Derivative of dirac delta function

[coaxmetal]

Homework Statement

show

x\frac{d}{dx}\delta(x)=-\delta)(x)

using the gaussian delta sequence (\delta_n) and treating \delta(x) and its derivative as in eq. 1.151.

Homework Equations

the gaussian delta sequence given in the book is
\delta_n=\frac{n}{\sqrt{\pi}}e^{-n^2x^2}

and eq 1.151 is just part of the definition of the delta function:
f(0)=\displaystyle\int_{-\infty}^{\infty}f(x)\delta(x)dx

The Attempt at a Solution

thus far, I have tried substitution the derivative of \delta_n(x) for the derivative of the delta function, and then taking the limit as n goes to infinity, but that got me nowhere. I have also tried integrating both sides to see where it got me, but that was nowhere useful. The problem is I just don't understand how the derivative of the delta function works on its own.

 

[Kreizhn]

The delta function is actually a distribution, and is not differentiable in the classical sense. In order to consider such differentiation, we have to revert to generalized derivatives. This is done by assuming a certain level of differentiability on f and some vanishing conditions.

If f is a smooth function with compact support on a set D, the generalized derivative v' of a distribution v is any function w such that
\int_D v' f dx = - \int_D w f' dx
Since the delta function is a distribution, it only truly makes sense to characterize its derivative under integration.

 

[Kreizhn]

You may want to take a look at the following two pdfs:

This link talks about generalized derivatives
http://links.uwaterloo.ca/amath731docs/sobolev.pdf

This link talks about the Dirac delta distribution
http://links.uwaterloo.ca/amath731docs/delta.pdf

 

[Kreizhn]

Okay, I realize that what I gave you may not be entirely helpful since you have to use the Gaussian sequence.

We know that
\displaystyle \lim_{n\to\infty} \delta_n = \delta [/itex]<br /> and that it only makes sense to consider the delta function under the integral. So try evaluating<br /> \lim_{n\to\infty} \int_{-\infty}^\infty x \frac{d}{dx} \delta_n \ dx [/itex]&lt;br /&gt; then show that the value you get is equivalent to&lt;br /&gt; \int_{-\infty}^\infty -\delta(x) \ dx

 

[coaxmetal]

your first answer, and the fact that it only makes sense under integration, actually got me doing just what you suggested, so thanks. THanks for the links too, I was looking for good resources on the delta function. I think I have it now.