Laplace transforms - second shift theorem

Question

Hi, i'm struggling with understanding how this expansion came about for determining the LT of f(t). Is anyone familiar with this? thank you

Answer 1

If I've understood your comment correctly, then I think I see the confusion.

Recall that the second shifting theorem says that if $\mathcal{L}\{ f(t)\} = F(s)$ then $\mathcal{L}\{ f(t-a)u(t-a)\} = e^{-as}F(s)$

Now, let's dissect taking the Laplace transform of $\frac{1}{2}t^2u(t-1)$. Note that our current function is $f(t) = \frac{1}{2}t^2$. So then $f(t-1) = \frac{1}{2}(t-1)^2$. As such, to get $f(t)$ in the desired form, we will conveniently add $0$ in a number of different ways: $$f(t)=f(t) + (t-t) + (\frac{1}{2}-\frac{1}{2}) = \frac{1}{2}(t^2-2t+1) + t - \frac{1}{2} = f(t-1) + t-\frac{1}{2} + (\frac{1}{2}-\frac{1}{2}) $$ $$= f(t-1) + (t-1) + \frac{1}{2} $$ Now we have $f(t)$ in the desired form to apply the shifting theorem.