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Laplace transform

  1. Jul 3, 2011 #1
    Okay so im brushing up my Laplace transforms as an independent study and I come across this proof for the "division by t theorem". The idea proof it self I have no problems with except for the limits of the first integration, It feels like they just arbitrarily choose the limits to be from <s,inf> to so convienently get rid of the negative sign. Is there any explaination for the limits of integration?

    here is the problem:


    here are the related theorems:

  2. jcsd
  3. Jul 4, 2011 #2
    Really like no answers =[???
  4. Jul 4, 2011 #3


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    It's just an application of the fundamental theorem of calculus, which tells you that
    [tex]\int_a^x h(u)\,du = H(x)-H(a)[/tex]
    where H(x) is a function that satisfies H'(x)=h(x). They chose [itex]a=\infty[/itex] to get rid of H(a).
  5. Jul 4, 2011 #4
    I actually want to know why you can pick integration limits it must have some meaning
  6. Jul 4, 2011 #5


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    If you don't use limits, an integral gives you a set of functions, where two elements of the set differ by a constant. In this problem, you only want one function so you need to use limits.

    You might find this derivation (from Arfken) more satisfying.

    Let [itex]f(s) = \mathcal{L}[F(t)][/itex]. Then
    \int_s^b f(u)\,du &= \int_s^b \int_{0}^\infty F(t)e^{-ut}\,dt \, du \\
    & = \int_{0}^\infty \int_s^b F(t)e^{-ut}\,du \, dt \\
    & = \int_{0}^\infty F(t) \frac{e^{-st}-e^{-bt}}{t} \, dt
    Taking the limit as [itex]b \to \infty[/itex] gives
    [tex]\int_s^\infty f(u)\,du =
    \lim_{b \to \infty} \int_{0}^\infty F(t) \frac{e^{-st}-e^{-bt}}{t} \, dt =
    \int_{0}^\infty \frac{F(t)}{t} e^{-st} \, dt = \mathcal{L}\left[\frac{F(t)}{t}\right][/tex]
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