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Alternative form of the 2nd translation theorem proof

  1. Mar 3, 2013 #1
    hi guys, this is literally my first post here on physicsforums so i apologize in advance that my latex formatting sucks.

    1. The problem statement, all variables and given/known data
    prove the alternative form of the 2nd translation theorem of the laplace transform:
    [itex]L[f(t)u(t-a)]=e^{-sa}L[f(t+a)] [/itex]

    where [itex]u(t-a)[/itex] is the unit step function and [tex]L[f(t)]=\int^\infty_0e^{-st}f(t)dt[/tex] is the definition of the laplace transform of [tex]f(x)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]L[f(t)u(t-a)]=\int^\infty_0 e^{-st}f(t)u(t-a)\,dt [/itex]
    [tex]=\int^a_0 e^{-st}f(t)u(t-a)\,dt + \int^\infty_a e^{-st}f(t)u(t-a)\,dt [/tex]
    the first integral is 0 since the unit step function 0 for any t < a, and [tex]u(t-a)=1[/tex] for t >= a so the 2nd integral becomes, [tex]\int^\infty_a e^{-st}f(t)\,dt[/tex]
    then since the laplace transform is an integral from 0 to [tex]\infty[/tex] we need to make the substitutions [tex]u = t-a, \ t = u+a,\ du = dt[/tex] to change the lower limit of integration from a to 0. then the integral becomes, [tex]\int^\infty_0 e^{-s(u+a)}f(u+a)\,du[/tex]
    then you pull the [tex]e^{-sa}[/tex] outside the integral to give [tex]e^{-sa}\int^\infty_0 e^{-su}f(u+a)\,du[/tex] and this is where i'm stuck. I don't know what to do here since the definition of the laplace transform is [tex]L[f(u)]=\int^\infty_0e^{-su}f(u)du[/tex] and i can't make any more substitutions without changing the limits of integration.
     
  2. jcsd
  3. Mar 4, 2013 #2
    What's [itex] L[f(t+a)] [/itex] then?
     
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