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Integration with Dirac Delta

  1. Jun 27, 2017 #1

    Ado

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

    I need to integrate this expression :

    P(k, w) = A * δ(w-k*v) * f(k, w)

    A is constant and δ, Dirac Delta.


    2. Relevant equations

    There is double integration :

    I = ∫0 dk ∫0 P(k,w) dw
    = A ∫∫0 δ(w-k*v) * f(k, w) dw dk

    3. The attempt at a solution

    I'm confused with Delta Dirac for calculating this integral.

    Let us proceed the w-variable integration firstly. Since the δ-term impose w = k*v, we need to calculate :
    0 f(k=w/v, w) dw
    But after that, there is no longer any k-dependence...

    Of course I'm wrong.. can you explain me how to proceed ??

    Thanks in advance !
     
  2. jcsd
  3. Jun 27, 2017 #2

    Charles Link

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    I believe you could do the integration in either order, ## dk ## or ## d \omega ## first. The delta function should get removed with the first integration, and both methods should give the same answer. When doing the ## dk ## integration first, I believe the factor ## v ## will affect the delta function, and the substitution ## u=vk ## would be useful. ## \\ ## Suggestion: Try something such as ## \int\limits_{0}^{+\infty} \int\limits_{0}^{+\infty} \exp(-\omega^2) \, \delta(\omega-vk) \, d \omega \, dk ##. When you do the ## dk ## integration first, nothing happens to the ## \omega ## in the exponential, since any ## k ## term would get converted, but ## \omega ## terms are unaffected. And yes, to get agreement with the answer where you do the ## d \omega ## first, you need the substitution ## u=vk ## when doing ## dk ## first. With this substitution, ## dk=\frac{du}{v} ##, and the delta function with the ## du ## integration integrates to unity. ## \\ ## The process when the ## d \omega ## integration is performed first is a little more straightforward. ## \\ ## (Note: Here we let ## f(k, \omega)=\exp(-\omega^2) ## with no ## k ## dependence).
     
    Last edited: Jun 27, 2017
  4. Jul 30, 2017 #3

    rude man

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    You're forcing w to be a function of k but this is not right. k and w are both variables. Go
    I = A ∫ ∫ f(k,w) δ(w - kv) dw dk
    Sampling characteristic of the delta function leads to
    I = A f(k, w=kv)dk
    provided (lower limit of integration) < kv < (upper limit of integration) over w.
     
  5. Aug 10, 2017 #4

    Ado

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    You are right rude man, thank's for your help :)
     
  6. Aug 10, 2017 #5

    rude man

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    Most welcome!
     
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