Inverse Laplace transform of 1/(s²+1)²

[nito18]

Homework Statement

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 :)

 

[LCKurtz]

Homework Statement

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 :)

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.

 

[nito18]

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.

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 don't know

 

[HallsofIvy]

use a trig "difference formula": $sin(t- \tau)= sin(t)cos(\tau)- cos(t)sin(\tau)$ so that your integral becomes
$$\int sin(\tau)sin(t- \tau) d\tau= \int sin(\tau)[cos(\tau)sin(\tau)- cos(t)sin(\tau)]d\tau$$
$$= sin(t)\int sin(\tau)cos(\tau) d\tau- cos(t)\int sin^2(\tau)d\tau$$

 

[LCKurtz]

yes, sorry. is:

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

so

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

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

 

[nito18]

use a trig "difference formula": $sin(t- \tau)= sin(t)cos(\tau)- cos(t)sin(\tau)$ so that your integral becomes
$$\int sin(\tau)sin(t- \tau) d\tau= \int sin(\tau)[cos(\tau)sin(\tau)- cos(t)sin(\tau)]d\tau$$
$$= sin(t)\int sin(\tau)cos(\tau) d\tau- cos(t)\int sin^2(\tau)d\tau$$

thanks :)