I have found seccond shifting rules with and without the Heaviside function. Even my professor taught us the one without Heaviside.
For example if $f(t)=2e^-e(t-3)$
1)Rewrite this expression as $2L\{e^{-3t}e^9\}$ and get the answer $\frac{(2e^9)}{(S+3)}$
2)Use the second shifting theorem: $L\{f(t-a)\}=e^{-at}F(S)$ and get the answer $\frac{2e^{-3s}}{S+3}$
As far as i know, every function has only one Laplace transform and vice-versa.
Is the theorem without Heaviside function wrong? Can you explain in simple terms, how does exactly the heaviside function affect the Laplace transform? Why is multiplication with Heaviside function necessary?