Laplace Transform of Exponential/Real Argument/Proof 3

Theorem

Let $\laptrans f$ denote the Laplace transform of a function $f$.

Let $e^x$ be the real exponential.

Then:

$\map {\laptrans {e^{a t} } } s = \dfrac 1 {s - a}$

where $a \in \R$ is constant, and $\map \Re s > \map \Re a$.

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

under suitable conditions.

Then:

$\ds \map f t$

$=$

$\ds e^{a t}$

$\ds \leadsto \ \ $

$\ds \map {f'} t$

$=$

$\ds a e^{a t}$

$\ds \map f 0$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds \laptrans {a e^{a t} }$

$=$

$\ds s \laptrans {e^{a t} } - 1$

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

$\ds \leadsto \ \ $

$\ds a \laptrans {e^{a t} }$

$=$

$\ds s \laptrans {e^{a t} } - 1$

$\ds \leadsto \ \ $

$\ds \laptrans {e^{a t} }$

$=$

$\ds \dfrac 1 {s - a}$

rearranging

$\blacksquare$

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