My introductory Laplace transform textbook says the following:
The improper integral is defined in the obvious way by taking the limit:
$$\lim_{R \to \infty} \int_a^R F(x) \ dx = \int_a^\infty F(x) \ dx$$
provided $F(x)$ is continuous in the interval $a \le x \le R$ for every $R$, and the limit on the left exists. The parameter $x$ is defined to take the increasing values from $a$ to $\infty$. The lower limit $a$ is normally $0$ in the context of Laplace transforms. The condition $|F(x)| \le Me^{\alpha x}$ is termed "$F(x)$ is of exponential order" and is, speaking loosely, quite a weak condition. All polynomial functions and (of course) exponential functions of the type $e^{kx}$ ($k$ constant) are included as well as bounded functions. Excluded functions are those that have singularities such as $\ln(x)$ or $1/(x - 1)$ and functions that have a growth rate more rapid than exponential, for example $e^{x^2}$.
(Bolding added by me for emphasis.)
My understanding is that the bolded part is false -- that is, $\ln(x)$ does have a Laplace transform:
$$\mathcal{L}\{ \ln(t) \} = - \dfrac{1}{s} [ \ln(s) + \gamma],$$
where $\gamma$ is the Euler-Mascheroni constant.
Did the author make an error here?
I would appreciate it if people could please take the time to clarify this.