Laplace Transform of 1/Proof 2

Theorem

Let $f: \R \to \R$ be the function defined as:

$\forall t \in \R: \map f t = 1$

Then the Laplace transform of $\map f t$ is given by:

$\laptrans {\map f t} = \dfrac 1 s$

for $\map \Re s > 0$.

$(1): \quad \laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$

under suitable conditions.

Then:

$\ds \map f t$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds \map {f'} t$

$=$

$\ds 0$

$\ds \map f 0$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds \laptrans 0$

$=$

$\ds 0$

$\ds $

$=$

$\ds s \laptrans 1 - 1$

from $(1)$, substituting for $\map f t$ and $\map f 0$

$\ds \leadsto \ \ $

$\ds s \laptrans 1$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds \laptrans 1$

$=$

$\ds \dfrac 1 s$

$\blacksquare$

1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $15 \ \text{(a)}$