Function with no Laplace transform

Counterexample

The function $f(t) = e^{t^2}$ on $[0, \infty)$ is continuous, but its Laplace integral $\int_0^\infty e^{t^2} e^{-st}\, dt$ converges for no $s \in \C$. Check: suppose it converged at some $s_0 \in \C$. By the half-plane theorem it would then converge at every real $s > \operatorname{Re} s_0$. But for real $s$ and $t \ge \max(s, 0)$, the integrand satisfies $e^{t^2 - st} = e^{t(t - s)} \ge 1$, so $\int_0^T e^{t^2 - st}\, dt \to \infty$ as $T \to \infty$. This is a contradiction, so the abscissa of convergence of $f$ is $+\infty$. The function is not of exponential order: $e^{t^2} / e^{at} \to \infty$ for every real $a$.