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Delta dirac integral

  • Thread starter Jillds
  • Start date
  • #1
22
1

Homework Statement


Solve the integral ## \int_0^{3\pi} \delta (sin \theta) d\theta##

Homework Equations




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?
 

Answers and Replies

  • #2
205
16
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...
 
  • #3
22
1
Thanks very much for the explanation. Yes, I understand.
 

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