Laplace Transform — Table and Properties

A Laplace transform table is a lookup of the standard pairs you reach for most: $1 \mapsto 1/s$, $e^{at} \mapsto 1/(s-a)$, $\sin(at) \mapsto a/(s^2+a^2)$. Pairing it with a few properties is the fastest way to handle common problems without recomputing the defining integral each time. In most calculus, differential equations, and engineering courses the default is the one-sided transform for $t \ge 0$:

where $s$ is generally complex and the formula only makes sense where the integral converges. The workflow: match $f(t)$ to a table row, then apply a small set of rules for sums, shifts, and derivatives.

Why the table works at all

The kernel $e^{-st}$ converts time-domain growth, decay, and oscillation into algebraic expressions in $s$ — and algebra is usually easier to push around than derivatives or integrals. So the table is not just memorization; it is a pattern-matching device. Once you see the pattern, the computation often collapses to a single line. That is also why a few base rows plus a handful of properties cover so many problems.

Common pairs

The entries assume the one-sided transform, and the convergence condition is part of the answer, not optional.

If you only keep a few rows, keep $1$, $e^{at}$, $\sin(at)$, and $\cos(at)$. Many textbook problems reduce to those plus one property.

The properties you actually use

Linearity. If the transforms exist,

This is what lets you split a sum into smaller parts.

Exponential shift in time. If $\mathcal{L}\{f(t)\} = F(s)$, then

Multiplying by an exponential in $t$ shifts the expression in $s$ — the property behind many lookups.

Derivative rule. Under the usual hypotheses,

A derivative becomes algebra plus the initial value, which is why these transforms suit initial value problems.

Multiplication by $t$. If $F(s)$ is differentiable where you need it,

Worked example: $\mathcal{L}\{e^{-2t}\cos(3t)\}$

Find the transform of $f(t) = e^{-2t}\cos(3t)$. Start from the base row:

Now apply the exponential shift. Here $e^{-2t}$ means $a = -2$, so replace $s$ by $s + 2$:

with the condition becoming $\operatorname{Re}(s) > -2$. Once you know the base pair and the shift rule, there is no need to revisit the integral.

Try the reverse, then check forward

The same table reads backward for inverse transforms. Given

recognize the shifted-cosine pattern and read it as $e^{-2t}\cos(3t)$ — then confirm by transforming that forward and landing back on the same expression. In solved examples, identifying the pattern first and justifying it with the table and shift rule is usually the quickest route.

Calculation pitfalls

1. Sign in the shift rule. For $e^{at}f(t)$ the result is $F(s-a)$, so $e^{-2t}f(t)$ gives $F(s+2)$.
2. Dropping convergence conditions. For real $a$, $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ needs $\operatorname{Re}(s) > a$.
3. Forgetting the initial value. $\mathcal{L}\{f'(t)\}$ is not just $sF(s)$.
4. Almost-matching rows. A small change in sign or shift can change the answer completely.
5. Mixing one-sided and two-sided transforms. Most introductory tables use the one-sided version from $t = 0$.

FAQ

Laplace tables shine when the problem is posed for $t \ge 0$ and initial conditions matter — turning derivatives into algebraic terms in differential equations, describing inputs and transfer functions in circuits and control, and capturing decay and oscillation compactly in signals and systems. The convergence region still matters: if the transform does not converge where you need it, the table entry alone is not enough. To practice, transform $e^{4t}\sin(2t)$ starting from the sine row and applying the shift, then try $e^{-t}\sin(5t)$ and watch how the sign flips the shift.