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Reflection Rule of a Fourier Transform

  1. Oct 14, 2009 #1
    I feel a bit dumb, but could someone help me see this:

    [tex]G(s):= \int_{-\infty}^{\infty}f(-x)e^{-2\pi isx}dx = \int_{-\infty}^{\infty}f(u)e^{-2\pi i(-s)u}du = F(-s)[/tex]
  2. jcsd
  3. Oct 14, 2009 #2


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    You're about to feel even dumber! (But actually be smarter.)

    Let u= -x. Then [itex]-2\pi isx= -2\pi i(-s)u[/itex], f(-x) becomes f(u), of course.

    As x goes to [itex]\infty[/itex], u goes to [itex]-\infty[/itex] and vice-versa so the limits of integration are switched. That's the reason the "-" in front of the first integral disappears.
  4. Oct 14, 2009 #3
    I feel dumber for not realizing this, and even more dumbest for this sentance. A little smarterest though for the learning...

  5. Oct 14, 2009 #4


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    Hey, you've got a long way to go before you are as dumberized as I am!
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