Laplace Transform of Exponential/Real Argument/Proof 2

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 $\map \Re s > \map \Re a$.

Proof

\(\ds \map {\laptrans {e^{a t} } } s\)

\(=\)

\(\ds \map {\laptrans {1 \times e^{a t} } } s\)

\(\ds \)

\(=\)

\(\ds \map {\laptrans 1} {s - a}\)

Laplace Transform of Exponential times Function

\(\ds \)

\(=\)

\(\ds \frac 1 {s - a}\)

Laplace Transform of Constant Mapping

$\blacksquare$