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Math Help - evaluating sine (or cos) at infinity

  1. #1
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    evaluating sine (or cos) at infinity

    so ive hit a snag in a problem (the topic is actually fourier integral transformations)
    i was wondering whether there was some approach to evaluating
    \int_0^{\infty}sin(x) dx

    in my mind sin(\infty) could be either -1, 0 or 1. is there a orthodox way of approaching this?
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  2. #2
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    Quote Originally Posted by walleye View Post
    so ive hit a snag in a problem (the topic is actually fourier integral transformations)
    i was wondering whether there was some approach to evaluating
    \int_0^{\infty}sin(x) dx

    in my mind sin(\infty) could be either -1, 0 or 1. is there a orthodox way of approaching this?

    There is a very orthodox solution as well: your integral diverges, since it involves evaluating the limit of the primitive function of sin x, i.e. -cos x, when x --> oo, and this limit just doesn't exist.

    Tonio
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  3. #3
    MHF Contributor chisigma's Avatar
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    The limit...

    \lim_{t \rightarrow \infty} \int_{0}^{t} \sin \tau\cdot d\tau (1)

    ... doesn't exist but exists the limit...

    \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} \sin \tau\cdot d\tau = \langle \sin t \rangle = 0 (2)

    In general if f(*) is the sum of a finite number of periodic functions not necessarly with the same period the limit...

    \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} f(\tau)\cdot d\tau = \langle f(t) \rangle (3)

    ... exists and \langle f(*) \rangle is called [improperly] 'mean value of f(*)'. It is...

    \langle f(t) \rangle = \sum_{n} \langle f_{n}(t) \rangle (4)

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    The limit...

    \lim_{t \rightarrow \infty} \int_{0}^{t} \sin \tau\cdot d\tau (1)

    ... doesn't exist but exists the limit...

    \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} \sin \tau\cdot d\tau = \langle \sin t \rangle = 0 (2)


    Welll...yes, in a rather trivial way: since INT{0,t} sin(s) ds = -cos(t) + 1,
    then lim (1/t)[1 - cos(t)] = 0 when t --> oo since 1 - cos t is bounded and (1/t) --> 0 .
    This though doesn't seem to appropiately address the OP's question.

    Tonio


    In general if f(*) is the sum of a finite number of periodic functions not necessarly with the same period the limit...

    \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} f(\tau)\cdot d\tau = \langle f(t) \rangle (3)

    ... exists and \langle f(*) \rangle is called [improperly] 'mean value of f(*)'. It is...

    \langle f(t) \rangle = \sum_{n} \langle f_{n}(t) \rangle (4)

    Kind regards

    \chi \sigma
    ..............
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