Laplace Symbol and Definition

The script letter $\mathcal{L}$ is the Laplace transform operator. It takes a function of time $t$ and returns a function of the complex variable $s$:

$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t)\,dt$

Read $\mathcal{L}\{f(t)\}$ as "the Laplace transform of $f$ of $t$". The braces are function-application brackets, not a set. By convention the time-domain function is lower case and its transform upper case, so $y(t) \leftrightarrow Y(s)$.

How to write it: in LaTeX, \mathcal{L}; in Unicode, the script capital ℒ (U+2112); by hand, a looped script L. The inverse operator is $\mathcal{L}^{-1}\{F(s)\} = f(t)$.

The lower limit is $0$, so the transform only ever sees $t \ge 0$.

When the integral converges

The improper integral converges if $f$ is piecewise continuous on $[0,\infty)$ and of exponential order, meaning $|f(t)| \le Me^{ct}$ for some constants. Then $F(s)$ exists for all $\operatorname{Re}(s) > c$. That inequality is why every entry below carries a condition on $s$: $\mathcal{L}\{e^{3t}\}$ exists only for $s > 3$, and $e^{t^2}$ has no transform at all.

Linearity

$\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$

Standard transforms

First shifting theorem: $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ — multiplying by an exponential in $t$ shifts the transform in $s$.

• Writing $\mathcal{L}\{f(t)\} = F(t)$: the transform lives in the $s$-domain. Mixing $t$ and $s$ in one expression is almost always an error.
• Dropping the condition on $s$: $\frac{1}{s-a}$ is only the transform of $e^{at}$ where $s > a$; outside that half-plane the defining integral diverges.
• Treating $\mathcal{L}$ as a multiplication: $\mathcal{L}\{f\}\mathcal{L}\{g\} \neq \mathcal{L}\{fg\}$. The product of transforms corresponds to a convolution in time.
• Assuming every function has a transform: exponential order is a real hypothesis, and $e^{t^2}$ fails it.
• Forgetting the transform ignores $t < 0$: implicitly every input is multiplied by the unit step $u(t)$.