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Proving a property of the dirac delta function

  1. Oct 19, 2013 #1
    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. jcsd
  3. Oct 19, 2013 #2

    tiny-tim

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

    what are δ(∞ - a) and δ(-∞ - a) ?

    can't you apply the delta definition directly to it?
     
  4. Oct 19, 2013 #3
    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.
     
  5. Oct 19, 2013 #4

    tiny-tim

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    uhh? the definition is …
    and you want to find …
     
  6. Oct 19, 2013 #5
    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...
     
  7. Oct 19, 2013 #6

    vela

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    You do realize that f' is just another function, right?
     
  8. Oct 19, 2013 #7
    Yes, I think its quite obvious i am aware of that since I used that fact when I integrated by parts...
     
  9. Oct 19, 2013 #8
    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
     
  10. Oct 19, 2013 #9

    vela

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    Let g(x)=f'(x). Now the integral becomes
    $$\int f'(x)\delta(x-a)\,dx = \int g(x)\delta(x-a)\,dx = ?$$
     
  11. Oct 20, 2013 #10
    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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