Laplace Transform of Function of t minus a/Proof 2

Theorem

Let $f$ be a function such that $\laptrans f$ exists.

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

Let $a \in \C$ or $\R$ be constant.

Let $g$ be the function defined as:

$\map g t = \begin{cases} \map f {t - a} & : t > a \\ 0 & : t \le a \end{cases}$

Then:

$\laptrans {\map g t} = e^{-a s} \map F s$

Proof

 $\ds \laptrans {\map g t}$ $=$ $\ds \int_0^\infty e^{-s t} \map g t \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^a e^{-s t} \map g t \rd t + \int_a^\infty e^{-s t} \map g t \rd t$ $\ds$ $=$ $\ds \int_0^a 0 \times e^{-s t} \rd t + \int_a^\infty e^{-s t} \map f {t - a} \rd t$ Definition of $\map g t$ $\ds$ $=$ $\ds \int_a^\infty e^{-s t} \map f {t - a} \rd t$ $\ds$ $=$ $\ds \int_0^\infty e^{-s \paren {u + a} } \map f u \rd u$ Integration by Substitution: $t = u + a$ $\ds$ $=$ $\ds e^{-a s} \int_0^\infty e^{-s u} \map f u \rd u$ $\ds$ $=$ $\ds e^{-a s} \map F s$ Definition of Laplace Transform

$\blacksquare$