Laplace Transform of Heaviside Step Function times Function

Theorem
Let $f\left({t}\right): \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.

Let $f$ be piecewise continuous on any closed interval of the form $\left[{0 \,.\,.\, b}\right]$, $b > 0$.

Let $\mu_c \left({t}\right)$ be the Heaviside step function.

Let $\mathcal L\left\{{f\left({t}\right)}\right\} = F\left({s}\right)$ be the Laplace transform of $f$.

Then:


 * $\displaystyle \mathcal L \left\{{\mu_c \left({t}\right) \, f \left({t - c}\right)}\right\} = e^{-s c} F \left({s}\right)$

for $\operatorname{Re}\left({s}\right) > a$.

Proof
Let $u = t - c$.

Then $\dfrac {\mathrm d u}{\mathrm d t} = 1$.

Then $u \to 0^-$ as $t \to c^+$.

Also, $u \to +\infty$ as $t \to +\infty$.

So: