"Find the inverse Laplace Transform of $H(s) = \frac{1}{s(s^2+4s+5)}$"
My Work So Far
Rewrite as the product of two familiar fractions:
$H(s) = \frac{1}{s}\frac{1}{s^2+4s+5}$
$H(s) = \frac{1}{s}\frac{1}{(s+2)^2+1}$
Now we see that $F(s) = \frac{1}{s} \to f(t) = 1$
And
$G(s) = \frac{1}{(s+2)^2+1} \to g(t) = e^{-2t}sin(t)$
Via the Convolution Theorem, $L^{-1}${$F(s)G(s)$} $= f(t)*g(t)$
Great! Now all I need to to is set up the integral for $1*e^{-2t}sin(t)$ and solve it, right?
$\int_0^te^{-2τ}sin(τ)dτ$
...Uh oh. This will give an integration by parts. But the remaining integral will ALSO give an integral by parts. And its remaining integral will ALSO give an integral by parts...so, somewhere, I must have gone wrong. But where would that be?
From the first IBP, we get
$$\tag 1 \displaystyle \int_0^t e^{-2v} \sin v~dv = \dfrac{1}{2}\int_0^t e^{-2v} \cos v~dv -\dfrac{1}{2} e^{-2t} \sin t$$
From the second IBP, we get
$$\tag 2 \displaystyle \int_0^t e^{-2v} \cos v~dv = -\dfrac{1}{4} \int_0^t e^{-2v} \sin v~dv -\dfrac{1}{4}e^{-2t} \cos t + \dfrac{1}{4} $$
From $(1)$ and $(2)$, we now have
$$\displaystyle \int_0^t e^{-2v} \sin v~dv = -\dfrac{1}{4} \int_0^t e^{-2v} \sin v~dv -\dfrac{1}{4}e^{-2t} \cos t + \dfrac{1}{4} -\dfrac{1}{2} e^{-2t} \sin t$$
Moving the integral on the RHS to the LHS, we have
$$\displaystyle \dfrac{5}{4}\int_0^t e^{-2v} \sin v~dv = -\dfrac{1}{4}e^{-2t} \cos t + \dfrac{1}{4} -\dfrac{1}{2} e^{-2t} \sin t$$
Simplifying, we get
$$\displaystyle \int_0^t e^{-2v} \sin v~dv = \dfrac{4}{5}\left(-\dfrac{1}{4}e^{-2t} \cos t + \dfrac{1}{4} -\dfrac{1}{2} e^{-2t} \sin t\right)$$