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Properties of Dirac delta function

  1. Feb 1, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove that [tex]\delta'(y)=-\delta'(-y)[/tex].

    2. Relevant equations

    3. The attempt at a solution

    I'm having trouble getting the LHS and the RHS to agree. I've used a test function [tex]f(y)[/tex] and I am integrating by parts.

    For the LHS, I have
    [tex]\int_{-\infty}^{\infty} f(y)\delta'(y)dy = \int_{-\infty}^{\infty} \frac{d}{dy}[f(y)\delta(y)]dy - \int_{-\infty}^{\infty} \delta(y)\frac{df(y)}{dy}dy = 0 - f'(0) = -f'(0)[/tex]

    For the RHS, I have
    [tex]-\int_{-\infty}^{\infty} f(y)\delta'(-y)dy = \int_{\infty}^{-\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} f(-t)\delta'(t)dt = -\int_{-\infty}^{\infty} \frac{d}{dt} [f(-t)\delta(t)]dt + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = 0 + \int_{-\infty}^{\infty} \frac{df(-t)}{dt} \delta(t)dt = f'(0)[/tex].

    I seem to be off by a minus sign, but I can't figure out where. Any help would be appreciated.
  2. jcsd
  3. Feb 1, 2008 #2


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    df(-t)/dt=-df(t)/dt at t=0.
  4. Feb 1, 2008 #3
    Yes, I suppose it does. Thanks!
  5. Jun 5, 2011 #4
    A proof of the above statement would be more helpful.
  6. Jun 5, 2011 #5


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    Chain rule.
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