# Derivative of dirac delta function

1. Jan 25, 2010

### coaxmetal

1. The problem statement, all variables and given/known data

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.

2. Relevant 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:
$$f(0)=\displaystyle\int_{-\infty}^{\infty}f(x)\delta(x)dx$$

3. 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.

2. Jan 25, 2010

### 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.

3. Jan 25, 2010

### Kreizhn

4. Jan 25, 2010

### 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
$$\displaystyle \lim_{n\to\infty} \delta_n = \delta [/itex] and that it only makes sense to consider the delta function under the integral. So try evaluating [tex]\lim_{n\to\infty} \int_{-\infty}^\infty x \frac{d}{dx} \delta_n \ dx [/itex] then show that the value you get is equivalent to [tex] \int_{-\infty}^\infty -\delta(x) \ dx$$

5. Jan 25, 2010

### 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.