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Improper integral

  1. Feb 27, 2012 #1
    Is ∫eiωtdω = √(2[itex]\pi[/itex]) (from -∞ to ∞)

    It should be so if my books derivation of the fourier transform is correct, but they don't really explain why the above equality is correct?

    can anyone show me how they get to that result for that integral or does anyone perhaps have a good link?
  2. jcsd
  3. Feb 27, 2012 #2
    If you set I = integral and multiplied by the same integral, you would have I^2. When you solve that you get 2pi but then you take the square root.

    However, isn't is supposed to be equal to \frac{1}{\sqrt{2\pi}
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