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Integral containing a delta function and a step function

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    (a) Show that that δ(a-b)=∫δ(x-a)δ(x-b)dx

    (b) Show that ∂/∂x θ(x) = δ(x) where θ(x) is the heaviside step function (0 for x<0, 1 for x>0)

    (c) Show that ∫(-inf to inf) δ(x) f(θ(x))dx=∫(0 to 1) f(y)dy

    2. Relevant equations

    The definition of the delta function: ∫(-inf to inf) δ(x-y)f(x)=f(y)

    3. The attempt at a solution

    (a) Just made a change of variables and compared to the definition of the δ-function

    (b) ∫∂/∂x θ(x)dx = θ(x)|limits
    = 1 if limits enclose 0, 0 if not
    = ∫δ(x)dx with same limits

    (c) I used the result from part (b) to get ∫(-inf to inf) ∂/∂x θ(x) f(θ(x))dx
    then integrated by parts to get θ(x) f(θ(x))|(-inf to inf) -∫(-inf to inf) θ(x) ∂/∂x f(θ(x))
    = f(1) -∫(-inf to inf) θ(x) ∂/∂x f(θ(x))

    Can anyone tell me if i'm going about this the right way? thanks in advance :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2013 #2

    Office_Shredder

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    For part (b) you have to include the definition of the delta function in your calculation. In particular you must show that
    [tex] \int_{a}^{b} \frac{d \theta(x)}{dx } f(x) dx [/tex]
    is either f(0) or 0 depending on the values of a and b. You will have to do a similar procedure for part (a)
     
  4. Oct 9, 2013 #3
    Thanks for your reply
    My current solutions are now:

    (a) I am given that ∫(-inf to inf) δ(x-y)f(x)=f(y) I changed y→a and said that f(x)→x-b so that this reads ∫(-inf to inf) δ(x-a)δ(x-b)=δ(a-b) as required

    (c) ∫(-inf to inf) δ(x) f(θ(x))dx = ∫(-inf to inf) ∂/∂x θ(x) f(θ(x))dx
    = ∫(-inf to inf) ∂/∂x [∫f(θ(x))∂θ] dx
    = [∫f(θ(x))∂θ]|(x=-inf to x=inf)
    By letting y = θ(x)
    = ∫(0 to 1) f(y)dy

    Does this look ok?

    (b) So would I need to show that ∫(-inf to inf) [∂/∂x θ(x)] f(x) dx = f(0) ?
    (which is the definition of the delta function given in the question with δ→∂/∂x θ and y=0)

    trying integrating LHS by parts: f(x)θ(x)|(-inf to inf) - ∫(-inf to inf) θ(x-y) ∂/∂x f(x) dx
    f(x)|inf - ∫(-inf to inf) θ(x-y) ∂/∂x f(x) dx

    is any of this the correct method?
     
  5. Oct 9, 2013 #4

    Office_Shredder

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    For part (a), f(x) has to be an actual honest to goodness function. You can't plug in the delta "function" and expect it to work. The thing you want to consider is expressions of the form
    [tex] \int_{-infty}^{\infty} \int_{-\infty}^{\infty} \delta(x-a) \delta(x-b) dx f(a) da [/tex]
    and you want to find out that this is equal to f(b).

    For your part (b) solution, I don't understand where the y came in but other than that it looks correct. You should be able to integrate
    [tex] \int_{-\infty}^{\infty} \theta(x) f'(x) dx [/tex]
    in a very straightforward way.
     
  6. Oct 9, 2013 #5
    Thanks, that seems to have pointed my along the right track.

    The y in b) was a typo but I suppose I could include it all the way through for the most general case. I think I can do it ok now, I end up with f(x)|inf - ∫(-inf to inf) ∂/∂x f(x) dx = f(x)|inf - f(x)|inf + f(x)|0 = f(0) as required.

    For the integral you presented regarding part a) is this a matter of doing the a integral first since a and x are independent leaving ∫(-inf to inf) f(x) δ(x-b) dx = f(b) ?

    Also does my part c) solution look ok to you?
     
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