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Help! inverse laplace transform of 1/(x^2+1)^2

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi.

    I need help to resolve the inverse laplace transform of {1/((x^2)+1)^2}


    2. The attempt at a solution

    I have tried to do:

    {(1/((x^2)+1) * (1/((x^2)+1)}

    then, convolution, sen x

    But, isn't working

    Thanks for your help :)
     
  2. jcsd
  3. Aug 16, 2014 #2

    LCKurtz

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    Using usual notation you would use ##s## as the variable in transform space:$$
    \mathcal L^{-1}\frac 1 {(s^2+1)^2}$$Forum rules require you to show some effort. So what did you get for$$
    \mathcal L^{-1}\frac 1 {s^2+1}\text{?}$$Then show us what your convolution integral looks like and where you are stuck.
     
  4. Aug 17, 2014 #3
    yes, sorry. is:

    $$
    \mathcal L^{-1}\frac 1 {(s^2+1)^2}$$

    so

    $$
    \mathcal L^{-1}\frac 1 {(s^2+1)} * \frac 1 {(s^2+1) }$$

    then

    f(τ) = sen τ
    g(t-τ) = sen (t-τ)

    ∫ sen τ * sen (t-τ) dτ

    Integration by parts

    u= sen (t-τ)
    du = -cos(t-τ)

    dv= sen τ dτ
    v= -cos τ


    ∫ sen τ * sen (t-τ) dτ = u= -(sen (t-τ) * cos τ) - ∫ cos τ * cos(t-τ)

    Integration by parts, again... not working

    maybe, another theorem... I dont know
     
  5. Aug 17, 2014 #4

    HallsofIvy

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    use a trig "difference formula": [itex]sin(t- \tau)= sin(t)cos(\tau)- cos(t)sin(\tau)[/itex] so that your integral becomes
    [tex]\int sin(\tau)sin(t- \tau) d\tau= \int sin(\tau)[cos(\tau)sin(\tau)- cos(t)sin(\tau)]d\tau[/tex]
    [tex]= sin(t)\int sin(\tau)cos(\tau) d\tau- cos(t)\int sin^2(\tau)d\tau[/tex]
     
    Last edited: Aug 17, 2014
  6. Aug 17, 2014 #5

    LCKurtz

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    Since Halls has already responded to your problem, I will just point out that the above isn't what you mean. You take the convolution after the inverse, so you want$$
    \mathcal L^{-1}\frac 1 {(s^2+1)} * \mathcal L^{-1}\frac 1 {(s^2+1) }$$
     
  7. Aug 17, 2014 #6
    thanks :)
     
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