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Existence of the Laplace Transform

A function $ x(t)$ has a Laplace transform whenever it is of exponential order. That is, there must be a real number $ B$ such that

$\displaystyle \lim_{t\to\infty} \left\vert x(t)e^{-Bt}\right\vert=0 $

As an example, every exponential function $ Ae^{\alpha t}$ has a Laplace transform for all finite values of $ A$ and $ \alpha$. Let's look at this case more closely. The Laplace transform of a causal, growing exponential function

$\displaystyle x(t) = \left\{\begin{array}{ll} A e^{\alpha t}, & t\geq 0 \\ [5pt] 0, & t<0 \\ \end{array}\right., $

is given by
\begin{eqnarray*} X(s) &\isdef & \int_0^\infty x(t) e^{-st}dt = \int_0^\infty A... ...alpha \\ [5pt] \infty, & \sigma<\alpha \\ \end{array} \right. \end{eqnarray*}
Thus, the Laplace transform of an exponential $ Ae^{\alpha t}$ is $ A/(s-\alpha)$, but this is defined only for re$ \left\{s\right\}>\alpha$.
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