- #1

- 62

- 0

## Homework Statement

Prove this theorem regarding a property of the Dirac Delta Function:

$$\int_{-\infty}^{\infty}f(x)\delta'(x-a)dx=-f'(a)$$

(by using integration by parts)

## Homework Equations

We know that δ(x) can be defined as

$$\lim_{\sigma\to\infty}(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}})$$

and another proper of the Dirac Delta Function is

$$\int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a)$$

Also, δ(x-a)=0 when x is not equal to a, and δ(x-a) is undefined when x=a. (Not really sure how to interpret the second part of that statement quite yet, I understand that it is because the two sided limit of σ->0 doesn't exist, but I don't see how to picture that graphically)

## The Attempt at a Solution

Starting with

$$\int_{-\infty}^{\infty}f(x)\delta'(x-a)dx=-f'(a)$$

I tried integration by parts, taking u=f(x) so u'=f'(x), and then v'=δ'(x-a) so v=δ(x-a), translating our original integral to

$$f(x)\delta(x-a) (from -\infty to \infty)-\int_{-\infty}^{\infty}f'(x)\delta(x-a)dx$$

I'm really not sure where to go next. I dont see what else I can do with evaluating the first part from negative infinity to infinity, and I don't see anything more I can do with the remaining integral. All in all I don't see how integration by parts was helpful at all.

Last edited: