Laplace Transform of Constant Mapping/Proof 2

Theorem

Let $a \in \R$ be a real constant.

Let $f_a: \R \to \R$ or $\C$ be the constant mapping, defined as:

$\forall t \in \R: \map {f_a} t = a$

Let $\laptrans {f_a}$ be the Laplace transform of $f_a$.

Then:

$\laptrans {\map {f_a} t} = \dfrac a s$

for $\map \Re s > a$.

$\ds \laptrans {\map {f_a} t}$

$=$

$\ds \laptrans a$

$\ds $

$=$

$\ds a \, \laptrans 1$

$\ds $

$=$

$\ds a \int_0^{\to +\infty} e^{-s t} \rd t$

$\ds $

$=$

$\ds \intlimits {- \frac a s e^{-st} } {t \mathop = 0} {t \mathop \to +\infty}$

$\ds $

$=$

$\ds 0 - \paren {- \frac a s}$

$\ds $

$=$

$\ds \frac a s$

$\blacksquare$