Why are the limits of integration set as they are for the Laplace Transform?

For a physicist's or engineer's perspective, let's examine what is probably the most common application of Laplace transforms in the real world. Suppose that $x(t)$, the position of a particle at time $t$ is given by the differential equation, $$ m frac{d^2 x}{dt^2} + gamma frac{dx}{dt} + k x = F(t),$$ where $F(t)$ is the force applied on the particle at time $t$. ($m$, $gamma$ and $k$ represent the particle's mass, the friction and the strength of the restorative force respectively.)

Furthermore, we suppose that when $t < 0$, the force is zero and the system is stationary. The force is only "turned on" at time $t = 0$. Therefore, $x(t)$ and $F(t)$ are non-zero only for $t geq 0$, we only really need to solve the differential equation for $t geq 0$. (The initial conditions at $t = 0$ are $x(0) = 0$ and $frac{dx}{dt}(0) = 0$.)

We can solve this equation using Laplace transforms. Taking Laplace transforms of both sides, the differential equation turns into an algebraic equation which is easier to solve. Once we have solved this algebraic equation, we invert the Laplace transform, giving the final answer, $$ x(t) = mathcal L^{-1} left[ frac{mathcal L[F](s)}{ms^2 + gamma s + k}right].$$

But the thing I want to draw your attention to is the fact that, in the context of this physics problem, the $x(t)$ and the $F(t)$ are only non-zero when $t geq 0$. So when you define their Laplace transforms, it is only natural to integrate over $t geq 0$: $$ mathcal L[x](s) := int_0^infty e^{-ts} x(t) dt, mathcal L[F](s) := int_0^infty e^{-ts} F(t) dt.$$ Integrating over the whole of $mathbb R$ is pointless, seeing that $x(t)$ and $F(t)$ are zero when $t < 0$ anyway! So in the context of this physics problem, it is natural to define the Laplace transforms using integrals over $[0, infty)$.