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

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

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

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

is given by

$$\begin{eqnarray*} X(s) &\isdef & \int_0^\infty x(t) e^{-st}dt = \int_0^\infty A e^{\alpha t} e^{-st}dt = A \int_0^\infty e^{(\alpha - s) t} dt \\ &=& \left.\frac{A}{\alpha-s}e^{(\alpha - s) t} \right\vert _{0}^{\infty} = \frac{A}{\alpha-s}e^{(\alpha - \sigma - j\omega) \infty} - \frac{A}{\alpha-s}\\ &=& \left\{\begin{array}{ll} \frac{A}{s-\alpha}, & \sigma>\alpha \\ [5pt] \hbox{(indeterminate)}, & \sigma=\alpha \\ [5pt] \infty, & \sigma<\alpha \\ \end{array} \right. \end{eqnarray*}$$

Thus, the Laplace transform of an exponential $Ae^{\alpha t}$ is $A/(s-\alpha)$ , but this is defined only for re$\left\{s\right\}>\alpha$ .