The Heaviside Step Function
Background
The Heaviside step function, also called the unit step function, is defined as follows
\[\theta(t) = \begin{cases} 1 &\quad\text{if } t > 0 \\ 0 &\quad\text{if } t \leq 0 \end{cases} \qquad\qquad \theta(t-c) = \begin{cases} 1 &\quad\text{if } t > c \\ 0 &\quad\text{if } t \leq c \end{cases}\]The Laplace Transform Proof
Given a function $f(t)$ with Laplace Transform $F(s)$ and any constant $c \in \mathbb{R}$
\[\begin{align} \mathcal{L}\{ \theta(t-c) f(t-c) \} &= \int_{0}^{\infty} \theta(t-c) f(t-c) e^{-st} \ dt \\[10pt] &= \int_{0}^{c} 0 \cdot f(t-c) e^{-st} \ dt + \int_{c}^{\infty} f(t-c) e^{-st} \ dt \\[10pt] &= \int_{c}^{\infty} f(t-c) e^{-st} \ dt \\[10pt] &\text{let } u = t-c \implies du = dt \\[10pt] &= \int_{0}^{\infty} f(u) e^{-s(u+c)} \ du \\[10pt] &= e^{-cs} \int_{0}^{\infty} f(u) e^{-su} \ du \\[10pt] &= e^{-cs} F(s) \end{align}\]This turns out to be very useful because so long as we don’t care about the negative values of $f(t)$, translation is very simple. Since in applications, $t$ represents time, this is not an outlandish assumption.