Laplace Transform of Heaviside Step Function times Function

Theorem
Let $\map f t: \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.

Let $f$ be piecewise continuous with one-sided limits on any closed interval of the form $\closedint 0 b$ where $b > 0$.

Let $\map {u_c} t$ be the Heaviside step function.

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.

Then:


 * $\laptrans {\map {u_c} t \, \map f {t - c} } = e^{-s c} \map F s$

for $\map \Re s > a$.

Proof
Let $u = t - c$.

Then $\dfrac {\rd u} {\rd t} = 1$.

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

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

So: