Laplace Transform Exists if Function Piecewise Continuous and of Exponential Order

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 exists.

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

Indeed:

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

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