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Math Help - Laplace Transform of log(t)

  1. #1
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    Laplace Transform of log(t)

    Hi everybody!
    I'm trying to find the Laplace transform of log(t) (Our lecturer promised to give a lot of respect to the one who solve that ...)

    Can you guys please help me and give me some direction?
    Thanks!

    Red_Fox
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  2. #2
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    Quote Originally Posted by Red_Fox
    Hi everybody!
    I'm trying to find the Laplace transform of log(t) (Our lecturer promised to give a lot of respect to the one who solve that ...)

    Can you guys please help me and give me some direction?
    Thanks!

    Red_Fox
    Have you consided looking it up in a table of Laplace transforms?

    RonL
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  3. #3
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    well... where can I find one? I mean, I didn't find it in the elementry Laplace Transform table!
    And by the way, I would like to know the way to find it.

    P.S I've found something: (-y+lns)/s , when y is Euler constant (limiting difference between the harmonic series and the natural logarithm).
    Maybe you know how they reached this??

    Thanks (:
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  4. #4
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    Quote Originally Posted by Red_Fox
    well... where can I find one? I mean, I didn't find it in the elementry Laplace Transform table!
    And by the way, I would like to know the way to find it.

    P.S I've found something: (-y+lns)/s , when y is Euler constant (limiting difference between the harmonic series and the natural logarithm).
    Maybe you know how they reached this??

    Thanks (:
    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-st}ln(t)dt

    let t'=st then:

    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}ln(\frac{t'}{s}) \frac{1}{s} dt'

    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}(ln(t')-ln(s)) \frac{1}{s} dt'

    {\cal{L}}\ ln(t)= \frac{1}{s}  \int_0^{\infty}e^{-t'}ln(t') dt' -\frac{ln(s)}{s} \int_0^{\infty} e^{-t'}dt'

    Now if I recall correctly the first integral on the RHS is -\gamma,
    and you should be able to finish the derivation from here.

    So not so difficult after all?

    By the way my handbook says it is -\frac{(\gamma+ln(s))}{s}.

    RonL
    Last edited by CaptainBlack; December 19th 2005 at 10:33 AM.
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  5. #5
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    Thanks!

    Wow (: it's so simple after t'=st!
    By the way , it is very interesting how to reach from this integral (with the e and the ln) to the limiting difference between the harmonic series and the natural logarithm. see also [HTML]http://en.wikipedia.org/wiki/Euler-Mascheroni_constant[/HTML]

    Thank You Very Much!!!
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  6. #6
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    Quote Originally Posted by CaptainBlack
    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-st}ln(t)dt

    let t'=st then:

    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}ln(\frac{t'}{s}) \frac{1}{s} dt'

    {\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}(ln(t')-ln(s)) \frac{1}{s} dt'

    {\cal{L}}\ ln(t)= \frac{1}{s}  \int_0^{\infty}e^{-t'}ln(t') dt' -\frac{ln(s)}{s} \int_0^{\infty} e^{-t'}dt'

    Now if I recall correctly the first integral on the RHS is -\gamma,
    and you should be able to finish the derivation from here.

    So not so difficult after all?

    By the way my handbook says it is -\frac{(\gamma+ln(s))}{s}.

    RonL

    you have one little mistake when you put t'
    you forgot s is a complex variable
    so you changed the intire intigration
    you need to prove it still works under complex variabels

    can you help me with that?
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  7. #7
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    Quote Originally Posted by mooshazz
    you have one little mistake when you put t'
    you forgot s is a complex variable
    so you changed the intire intigration
    you need to prove it still works under complex variabels

    can you help me with that?
    Its not a mistake, I am using a definition of the Laplace transform
    for s real.

    At least in some places the Laplace transform is defined for s real,
    with appropriate continuation being used to extend it to most of
    \mathbb{C} when required.

    See attachment for a definition restricted to s-real (I know its
    intended for engineers but here it is anyway)

    RonL
    Attached Thumbnails Attached Thumbnails Laplace Transform of log(t)-laplace1.jpg  
    Last edited by CaptainBlack; December 22nd 2005 at 11:37 AM.
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  8. #8
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    Quote Originally Posted by CaptainBlack
    At least in some places the Laplace transform is defined for s real,
    with appropriate continuation being used to extend it to most of
    \mathbb{C} when required.

    See attachment for a definition restricted to s-real (I know its
    intended for engineers but here it is anyway)

    RonL

    but as a mathmatician i need to prove that for complex s
    how can i do it?
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  9. #9
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    Quote Originally Posted by mooshazz
    but as a mathmatician i need to prove that for complex s
    how can i do it?
    This is a bit sketchy as I don't have the relevant reference books to
    hand, but as you are a mathematician you should be able to fill in the
    detail.

    1. Show that in this case the Laplace transform is analytic in the right half
    plane. You should have a set of conditions for a functions to guarantee
    that its Laplace transform has this property.

    2. Then the value of the Laplace transform at all points in the right half
    plane is uniquely defined by its values on an arc contained in the right half
    plane.

    3. Any segment of the positive real axis supplies such an arc. The given
    Laplace transform with s complex is a function analytic on the right
    half plane, which agrees with the real Laplace transform on the real axis,
    and so is the complex Laplace transform.

    Does that make any sort of sense?

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlack
    This is a bit sketchy as I don't have the relevant reference books to
    hand, but as you are a mathematician you should be able to fill in the
    detail.

    1. Show that in this case the Laplace transform is analytic in the right half
    plane. You should have a set of conditions for a functions to guarantee
    that its Laplace transform has this property.

    2. Then the value of the Laplace transform at all points in the right half
    plane is uniquely defined by its values on an arc contained in the right half
    plane.

    3. Any segment of the positive real axis supplies such an arc. The given
    Laplace transform with s complex is a function analytic on the right
    half plane, which agrees with the real Laplace transform on the real axis,
    and so is the complex Laplace transform.

    Does that make any sort of sense?

    RonL
    not even a bit
    i didn't understand any of the stages

    at st. 1 did you ment the result of the transform or the interior of the integral

    at st. 2 i didn't get anything
    and so in stage 3
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  11. #11
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    Quote Originally Posted by mooshazz
    not even a bit
    i didn't understand any of the stages

    at st. 1 did you ment the result of the transform or the interior of the integral

    at st. 2 i didn't get anything
    and so in stage 3
    Stage 2 is key. Roughtly it is the result that a function analytic on a region A \subseteq \mathbb C
    is uniquely determined by its values on a curve  \Gamma \subset A. If you haven't covered something
    like that in Complex Analysis this method is not available to you and you will
    have to try evaluating the integral after the change of variable as a line
    integral over the line s.(0, \infty) and see if that works.

    RonL
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  12. #12
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    as a hebrew speaker it's a bit difficult for me with mathmatical english
    can you explain me those stages in the language of signs and numbers?
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