Laplace Transform of Derivative DeltaDirac(t-1)

Question

$\mathscr{L}\{\delta'(t-1)\}=F(s) $

Im having a hard time thinking this.

My first thought was doing $F(s) =e^{-s} s$

But when i do the inverse laplace transform of this i get

$f(t)= \delta'(t-1) Heaviside(t-1)$

Answer 1

I also think that the Laplace transform of $\delta'(t-1)$ is $s e^{-s}.$ Formally we have: $$ \mathscr{L}\{\delta'(t-1)\} = \int_0^\infty \delta'(t-1) \, e^{-st} \, dt \\ = \underbrace{\left[ \delta(t-1) \, e^{-st} \right]_0^\infty}_{=0} - \int_0^\infty \delta(t-1) \, (-s) e^{-st} \, dt \\ = s \int_0^\infty \delta(t-1) \, e^{-st} \, dt = s e^{-s\cdot 1} = s e^{-s} $$