The Laplace Transform: Definition and Existence

The Laplace transform is the standard method for solving linear constant-coefficient ODEs with discontinuous or impulsive forcing — switched circuits, hammer blows, step inputs — because it converts differentiation into multiplication by $s$, turning an ODE into an algebraic equation (foreshadowing Transforms of Derivatives). Once solved algebraically, recovering the time-domain solution uses Inverse Laplace Transform and Partial Fractions. In electrical engineering, transfer functions and circuit impedances are phrased directly in the $s$-domain. In control theory, stability of a feedback system is read off from the locations of poles of $F(s)$ relative to the imaginary axis, a direct descendant of the exponential-order threshold $s>c$ proven here. In probability theory, the closely related Laplace transform of a density function generalizes the moment generating function and underlies large-deviation theory. In numerical analysis and applied PDE (heat and wave equations), Laplace transforms in time convert PDEs into ODEs in a spatial variable, solvable classically. Understanding exactly when $F(s)$ exists, as proven rigorously here rather than assumed, is what protects all of these downstream applications from silently manipulating a divergent integral, and it is the same convergence machinery — comparison tests for improper integrals building on Improper integrals — used to justify related transforms such as the Fourier and Mellin transforms elsewhere in analysis.

Ask the tutor: How do I find the exponential order constants $M$, $c$, $T$ for a given piecewise function, and use them to determine the exact domain of convergence of its Laplace transform?

Think of $e^{-st}$ as a dial that squashes the graph of $f(t)$ toward zero as $t$ grows, with the squashing rate controlled by $s$. If $f$ itself grows no faster than some exponential $Me^{ct}$, then choosing $s>c$ makes the product $e^{-st}f(t)$ decay like $e^{-(s-c)t}$, and the "area under the curve" from $0$ to $\infty$ is finite. The transform $F(s)$ is literally that total signed area, viewed as a function of the dial setting $s$. The picture below shows a function $f(t)=e^{2t}$ (exponential order $c=2$) alongside the damping factor $e^{-st}$ for $s=5$. Their product decays, and the shaded region under the product curve is the value of $L\{f\}(5)$.

The reason exponential order matters is that not every function is this well-behaved: a function like $e^{t^2}$ outgrows every exponential $e^{ct}$ no matter how large $c$ is, so no choice of damping rate $s$ can rescue convergence — its transform simply does not exist. Exponential order is precisely the borderline condition that separates "transformable" from "not transformable," and the existence theorem below makes this precise with a genuine convergence proof, not just a heuristic.

Example 1: A piecewise-defined forcing function. Let

This $f$ has a jump discontinuity at $t=2$ (since $\lim_{t\to 2^-}f(t)=6$ but $\lim_{t\to2^+}f(t)=5e^8\approx 14904.8$), but it is piecewise continuous on every $[0,A]$: on $[0,2)$ it's the polynomial $3t$, on $[2,\infty)$ it's the continuous $5e^{4t}$, and both one-sided limits at $t=2$ are finite. For $t\ge 2$, $|f(t)| = 5e^{4t} \le 5e^{4t}$, so $f$ is of exponential order $c=4$ with $M=5$, $T=2$ (and on $[0,2)$, $3t \le 6 \le 5e^{4t}$ trivially since $5e^{4t}\ge 5$). By the existence theorem, $F(s)=L\{f\}(s)$ exists for all $s>4$. We compute it directly, splitting at $t=2$:

For the first integral, integrate by parts with $u=3t$, $dv=e^{-st}dt$, so $du=3\,dt$, $v=-\frac{1}{s}e^{-st}$:

For the second integral, with $s>4$ so that $s-4>0$:

Combining,

As a check, note every term decays appropriately as $s\to\infty$ (each has a factor of $e^{-2s}$ or is $O(1/s^2)$ or $O(e^{-2s}/(s-4))$), consistent with $F(s)\to 0$, as the theorem predicts.

Example 2: A transform that fails to converge below its exponential-order threshold. Let $f(t) = 7e^{3t}\cos(t)$. Since $|\cos t|\le 1$, we have $|f(t)| \le 7e^{3t}$ for all $t\ge 0$, so $f$ has exponential order $c=3$ with $M=7$, $T=0$. Also $f$ is continuous (hence piecewise continuous) on every $[0,A]$. The theorem guarantees $F(s)$ exists for $s>3$. Compute directly using the standard formula $\int_0^\infty e^{-at}\cos(bt)\,dt = \frac{a}{a^2+b^2}$ for $a>0$ (itself derivable by two integrations by parts, or by taking the real part of $\int_0^\infty e^{-(a-ib)t}dt = \frac{1}{a-ib}$ when $a>0$):

This requires $s-3>0$, i.e., $s>3$, exactly matching the threshold from exponential order. Now check $s=3$ directly: the integral becomes $7\int_0^\infty \cos t\,dt$, whose partial integrals $7\sin A$ oscillate between $-7$ and $7$ forever and never settle to a limit as $A\to\infty$ — so $F(3)$ does not exist. This shows the threshold $s>c$ in the theorem is generally sharp: it is not merely a convenient sufficient bound but often the actual boundary of convergence, and the theorem correctly makes no claim about $s\le c$.