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Proving the scaling property of the Delta function

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]\delta(at)=\frac{1}{abs(a)}\delta(t)[/tex]

    Hint: Show that [tex]\int\phi(t)\delta(at)dt=\frac{1}{abs(a)}\phi(0)[/tex]

    (the limits of integration are from -inf to +inf btw, I couldn't find how to put them in..)

    2. Relevant equations

    3. The attempt at a solution

    Ok. I understand that the integral is only defined for at = 0, i.e., t = 0. But if I follow this logical then I evaluate the integral to be phi(0), not phi(0)/a. Where does the a come from?

    And I'm not entirely sure how proving that relationship helps with the proof of the scaling function...
  2. jcsd
  3. Mar 17, 2010 #2


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    That trick only works when the argument to the delta is the bare integration variable.
    So try substituting u = a t, so you get something like
    [tex]\int_{-\infty}^{\infty} f(u) \delta(u) \, du = f(0)[/tex]
    (and click the formula to see how I did the boundaries)
  4. Apr 8, 2010 #3
    Ok I've substituted x = at, now I get

    \int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)d\frac{x}{a} = \frac{1}{a}\int_{-\infty}^{\infty}\phi(\frac{x}{a})\delta(x)dx = \frac{1}{a}\phi(\frac{0}{a}) = \frac{1}{a}\phi(0)

    I don't quite see how I'm allowed to take the 1/a out of the d/dx operater and put it at the front though. And why should it be the abs(a) not a?

    And I still can't see how this helps prove anything.

    Cheers, Benjamin
  5. Apr 8, 2010 #4


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    You are allowed to take the a outside, because it is just a constant: du = a dt, so dt = (1/a) du, and you can put constants outside the integration as in
    [tex] \int \tfrac12 x^2 \, dx = \tfrac12 \int x^2 \, dx[/tex]

    The absolute value of a comes from transforming the integration boundaries. If a < 0, then -infinity goes to +infinity in the transformed integral, and vice versa. However, in the final formula you simply want -infinity to be the lower boundary, and +infinity as the upper boundary. So if a < 0 you have to swap them, at the cost of a minus sign, whereas for a > 0 you don't. Do you have to multiply the whole thing by sgn(a), and then you can shorten sgn(a) / a to 1/abs(a).
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