Why does the Laplace transform of $t^2 \exp(at)$ exist? My book states a theorem :
"Let $f(t)$ be a function piecewise continuous on $[0, A]$ for $ A > 0$ and have an exponential order at infinity with $|f(t)| \leq M \exp(at)$. Then the Laplace transform is defined for $s > a$." 
But the Laplace transforms of $t \exp(at)$ and $t^n \exp(at)$ are defined and found in most tables of Laplace transforms. How can this be when they both grow faster than $M\exp(at)$ for $t > 0$, which means they are not of exponential order.
One of the conditions for existence of the transform is that $f(t)$ be of exponential order, but this is clearly not the case.
What am I overlooking here? Both my intuition and the theorem seem to be saying that these Laplace transforms should not be defined.
 A: Let's show that in general, for $a \in \mathbf{C}$ and a polynomial $p$, that $p(t) \mathrm{e}^{at}$ is of exponential type $\Re(a) + \varepsilon$ for any $\varepsilon > 0$. Recall that a function $f\colon [0, \infty) \to \mathbf{C}$ is of exponential type $A$ if there exists a constant $C$ such that
$$|f(t)| \le C\mathrm{e}^{At}$$
for all $t \ge 0$. Suppose $\varepsilon > 0$. By l'Hopital's rule,
$$\lim_{t\to\infty}p(t)/\mathrm{e}^{\varepsilon t} = 0.$$
This means there exists a number $T$ such that $|p(t)/\mathrm{e}^{\varepsilon t}| < 1$ if $t > T$. Now, the function $t\mapsto p(t)/\mathrm{e}^{\varepsilon t}$ is continuous on the compact interval $[0, T]$, and is hence bounded there. That is, there exists $B$ such that $|p(t)/\mathrm{e}^{\varepsilon t}| \le B$ for all $t \in [0, R]$. Thus, we may say that $|p(t)/\mathrm{e}^{\varepsilon t}| \le B + 1$ for all $t \ge 0$, and so 
$$|p(t)| \le (B + 1)\mathrm{e}^{\varepsilon t} \quad \text{ for all $t\ge0$}.$$
Hence, $|p(t)\mathrm{e}^{at}| \le (B + 1)\mathrm{e}^{\varepsilon t}|\mathrm{e}^{at}| = (\underbrace{B + 1}_{C})\mathrm{e}^{(\overbrace{\Re(a) + \varepsilon}^{A})t}$ for all $t \ge 0$, as required.
