Laplace transform tends to zero at infinity

Remarks

For $\operatorname{Re} s = \sigma \ge \sigma_0$, $\abs{F(s)} \le \int_0^\infty \abs{f(t)} e^{-\sigma t}\, dt$. The integrand decreases to $0$ for each $t > 0$ as $\sigma \to \infty$ and is bounded by $\abs{f(t)} e^{-\sigma_0 t}$, so the right side tends to $0$ by dominated convergence. The hypothesis says that $\sigma_0$ lies to the right of the abscissa of absolute convergence, or on it. Example: $\mathcal{L}\{e^{at}\}(s) = 1/(s - a)$ tends to $0$ as $\operatorname{Re} s \to \infty$. Consequence: functions such as $1$, $s$, $e^{s}$ or a rational function $P/Q$ with $P \ne 0$ and $\deg P \ge \deg Q$ are not Laplace transforms of any such $f$. The constant $1$ is the transform of the Dirac delta, which is a measure and not a function.

Going deeper

For a student

The test is used in partial fraction inversion: only proper rational functions, with numerator degree less than denominator degree, are inverted to ordinary functions. Improper parts correspond to the Dirac delta and its derivatives, which belong to the Laplace transform of distributions. The initial value theorem refines the statement: if $f(t) \to c$ as $t \to 0^+$, then $sF(s) \to c$, so $F(s)$ behaves like $c/s$.