Laplace Transform of 1/Proof 1
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Theorem
Let $f: \R \to \R$ be the function defined as:
- $\forall t \in \R: \map f t = 1$
Then the Laplace transform of $\map f t$ is given by:
- $\laptrans {\map f t} = \dfrac 1 s$
for $\map \Re s > 0$.
Proof
\(\ds \laptrans {\map f t}\) | \(=\) | \(\ds \laptrans 1\) | Definition of $\map f t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^{\to +\infty} e^{-s t} \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \rd t\) | Definition of Improper Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{L \mathop \to \infty} \intlimits {-\frac 1 s e^{-s t} } 0 L\) | Primitive of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{L \mathop \to \infty} \paren {-\frac 1 s e^{-s L} - \paren {-\frac 1 s} }\) | Exponential of Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{L \mathop \to \infty} \paren {\frac {1 - e^{-s L} } s}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 s\) | Complex Exponential Tends to Zero |
$\blacksquare$
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: $1 \ \text{(a)}$
- For a video presentation of the contents of this page, visit the Khan Academy.