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Simplifying the integral of dirac delta functions

  1. Oct 21, 2009 #1
    hello all,
    i am unaware of how to handle a delta function. from what i read online the integral will be 1 from one point to another since at zero the "function" is infinite. overall though i dont think i know the material well enough to trust my answer. and help on how to take the integral of a dirac delta function would be much appreciated. the kind of problems im dealing with are similar to
    [tex]\int[/tex]e^(-x^2) [tex]\delta[/tex](x+1)[1-cos(5 (pi/2) x)] dx from -infinity to 0
    for this i solved the delta function for when x+1=0 since that is the only time when the function will have a value. i then subbed -1 for all x's and took the integral, assuming the delta function became 1.
    thank you very much for help
     
  2. jcsd
  3. Oct 21, 2009 #2

    tiny-tim

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    hello sleventh! :smile:

    (have a delta: δ :wink:)
    no, there's no integral in the result …

    δ is defined by the property ∫ δ(x) f(x) dx = f(0) for any f.

    Put F(x) = f(x-1).

    Then ∫ δ(x+1) f(x) dx = ∫ δ(x) f(x-1) dx = ∫ δ(x) F(x) dx = F(0) = f(-1). :wink:

    Do you see how the δ function gets rid of the integral? :smile:
     
  4. Oct 21, 2009 #3
    hey tiny-tim. thank you very much. that clears up a lot. so if you were to be taking the integral of the above problem from limits 0-∞ (thanks for the copy and paste) would the problem have the same solution as -∞ -0 because you are still subbing in 0 to the f(x-1) adaptation of the function?
     
  5. Oct 21, 2009 #4

    tiny-tim

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    Hi sleventh! :smile:

    Any limits have to strictly include the zero of what's inside the δ (which in this case is -1) …

    so it can be ∫-∞ or ∫-∞0 or ∫-20,

    but not ∫0 or even ∫-1 or ∫-∞-1 :wink:
     
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