# Proving a property of the dirac delta function

1. Oct 19, 2013

### skate_nerd

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

2. Relevant 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)

3. 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: Oct 19, 2013
2. Oct 19, 2013

### tiny-tim

hi skate_nerd!
(i'm not convinced that that part of the formula is valid , but anyway …)

what are δ(∞ - a) and δ(-∞ - a) ?

can't you apply the delta definition directly to it?

3. Oct 19, 2013

### skate_nerd

Thanks for the reply tiny tim. I see what you are saying about the first part of the formula I reached. However for the second part, if I applied the definition of the delta function I would have a very odd integral that I would really have no idea how to solve.

4. Oct 19, 2013

### tiny-tim

uhh? the definition is …
and you want to find …

5. Oct 19, 2013

### skate_nerd

Yeah but how does knowing that first definition help with that second integral? I don't see any way to use that to solve the second integral...

6. Oct 19, 2013

### vela

Staff Emeritus
You do realize that f' is just another function, right?

7. Oct 19, 2013

### skate_nerd

Yes, I think its quite obvious i am aware of that since I used that fact when I integrated by parts...

8. Oct 19, 2013

### skate_nerd

Am I just making this harder than it needs to be or what? I honestly am really stuck here and I've been thinking about this problem since last night

9. Oct 19, 2013

### vela

Staff Emeritus
Let g(x)=f'(x). Now the integral becomes
$$\int f'(x)\delta(x-a)\,dx = \int g(x)\delta(x-a)\,dx = ?$$

10. Oct 20, 2013

### skate_nerd

Ahhhhh wow okay that totally makes sense now. Thank you vela. I guess for some reason I was thinking that if you substituted something that way it would be like a change of variable but it actually isn't. I kept trying to use the definition
$$\lim_{\sigma\to0}(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}})$$
but integrating something like that is a lot more difficult than it sounds...

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