I am currently studying the textbook An Introduction to Laplace Transforms and Fourier Series, second edition, by Phil Dyke. Chapter 2.1 Real Functions describes Heaviside's unit step function as follows:
Sometimes, a function $F(t)$ represents a natural or engineering process that has no obvious starting value. Statisticians call this a time series. Although we shall not be considering $F(t)$ as stochastic, it is nevertheless worth introducing a way of "switching on" a function. Let us start by finding the Laplace transform of a step function the name of which pays homage to the pioneering electrical engineer Oliver Heaviside (1850 - 1925). The formal definition runs as follows.
Definition 2.1 Heaviside's unit step function, or simply the unit step function, is defined as
$$H(t) = \begin{cases} 0 & t < 0, \\ 1 & t \ge 0. \end{cases}$$
Since $H(t)$ is precisely the same as $1$ for $t > 0$, the Laplace transform of $H(t)$ must be the same as the Laplace transform of $1$, i.e., $1/s$. The switching on of an arbitrary function is achieved simply by multiplying it by the standard function $H(t)$, so if $F(t)$ is given by the function shown in Fig. 2.1 and we multiply this function by the Heaviside unit step function $H(t)$ to obtain $H(t)F(t)$, Fig 2.2 results. Sometimes it is necessary to define what is called the two sided Laplace transform
$$\int_{-\infty}^\infty e^{-st} F(t) \ dt,$$
which makes a great deal of mathematical sense. However, the additional problems that arise by allowing negative values of $t$ are severe and limit the use of the two sided Laplace transform. For this reason, the two sided transform will not be pursued here.
What are these "severe" problems that arise for the two-sided Laplace transform?
I would greatly appreciate it if people would please take the time to explain this.