# Laplace Transform of Heaviside Step Function

## Theorem

Let $\map {u_c} t$ denote the Heaviside step function:

$\map {u_c} t = \begin{cases} 1 & : t > c \\ 0 & : t < c \end{cases}$

The Laplace transform of $\map {u_c} t$ is given by:

$\laptrans {\map {u_c} t} = \dfrac {e^{-s c} } s$

for $\map \Re s > c$.

## Proof 1

 $\ds \laptrans {\map {u_c} t}$ $=$ $\ds \int_0^{\to +\infty} \map {u_c} t e^{-s t} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^c \map {u_c} t e^{-s t} \rd t + \int_c^{\to +\infty} \map {u_c} t e^{-s t} \rd t$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\ds$ $=$ $\ds \int_0^c 0 \times e^{-s t} \rd t + \int_c^{\to +\infty} 1 \times e^{-s t} \rd t$ Definition of Heaviside Step Function $\ds$ $=$ $\ds \int_c^{\to +\infty} e^{-s t} \rd t$ $\ds$ $=$ $\ds \lim_{L \mathop \to +\infty} \int_c^L e^{-s t} \rd t$ Definition of Improper Integral $\ds$ $=$ $\ds \lim_{L \mathop \to +\infty} \intlimits {\dfrac {e^{-s t} } {-s} } c L$ Primitive of $e^{a x}$ $\ds$ $=$ $\ds \lim_{L \mathop \to +\infty} \dfrac {e^{-s L} } {-s} - \dfrac {e^{-s c} } {-s}$ $\ds$ $=$ $\ds 0 + \dfrac {e^{-s c} } s$ simplification

Hence the result.

$\blacksquare$

## Proof 2

 $\ds \laptrans 1$ $=$ $\ds \dfrac 1 s$ Laplace Transform of 1 $\ds \leadsto \ \$ $\ds \laptrans {1 \times \map {u_c} t}$ $=$ $\ds \dfrac 1 s \times e^{-c s}$ Laplace Transform of Function of t minus a $\ds$ $=$ $\ds \dfrac {e^{-s c} } s$ simplification

Hence the result.

$\blacksquare$