Laplace Transform Exists if Function Piecewise Continuous and of Exponential Order
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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 $N \in \Z_{>0}$:
- $\ds \int_0^\infty e^{-s t} \map f t \rd t = \int_0^N e^{-s t} \map f t \rd t + \int_N^\infty e^{-s t} \map f t \rd t$
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:
\(\ds \size {\int_N^\infty e^{-s t} \map f t \rd t}\) | \(\le\) | \(\ds \int_N^\infty \size {e^{-s t} \map f t} \rd t\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_0^\infty e^{-s t} \size {\map f t} \rd t\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_0^\infty e^{-s t} M e^{\gamma t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac M {s - \gamma}\) | 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: $47$