I'm just now learning how to take the Laplace of a simple step function, but I have a question about the terms. I'll show my work so far and hopefully someone can step in and answer the question I pose at the end.
$\int_0^\infty u_c(t)f(t-c)e^{-st}dt$
$u_c$ is the unit step function.
Up until the point $c$, this function will evaluate to zero. So, I'll get the same value if I change my lower limit to $c$.
$\int_0^\infty u_c(t)f(t-c)e^{-st}dt = \int_c^\infty f(t-c)e^{-st}dt$
That takes care of the unit step function. Now, a substitution and a rewrite.
$x=t-c, dx=dt, t=x+c$
$\int_0^\infty f(x)e^{-s(x+c)}dx$
$=e^{-sc}\int_0^\infty e^{-sx}f(x)dx$
Now, I can see that the integral here is the Laplace of $f(x)$ but with an $x$ where a $t$ usually is.
Here's where my confusion is. The book then says that I can go ahead and switch $x$ for $t$ and say that the Laplace transform of my step function is
$=e^{-sc}\scr L\{f(t)\}$
I understand that the choice of symbol is arbitrary. However, $t$ still has a context in this problem, and it's equal to $x+c$. So, is it really correct to use $t$ again like this?
I mean, saying that
$\scr L\{f(t)\}=\scr L\{f(x)\}$ is fine if $x=t$. But $x=t-c$. So I'm confused.
Is it just a matter of assuming that $c$ is relatively small? That seems a little too convenient.