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Cosine evaluated at infinity

  1. Sep 21, 2015 #1
    okay so I have this integral:

    2∫cos(kx)dk

    The bounds are from zero to infinity, this doesn't converge but is there any other way to describe this?
     
  2. jcsd
  3. Sep 21, 2015 #2

    Mark44

    Staff: Mentor

    Like this? ##2\int_0^{\infty}\cos(kx) dk##
    Because the upper limit of integration is ∞, this is an improper integral. You can't just "plug in" ∞ when you evaluate the antiderivative -- you need to use limits to evaluate it.
     
  4. Sep 21, 2015 #3
    okay so the limit would be lim_k->∞ sin(kx)
    x


    How would I go about evaluating it? Is there a substitution trick?
     
  5. Sep 21, 2015 #4

    Mark44

    Staff: Mentor

    The limit doesn't exist because sin(kx) oscillates endlessly.
     
  6. Sep 21, 2015 #5
    okay gotcha. Thanks!
     
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