# Laplace Transform of Exponential/Real Argument/Proof 1

## 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

 $\ds \map {\laptrans {e^{a t} } } s$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t$ 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 $\ds$ $=$ $\ds \frac 1 {s - a}$

$\blacksquare$