Integral transforms, operational calculus Laplace transform: operational rules Proposition

Laplace transform of an integral

Builds on Laplace transform Function of exponential order Laplace transform of a derivative

Statement

Let $f \colon [0, \infty) \to \C$ be piecewise continuous and of exponential order $a$ , that is, $\abs{f(t)} \le M e^{a t}$ for all $t \ge 0$ and some constants $M$ and $a$ . Let $F$ be the Laplace transform of $f$ , and let $g(t) = \int_0^t f(\tau) \, d\tau$ . Then for every complex $s$ with $\operatorname{Re} s > \max(a, 0)$ , $\mathcal{L}\{g\}(s) = \frac{F(s)}{s}.$

Remarks

The function $g$ is continuous, $g(0) = 0$ , and $g' = f$ at each point where $f$ is continuous. The bound on $f$ gives a bound $\abs{g(t)} \le M' e^{b t}$ for every $b > \max(a, 0)$ , so $\mathcal{L}\{g\}$ exists on $\operatorname{Re} s > \max(a, 0)$ . The rule for derivatives then gives $F(s) = s \, \mathcal{L}\{g\}(s) - g(0) = s \, \mathcal{L}\{g\}(s)$ . Example: with $f = 1$ , one gets $\mathcal{L}\{t\}(s) = 1 / s^2$ ; repeating gives $\mathcal{L}\{t^n / n!\}(s) = 1 / s^{n + 1}$ .

Going deeper

For a student

Read backwards, the rule inverts transforms of the form $F(s) / s$ : the answer is the running integral of $\mathcal{L}^{-1}\{F\}$ . For example $1 / (s (s^2 + 1))$ inverts to $\int_0^t \sin \tau \, d\tau = 1 - \cos t$ . It is a special case of the convolution theorem, since $g = f * 1$ and $\mathcal{L}\{1\} = 1/s$ .

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Function concept
Function of exponential order concept
Laplace transform definition
Laplace transform of a derivative statement
Laplace transform of an integral statement

Tag	`VGYXWXV5M4`
URI	`m2://VGYXWXV5M4`
Kind	statement, proposition
Writing	W0 stub
MSC	44A10
Status	This copy of the site has no build data. Run `m2 status VGYXWXV5M4`.
About	Laplace transform
Uses	Laplace transform, Function of exponential order, Laplace transform of a derivative