is going wrong here..
integral u(x)v(x) dx = u(x)integral v(x) - integral( (derivative u(x)) (integral v(x)) )
where u(x) = something and v(x) = somethign else.

but anyways, i solved my way through the problem, if only i can replace delta by the defination of gaussian functions delta= n/(pi)^.5 *exp (-(n*x)^2) where n--> infinity.
i get my result

 

[Hurkyl]

i am not sure what is going wrong here..
integral u(x)v(x) dx = u(x)integral v(x) - integral( (derivative u(x)) (integral v(x)) )
where u(x) = something and v(x) = somethign else.

You use a different notation for IBP than I've ever seen. I'm used to:
$$\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$$
So your v is my v'.


Anyways, it's clear why your attempts at integration by parts isn't working: you have no idea what delta''(x) is, so it doesn't help to differentiate delta'(x).