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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.

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2 Answers

The notation is poor, but unfortunately is ingrained in text books.

The Laplace operator $\scr L$ takes a function and returns a function. It would be better to write $\scr L (f)$ (or $\scr L f$) or $\scr L (t \mapsto f(t))$ to denote the $s$-domain function. The value of the transformed function at some $s \in \mathbb{C}$ would be $ (\scr L f) (s)$, or even $\scr L (t \mapsto f(t))(s)$.

Then $\scr L (t \mapsto f(t))$ and $\scr L (x \mapsto f(x))$ are obviously the same functions, just with a different 'dummy' variable representing the input function.

So if we let $\phi(t) = u_c(t)f(t-c)$, what you have shown above is that \begin{eqnarray} (\scr L \phi) (s) = e^{-sc} (\scr L (x \mapsto f(x))) (s) = e^{-sc} (\scr L (t \mapsto f(t))) (s) = e^{-sc} (\scr L f) (s) \end{eqnarray}

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Strictly speaking, the second $t$ they refer to is different than the first $t$, but it doesn't matter. Here's why: when you get to this step $$e^{-sc}\int_0^\infty e^{-sx}f(x)dx,$$ just think of that latter part as $$\int_0^\infty e^{-s\bigstar}f(\bigstar)d\bigstar$$ which is (by definition) the same as $$\mathscr{L}\{f(\bigstar)\},$$ regardless of what $\bigstar$ is (it's just a dummy variable of integration).

For emphasis of what plays the role of the independent variable in the transform domain, we might write $$\mathscr{L}\{f(\bigstar)\}(s)\quad\text{ or }\quad F(s).$$

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