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Prove a Laplace Transform Equality

  1. Sep 2, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that http://www4f.wolframalpha.com/Calculate/MSP/MSP26931c1531g07285beh7000062h7f6g1ggd95eea?MSPStoreType=image/gif&s=5&w=98.&h=38. [Broken] =http://www4f.wolframalpha.com/Calculate/MSP/MSP6901c153574d0bdbh20000048829f0g4d1fi1d0?MSPStoreType=image/gif&s=5&w=69.&h=35. [Broken]

    3. The attempt at a solution

    Using the Def of Laplace i got to ∫[e ^ -st - e ^ -t(s-1)] / t dt ; of course with limits from 0 to infinite

    tryed then to do a variable change of w = st but i got inmmerse on a huge process of solving

    ∫{ [e ^ -w . e ^ (1/s-1)w] / w/s } dw/s , i know that the change was wrong from that point and i have been trying to solve it for parts or another change but havent got accross any luck with that so far

    Id appreciate some help or any advice in how to proceed from that point, its the only problem left to finish my homework and its driving me crazy :S

    Side Note: i haven´t seen yet Inverse Laplace, but a fellow Math Degree friend told me that using that i might solve it pretty easy.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Sep 2, 2015 #2

    SteamKing

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    I would first check your algebra in multiplying e-st by e-t / t
     
    Last edited by a moderator: May 7, 2017
  4. Sep 3, 2015 #3

    Zondrina

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    You don't have to use the formal definition of a Laplace transform because there is a theorem that will greatly simplify the calculation.

    Theorem: Suppose ##f(t)## is piecewise continuous and of exponential order. By exponential order, I mean there exists a constant ##\alpha## such that ##e^{- \alpha t} |f(t)| \to 0## as ##t \to \infty##.

    Suppose further ##\mathcal{L} \{ f(t) \} = F(s)## for ##s > c \geq 0##, and ##\displaystyle \lim_{t \to 0^+} \frac{f(t)}{t}## exists. Then:

    $$\mathcal{L} \left \{ \frac{f(t)}{t} \right \} = \int_s^{\infty} F(x) \space dx, \quad s > c$$
     
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