Laplace transform of the Heaviside step function

Example

Let $H$ be the Heaviside step function, with $H(t) = 0$ for $t < 0$ and $H(t) = 1$ for $t > 0$, and let $a \ge 0$. The delayed step $f(t) = H(t - a)$ is $0$ on $[0, a)$ and $1$ for $t > a$. Its Laplace transform is $$\mathcal{L}\{H(t - a)\}(s) = \frac{e^{-as}}{s} \quad (\operatorname{Re} s > 0).$$ Check: for $\operatorname{Re} s > 0$, $$\int_0^\infty H(t - a) e^{-st}\, dt = \int_a^\infty e^{-st}\, dt = \lim_{T \to \infty} \frac{e^{-as} - e^{-sT}}{s} = \frac{e^{-as}}{s}.$$ The value of $H$ at the single point $0$ does not change the integral. For $a = 0$ this is $\mathcal{L}\{1\} = 1/s$. A rectangular pulse $H(t - a) - H(t - b)$ with $0 \le a < b$ has transform $(e^{-as} - e^{-bs})/s$, by linearity.