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Solve a Laplace transform puzzle

  1. Jan 18, 2014 #1
    Hi~
    I recently solve a Laplace transform problem as following
    L[int{t,0}cosh(t'-1)U(t'-1)dt']=? U(t'-1) is the unit step function(=1 for t'>1, =zero otherwise)
    According the standard Laplace formula :
    (1)L[cosh(t-1)U(t-1)]=exp(-s)*s/(s^2-1);
    (2)L(int{t,0}f(t')dt')=F(s)/s. where F(s)=L(f(t))

    I conclude the answer to be:
    exp(-s)/(s^2-1)

    But from the direct integral formula:
    L[int{t,0}cosh(t'-1)U(t'-1)dt']=L[sinh(t-1)]=(1/2)*(exp(-1)/(s-1)-exp(1)/(s+1)).

    This two answers are not consistent!
    I totally have no idea what's wrong with the two methods?

    I believe the second method should be correct by basic definition of Laplace transformation.

    Thanks very much!
     
  2. jcsd
  3. Jan 18, 2014 #2
    How did you find the direct integral (the step at the first equal sign)?
     
  4. Jan 19, 2014 #3

    Ray Vickson

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    [tex] \int_0^t \cosh(t'-1) U(t'-1) \, dt' \neq \sinh(t-1) [/tex]
    because the integral vanishes if t < 1 but sinh(t-1) does not.
     
  5. Jan 19, 2014 #4

    HallsofIvy

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    [tex]\int_0^t cosh(t'- 1)U(t'-1)dt'= \int_1^t cos(t'- 1)dt'[/tex]
    and with the substitution u= t'- 1, du= dt', when t'= 1, u= 0 and when t'= t, u= t- 1 so that becomes
    [tex]\int_0^{t-1} cosh(u)du[/tex].
     
  6. Jan 20, 2014 #5
    Thanks Ray Vickson and Hallsoflvy for nice answering.
    I think I make mistake to deal with the unit step function in the integral form.
    It clear now for me!
     
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