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Integral tranforms and symmetry

  1. Jan 27, 2014 #1
    The fourier transform, wrt to angular frequency, needs of a factor (1/2π) for get f(t) or F(ω), actually, this factor is broken in 2 factors (1/√2pi) and each kernel, direct and inverse, receives one factor for keep the symmetry in equation.
    [tex]F(\omega)=\int_{-\infty }^{+\infty }\frac{e^{-i\omega t}}{\sqrt{2\pi}} f(t)dt[/tex]
    [tex]f(t)=\int_{-\infty }^{+\infty }\frac{e^{+i\omega t}}{\sqrt{2\pi}} F(\omega)d\omega[/tex]

    Why others transform, by definition, don't have its "conversion factor" "broken" in 2 and distributed in each kernel? Is wrong define the transforms in this way? Prejudice the calculus?
     
  2. jcsd
  3. Jan 27, 2014 #2

    Simon Bridge

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    You can break the any transformation any way you like - it's just not always useful to do so.
     
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