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Math Help - opposite furier transfom

  1. #1
    MHF Contributor
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    opposite furier transfom

    F^{-1}[\frac{\sin \omega}{\pi \omega}]=\frac{h(x^+)+h(x^-)}{2}
    F^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1

    first of all the fulf tansform is 1 when its |x|<1 0 other wise
    so why they chose x=0
    ??
    from this oppisite transform how can they say that
    \int_{-\infty}^{\infty}[\frac{\sin \omega}{\pi \omega}]=1
    ?

    its only a tranform how can they say that the integral gives the same result straigh away
    ?
    Last edited by transgalactic; July 17th 2010 at 08:59 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by transgalactic View Post
    F^{-1}[\frac{\sin \omega}{\pi \omega}]=\frac{h(x^+)+h(x^-)}{2}
    F^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1

    first of all the fulf tansform is 1 when its |x|<1 0 other wise
    so why they chose x=0
    ??
    from this oppisite transform how can they say that
    \int_{-\infty}^{\infty}[\frac{\sin \omega}{\pi \omega}]=1
    ?

    its only a tranform how can they say that the integral gives the same result straigh away
    ?
    I'm sorry but most of your post is incomprehensible.

    But to see why the final integral is 1 from:

    \mathfrak{F}^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1

    just write out the integral for this inverse FT.

    CB
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