What is the inverse laplace transform of $\frac{s}{s+1}$?
My work was:
$$
X(s)=\frac{s}{s+1}\\
X(s)=s\frac{1}{s+1}\\
x(t)=\frac{d}{dt}e^{-t}=-e^{-t}
$$
My only issue is that when I check my answer with wolfram alpha, it says that the inverse laplace transform of $\frac{s}{s+1}$ is actually $-e^{-t}+\delta(t)$. What is the correct way to find the inverse transform?
Best Answer
HINT: $$\frac s{s+1}=1-\frac1{s+1}$$
We know, $\mathcal{L}(e^{at})=\frac1{s-a}$
and we can prove $\mathcal{L}(\delta(t-b))=e^{-bs}$, put $b=0$