I have to find the Laplace inverse of the following function:
$$\mathcal{L} ^ {-1} \left\{ (s+1)^2 \right\} = f(t)$$
So, far, I have expanded the term and while I understand that $\mathcal{L}^{-1}\left\{1\right\}$ is $\delta(t)$. I am however not able to find the laplace inverse of the other two terms. I did think they can be the derivative of $\delta(t)$ and Wolfram alpha also confirmed my suspicions. However, if I use basic Laplace transform rules, then I get a term of $\delta(0)$ if I try to find out the laplace of derivative of $\delta(t)$ which I am not sure as to how to eliminate.
The integral of the delta function from zero to infinity is undefined. Thus the usual definition of the Laplace transform for ordinary functions doesn't work for distributions.
Instead, for distributions essentially we are computing $$\mathcal L[f] = \int_{-\infty}^\infty f(t) e^{-p t} dt,$$ but we're working only with distributions $f$ whose support is contained in $[0, \infty)$. From which we obtain $$\mathcal L[f'] = (f', e^{-p t}) = - (f, (e^{-p t})') = p (f, e^{-p t}) = p \mathcal L[f].$$ This holds provided we correctly compute $1' = \delta$ or $(\cos t)' = -\sin t + \delta(t)$, because we're in fact working with $1 \cdot H(t)$ and $\cos t \cdot H(t)$. It follows that $\mathcal L[\delta'] = p$.
The Laplace transform can also be defined in terms of the Fourier transform as $$\mathcal L_{t \to p}[f] = \mathcal F_{t \to \omega}[f(t) e^{-\sigma t}], \\ p = \sigma + i \omega, \sigma > 0.$$