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

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

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


Sources