# Laplace Transform of Exponential/Real Argument

## 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 1

 $\ds \map {\laptrans {e^{a t} } } s$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^{\to +\infty} e^{\paren {a - s} t} \rd t$ Exponential of Sum $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \int_0^L e^{\paren {a - s} t} \rd t$ Definition of Improper Integral $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \intlimits {\frac 1 {a - s} e^{\paren {a - s} t} } 0 L$ Primitive of Exponential Function, Integration by Substitution $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \frac 1 {a - s} \paren {e^{\paren {a - s} L} - e^{\paren {a - s} 0} }$ $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \frac 1 {s - a} \paren {1 - e^{\paren {a - s} L} }$ Exponential of Zero and rearranging

Because $s > a$, we have that $a - s < 0$.

Hence:

$\ds \lim_{L \mathop \to \infty} \paren {a - s} L \to -\infty$

So:

 $\ds \lim_{L \mathop \to \infty} \frac 1 {s - a} \paren {1 - e^{\paren {a - s} L} }$ $=$ $\ds \frac 1 {s - a} \paren {1 - 0}$ Exponential Tends to Zero $\ds$ $=$ $\ds \frac 1 {s - a}$

$\blacksquare$

## Proof 2

 $\ds \map {\laptrans {e^{a t} } } s$ $=$ $\ds \map {\laptrans {1 \times e^{a t} } } s$ $\ds$ $=$ $\ds \map {\laptrans 1} {s - a}$ Laplace Transform of Exponential times Function $\ds$ $=$ $\ds \frac 1 {s - a}$ Laplace Transform of Constant Mapping

$\blacksquare$

## Proof 3

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

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

$\blacksquare$