Is $ℒ\{ t·γ(t-1) \} = ℒ\{ (t-1)·γ(t-1)+γ(t-1)\}$ ? I'm confused if these two are the same or if I can even write the second one like this .

Question

The step function $γ(t-1)$ is $0$ for all values less then $1$. Thus, my reasoning is that multiplying $(t-1)$ by $γ(t-1)$ and adding $γ(t-1)$ should be the same to the original $(ℒ\{ t·γ(t-1) \})$. In addition, I was wondering is there is any property for transforms of the form $ℒ\{ f(t)·γ(t-a) \}$? If so, how can we use these? I know that [ $ℒ\{ f(t-a)·γ(t-a) \}$ = $e^{-as}F(s)$ ], but I do not think this will be helpful for transforms in the form $ℒ\{ f(t)·γ(t-a) \}$. I just started learning Laplace transforms a day ago and my professor has not provided us anything in regards to this.