Say $f$ is a function $\mathbb R_{>0}\to \mathbb C$ and $a > 0$.
Say I want to calculate the Laplace transform of $f(x + a)$.
This shifts $f$ to the left so that $f(x+a)$ is no longer zero on the negative numbers thereby violating one of the conditions necessary to calculate the Laplace transform of it.
My idea on how to work around this problem is: Multiply $f(x+a)$ with the Heaviside function.
My questions are:
Is this really a possible work around or am I just making a bad mistake? It seems that $f(x)$ and $f(x+a)$ times Heaviside are no longer the same functions.
If multiplying with the Heaviside function is wrong, what is the correct way to argue that one can nevertheless calculate the Laplace transform of $f(x+a)$?
Not being zero on the negative numbers doesn't inherently matter, the Laplace transform restricts attention to $[0,\infty)$ by itself anyway. The problem is that some of the information about $f$ on $[0,\infty)$ has been "lost" to $[-a,0]$ and so the Laplace transform of $f(x+a)$ has no generic relationship to that of $f$. This is in contrast to $f(x-a) H(x-a)$, which has the entire graph of $f$ on $[0,\infty)$ and only the graph of $f$ on $[0,\infty)$, it has just been transported over to $[a,\infty)$.