# Laplace Transform Exists if Function Piecewise Continuous and of Exponential Order

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## Contents

## Theorem

Let $f$ be a real function which is:

- piecewise continuous in every closed interval $\closedint 0 N$
- of exponential order $\gamma$ for $t > N$

Then the Laplace transform $\map F s$ of $\map f t$ exists for all $s > \gamma$.

## Proof

For all $N \in \Z_{>0}$:

- $\displaystyle \int_0^\infty e^{-s t} \map f t \rd t = \int_0^N e^{-s t} \map f t \rd t + \int_N^\infty e^{-s t} \map f t \rd t$

We have that $f$ is piecewise continuous in every closed interval $\closedint 0 N$.

Hence the first of the integrals on the right hand side exists.

Also, as $\map f t$ is of exponential order $\gamma$ for $t > N$, so does the second integral on the right hand side.

Indeed:

\(\displaystyle \size {\int_N^\infty e^{-s t} \map f t \rd t}\) | \(\le\) | \(\displaystyle \int_N^\infty \size {e^{-s t} \map f t} \rd t\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \int_0^\infty e^{-s t} \size {\map f t} \rd t\) | |||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \int_0^\infty e^{-s t} M e^{\gamma t} \rd t\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac M {s - \gamma}\) | Laplace Transform of Exponential |

Thus the Laplace transform exists for $s > \gamma$.

$\blacksquare$

## Also see

- Laplace Transform of Reciprocal of Square Root for a real function which is not of exponential order but which does have a Laplace transform

## Sources

- 1965: Murray R. Spiegel:
*Theory and Problems of Laplace Transforms*... (previous) ... (next): Chapter $1$: The Laplace Transform: Sufficient Conditions for Existence of Laplace Transforms: Theorem $1 \text{-} 1$ - 1965: Murray R. Spiegel:
*Theory and Problems of Laplace Transforms*... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Miscellaneous Problems: $47$