I have invested some time now trying to understand how to use the Second Shift Theorem, mostly by doing the full integration first. What threw me off at first, I discovered, is that almost all books use the same dummy variable. Now I just want to know if the following method is fully correct, or if you would suggests improvements or point out misunderstandings:
$f(t)=\begin{cases}0 & 0 \leq t <2 \\ e^{at} & t \geq2 \end{cases}$
$\implies $
$\mathcal{L}[f(t)]=\mathcal{L}[u_2(t)(e^{a(t-2)})]$
Let $\tau=t-2$ and use Second Shift Theorem:
$\implies$
$=e^{-2s}\mathcal{L}[e^{a \tau}]=e^{-2s}\mathcal{L}[e^{a(t-2)}]=e^{-2s}e^{2a}\mathcal{L}[e^{at}]=e^{-2(s-a)}\frac{1}{s-a}$
I cannot shake off the feeling that I am taking a detour here somewhere.
When I do Laplace transforms, I find it easier and more rewarding to just integrate then to remember all the theorems. Additionally, solving the problem via the definition usually helps me obtain a better understanding of what is actually going on. \begin{align} \mathcal{L}\{f(t)\} & = \int_0^{\infty}f(t)e^{-st}dt\\ &= \int_0^20\cdot e^{-st}dt + \int_2^{\infty}e^{t(a-s)}dt\\ &=\int_2^{\infty}e^{t(a-s)}dt\tag{1}\\ &= -\frac{e^{-t(s-a)}}{s-a}\biggl|_2^{\infty}\tag{2}\\ &= \frac{e^{-2(s-a)}}{s-a} \end{align} In equation $(2)$, I pulled out a negative sign. Since in order for equation $(1)$ to converge, we need $t(a-s)<0\iff a<s$. Therefore, $a-s$ is already a negative quantity so let's write $-t(s-a)$.