I'm currently trying to find the inverse Laplace transform for:
$$\frac{1}{s + 1}e^{-s}$$
The answer says that the inverse Laplace transform is:
$$\mathscr{L}^{-1}\left( \frac{1}{s + 1}e^{-s} \right) = e^{t - 1}u(t - 1)$$
I'm aware that the Heaviside function's transform is:
$$ \begin{align} \mathscr{L}(u(t - a)) & = \frac{1}{s}e^{-as} \\ \mathscr{L}(f(t - a) u(t - a)) & = e^{-as}F(s) \end{align} $$
but I'm having trouble figuring out how the inverse transform was derived. Any tips are appreciated. Thanks in advance!
Hint: Apply the transform in reverse
$$\mathcal{L}^{-1}\left( e^{-as}F(s) \right) = f(t-a)u(t-a)$$
where $$F(s)=\frac{1}{s+1}$$
therefore
$$f(t)=\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}\left( \frac{1}{s+1}\right)=e^{-t}$$
Can you finish?