Laplace Transform of Exponential/Real Argument/Proof 1
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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
\(\ds \map {\laptrans {e^{a t} } } s\) | \(=\) | \(\ds \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t\) | 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {s - a}\) |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transforms of some Elementary Functions: $1 \ \text{(c)}$