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Math Help - Proving one of the laplace transform properties

  1. #1
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    Proving one of the laplace transform properties

    What is the proof of the following property? :

    If F of class A and L \left( F(t) \right)=f(s) then :

    \displaystyle L \left( \dfrac{F(t)}{t} \right) = \int_s^{\infty} f(u) \, du
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting G(t)=\frac{F(t)}{t} \implies F(t)= t\ G(t) You have...

    \displaystyle \mathcal{L} \{F(t)\}= -\frac{d}{ds} \mathcal {L} \{g(t)}\} \implies f(s)= -\frac{d g(s)}{ds} (1)

    Integrating (1) You obtain...

    \displaystyle g(s) = \mathcal{L} \{\frac{F(t)}{t}\} = \int_{s}^{\infty} f(u)\ du (2)

    Kind regards

    \chi \sigma
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    another statement is if \dfrac{f(t)}t has limit as t\to0^+ and f is exponentially bounded, then \mathcal L\left(\dfrac{f(t)}{t}\right)=\displaystyle\int_s^  \infty \widehat f(u)\,du.

    the proof is basically to write f(t)=t\cdot\dfrac{f(t)}t since f is well defined for t=0, then use the Laplace transform for t\cdot f(t), integrate between [s,\infty) and use that the Laplace transform of a function tends to 0 as s\to\infty.
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  4. #4
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    I did not get it ;|
    Can you give me a full proof?
    Am lost :/
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Liverpool View Post
    I did not get it ;|
    Can you give me a full proof?
    Am lost :/
    Is there a particular part about this that you don't get?
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  6. #6
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    Yes this one:


    Quote Originally Posted by chisigma View Post

    \displaystyle \mathcal{L} \{F(t)\}= -\frac{d}{ds} \mathcal {L} \{g(t)}\} \implies f(s)= -\frac{d g(s)}{ds} (1)

    Integrating (1) You obtain...

    \displaystyle g(s) = \mathcal{L} \{\frac{F(t)}{t}\} = \int_{s}^{\infty} f(u)\ du (2)
    How did we obtain the limits of the last integral?
    from s to infinity??
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  7. #7
    MHF Contributor chisigma's Avatar
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    I apologize for the fact that a [stupid...] 'competition stress' often produces incomplete or 'impossible to undestand' answers...

    Setting G(t)= \frac{F(t)}{t} \implies F(t)= t\ G(t) we have...

    \displaystyle g^{'}(s) = \frac{d}{ds} \int_{0}^{\infty} G(t)\ e^{-s t}\ dt= - \int_{0}^{\infty} t\ G(t)\ e^{-s t}\ dt = - \mathcal{L} \{F(t)\} (1)

    ... and the first step is terminated. The (1) can be written as an ordinary first order DE as follows...

    \displaystyle g^{'}(s) = - f(s) (2)

    ... with the 'condition' \lim_{s \rightarrow \infty} g(s)=0 and its solution is...

    \displaystyle g(s)= - \int_{\infty}^{s} f(u)\ du = \int_{s}^{\infty} f(u)\ du (3)

    Kind regards

    \chi \sigma
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  8. #8
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    Thanks!
    Now it is too clear
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