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Math Help - Laplace transform

  1. #1
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    Laplace transform

    I'm asked to find the Laplace transform of f(t)=\dfrac{sin(t)}{t}.

    I begin by applying the definition to the function:

    <br />
F(s) = \int_0^{\infty}e^{-st} \dfrac{sin(t)}{t}~dt<br />

    I'm getting the feeling that I need to use integration by parts, but I'm having a hard time identifying good choices for u and dv

    Please advise.

    note: I will continue to add more troublesome Laplace transform problems to this thread as I encounter them.
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  2. #2
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    A hint

    Note that
    F'(s)=\frac{d}{ds}\int_0^{\infty}e^{-st}\frac{\sin{t}}{t}~dt
    =\int_0^{\infty}\frac{d}{ds}\left(e^{-st}\frac{\sin{t}}{t}\right)dt
    =\int_0^{\infty}-t\cdot{e^{-st}}\frac{sin(t)}{t}dt
    =-\int_0^{\infty}e^{-st}\sin{t}dt
    =-\mathcal{L}[\sin{t}](s).
    Do you know the Laplace transform of the sine function?

    --Kevin C.
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  3. #3
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    i do indeed
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  4. #4
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    actually, are you sure this is correct?

    also, it just occurred to me that perhaps i should expand sin(t) via euler's formula and try it that way.

    maple is kicking out arctan(1/s)
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  5. #5
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    Quote Originally Posted by jmedsy View Post
    I'm asked to find the Laplace transform of f(t)=\dfrac{sin(t)}{t}.

    I begin by applying the definition to the function:

    <br />
F(s) = \int_0^{\infty}e^{-st} \dfrac{sin(t)}{t}~dt<br />

    I'm getting the feeling that I need to use integration by parts, but I'm having a hard time identifying good choices for u and dv

    Please advise.

    note: I will continue to add more troublesome Laplace transform problems to this thread as I encounter them.
    Theorem (simple to prove): LT\left[ \frac{f(t)}{t}\right] = \int_s^{+\infty} F(\tau) \, d \tau where F(s) = LT[f(t)].

    In your question f(t) = \sin (t).

    Quote Originally Posted by jmedsy View Post
    actually, are you sure this is correct?

    also, it just occurred to me that perhaps i should expand sin(t) via euler's formula and try it that way.

    maple is kicking out arctan(1/s)
    Note that \frac{\pi}{2} - \tan^{-1} (s) = \tan^{-1} \left( \frac{1}{s} \right).
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