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A Integration using delta function and step function

  1. Oct 7, 2016 #1
    I would like to evaluate the following integral:

    ##\displaystyle{\int_{-\infty}^{\infty} dp^{0}\ \delta(p^{2}-m^{2})\ \theta(p^{0})}##

    ##\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0})^{2}-\omega^{2}]\ \theta(p^{0})}##

    ##\displaystyle{= \int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)(p^{0}-\omega)]\ \theta(p^{0})}##

    ##\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta[(p^{0}+\omega)+(p^{0}-\omega)]\ \theta(p^{0})}##

    ##\displaystyle{= \frac{1}{2\omega}\int_{-\infty}^{\infty} dp^{0}\ \delta(2p^{0})\theta(p^{0})}##

    Have I made a mistake somewhere?

    The answer is supposed to be ##\frac{1}{\omega}## but I'm not sure how to proceed next.
     
  2. jcsd
  3. Oct 8, 2016 #2

    andrewkirk

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    I don't follow some of your steps so far. But if you want to be able to get rid of the integral, try a change of variable, making the variable over which we are integrating ##u=2p^0##.

    The integral can then be evaluated quite simply. However I then get the answer ##1/4\omega##, which is why I am wondering about your earlier steps.
     
  4. Oct 8, 2016 #3
    Would you recommend me to post this on the Relativity forum?

    This question is better suited for that forum.
     
  5. Oct 8, 2016 #4
    [tex]
    \delta(x^2 - \alpha^2) = \frac{1}{2|\alpha|} \left[\delta(x-\alpha) + \delta(x+\alpha) \right] \neq \frac{1}{2|\alpha|} \left[\delta( (x-\alpha) + (x+\alpha)) \right]
    [/tex] which is what you seem to have done.
    Are you sure the answer is supposed to be ##1/\omega## and not ##1/2\omega## though?
     
    Last edited: Oct 8, 2016
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