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Delta dirac integral

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the integral ## \int_0^{3\pi} \delta (sin \theta) d\theta##

    2. Relevant equations

    3. The attempt at a solution
    I can rewrite ## delta (sin \theta) ## as ##\sum_{n=-\infty}^{\infty} \frac{\delta(\theta - n\pi)}{|cos (n\pi)|}=\sum_{n=-\infty}^{\infty} \delta(\theta-n\pi)##

    So the integral becomes:
    ## \int_0^{3\pi} \delta (sin \theta) d\theta = \int_0^{3\pi} [\delta (\theta) + \delta (\theta - \pi) + \delta (\theta - 2\pi) + \delta (\theta - 3\pi)] d\theta ##

    I understand that the dirac-delta function is symmetrical and the integral is taken over the half, so that

    ##\int_0^{3\pi} \delta (\theta) d\theta = 1/2##

    My solution sheet says the others equal to 1, 1, and 1/2 again. But I'm scratching my head how that is. Can someone explain it to me?
  2. jcsd
  3. Feb 1, 2015 #2
    Hi. Your interval is [0;3π] so at 0 and 3π the delta function sits on the limit and, as you correctly put for the 0 case, gives you 1/2 (the same convention is sometimes taken for the Heaviside theta). But at π and 2π you are well within the interval so the delta gives 1 as it should...
  4. Feb 1, 2015 #3
    Thanks very much for the explanation. Yes, I understand.
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