Laplace Transform of Exponential/Real Argument

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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

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


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

Hence:

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


So:

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Proof 3

From Laplace Transform of Derivative:

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

under suitable conditions.


Then:

\(\displaystyle \map f t\) \(=\) \(\displaystyle e^{a t}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {f'} t\) \(=\) \(\displaystyle a e^{a t}\)
\(\displaystyle \map f 0\) \(=\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \laptrans {a e^{a t} }\) \(=\) \(\displaystyle s \laptrans {e^{a t} } - 1\) from $(1)$, substituting for $\map f t$ and $\map f 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \laptrans {e^{a t} }\) \(=\) \(\displaystyle s \laptrans {e^{a t} } - 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \laptrans {e^{a t} }\) \(=\) \(\displaystyle \dfrac 1 {s - a}\) rearranging

$\blacksquare$


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