# Laplace Transform of 1

## 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

 $\displaystyle \laptrans {\map f t}$ $=$ $\displaystyle \laptrans 1$ Definition of $\map f t$ $\displaystyle$ $=$ $\displaystyle \int_0^{\to +\infty} e^{-s t} \rd t$ Definition of Laplace Transform $\displaystyle$ $=$ $\displaystyle \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \rd t$ Definition of Improper Integral $\displaystyle$ $=$ $\displaystyle \lim_{L \mathop \to \infty} \intlimits {-\frac 1 s e^{-s t} } 0 L$ Primitive of Exponential Function $\displaystyle$ $=$ $\displaystyle \lim_{L \mathop \to \infty} \paren {-\frac 1 s e^{-s L} - \paren {-\frac 1 s} }$ Exponential of Zero $\displaystyle$ $=$ $\displaystyle \lim_{L \mathop \to \infty} \paren {\frac {1 - e^{-s L} } s}$ simplification $\displaystyle$ $=$ $\displaystyle \frac 1 s$ Complex Exponential Tends to Zero

$\blacksquare$

## Proof 2

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

under suitable conditions.

Then:

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

$\blacksquare$