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Proving a property of the dirac delta function

  • Thread starter skate_nerd
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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.
 
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Answers and Replies

  • #2
tiny-tim
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hi skate_nerd! :smile:
$$f(x)\delta(x-a) (from -\infty to \infty)-\int_{-\infty}^{\infty}f'(x)\delta(x-a)dx$$
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
(i'm not convinced that that part of the formula is valid :redface:, but anyway …)

what are δ(∞ - a) and δ(-∞ - a) ?

…, and I don't see anything more I can do with the remaining integral.
can't you apply the delta definition directly to it?
 
  • #3
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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
tiny-tim
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uhh? the definition is …
$$\int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a)$$
and you want to find …
$$\int_{-\infty}^{\infty}f'(x)\delta(x-a)dx$$
 
  • #5
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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
vela
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You do realize that f' is just another function, right?
 
  • #7
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Yes, I think its quite obvious i am aware of that since I used that fact when I integrated by parts...
 
  • #8
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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
vela
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Let g(x)=f'(x). Now the integral becomes
$$\int f'(x)\delta(x-a)\,dx = \int g(x)\delta(x-a)\,dx = ?$$
 
  • #10
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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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