Laplace Transform of 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 1

\(\ds \laptrans {\map f t}\) \(=\) \(\ds \laptrans 1\) 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

$\blacksquare$


Proof 2

From Laplace Transform of Derivative:

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

under suitable conditions.


Then:

\(\ds \map f t\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} t\) \(=\) \(\ds 0\)
\(\ds \map f 0\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \laptrans 0\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds s \laptrans 1 - 1\) from $(1)$, substituting for $\map f t$ and $\map f 0$
\(\ds \leadsto \ \ \) \(\ds s \laptrans 1\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \laptrans 1\) \(=\) \(\ds \dfrac 1 s\)

$\blacksquare$


Sources

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