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Laplace transform of a piecewise function

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data

    f(t) = e^t when 0≤t<1
    and 0 when t≥1



    2. Relevant equations
    Laplace transformations

    3. The attempt at a solution

    so the Laplace integral becomes


    from 0 to 1 ∫e^(st^2)dt + 0

    how do I integrate this?
     
  2. jcsd
  3. Apr 11, 2015 #2
    Clarification:

    Laplace transform or Lagrange Transform...?
     
  4. Apr 11, 2015 #3
    My bad, Laplace :).
     
  5. Apr 11, 2015 #4
    Alright, let's do this:

    From 0 to 1 we have one function, and from 1 onward we have another. Split up our integral as so:

    $$\int_0^1 e^{-st} e^{t}dt + \int_{1}^{\infty} e^{-st}(0)dt \implies \int_0^1 e^{t(1-s)}dt$$
     
  6. Apr 11, 2015 #5
    Wait a second, doesn't the part that goes from 1 to +infinity get canceled out because the integral becomes

    ∫ e^(-st) (0) dt = 0

    ?
     
  7. Apr 11, 2015 #6
    Excuse my reading comprehension, I thought it said f(t) = 1. Corrected.
     
  8. Apr 11, 2015 #7
    Ah now I see what I did wrong! DUH! Such a stupid mistake.

    Thank you sir!
     
  9. Apr 11, 2015 #8

    SteamKing

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    How did you get an integrand of est2 ?

    Remember, ex ⋅ ey = e(x + y), not exy :wink:
     
  10. Apr 11, 2015 #9
    I was just asking myself the same thing. I think I need to take a break. I've been doing math since 7.30 this morning. It's 1.30 pm now :).
     
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