Laplace Transform — Table and Properties

A Laplace transform table gives you the standard pairs you use most often, such as $1 \mapsto 1/s$, $e^{at} \mapsto 1/(s-a)$, and $\sin(at) \mapsto a/(s^2+a^2)$. It is the fastest way to handle common Laplace transform problems without recomputing the defining integral each time.

In most calculus, differential equations, and engineering courses, the default is the one-sided Laplace transform for $t \ge 0$:

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

Here $s$ is usually a complex variable, and the formula only makes sense where the integral converges.

The workflow is simple: match the function to a table row, then use a small set of properties for sums, shifts, or derivatives.

Laplace Transform Table: Common Pairs

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

$f(t)$	$\mathcal\{L\}\{f(t)\}(s)$	Condition
$1$	$\frac\{1\}\{s\}$	$\operatorname\{Re\}(s) > 0$
$t$	$\frac\{1\}\{s^2\}$	$\operatorname\{Re\}(s) > 0$
$t^n$	\frac\{n!\}\{s^\{n+1\}}	$n$ is a nonnegative integer, $\operatorname\{Re\}(s) > 0$
$e^\{at\}$	$\frac\{1\}\{s-a\}$	for real $a$, $\operatorname\{Re\}(s) > a$
$\sin(at)$	$\frac\{a\}\{s^2 + a^2\}$	for real $a$, $\operatorname\{Re\}(s) > 0$
$\cos(at)$	$\frac\{s\}\{s^2 + a^2\}$	for real $a$, $\operatorname\{Re\}(s) > 0$
$\sinh(at)$	$\frac\{a\}\{s^2 - a^2\}$	for real $a$, $\operatorname{Re}(s) >
$\cosh(at)$	$\frac\{s\}\{s^2 - a^2\}$	for real $a$, $\operatorname{Re}(s) >

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

Laplace Transform Properties You Actually Use

The table gets most of its power from a few rules. These are the ones students use over and over.

Linearity

If the transforms exist, then

$$\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$$

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

Exponential Shift In Time

If $\mathcal{L}\{f(t)\} = F(s)$, then

$$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$$

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

Derivative Rule

Under the usual hypotheses for the one-sided transform,

$$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$

This is why Laplace transforms are so useful for initial value problems: the derivative turns into algebra plus the initial value.

Multiplication By $t$

If $F(s)$ is differentiable in the region you need, then

$$\mathcal{L}\{t f(t)\} = -\frac{d}{ds}F(s)$$

This helps when the time-domain function has a factor of $t$ multiplying something simpler.

Why A Laplace Transform Table Works

The kernel $e^{-st}$ turns time-domain growth, decay, and oscillation into algebraic expressions in $s$. That matters because algebra is often easier to manipulate than derivatives or integrals.

So the table is not just something to memorize. It is a pattern-matching tool: once the pattern is clear, the computation often collapses to one line.

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

Find the Laplace transform of

$$f(t) = e^{-2t}\cos(3t)$$

Start with the base table entry

$$\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2 + 9}$$

Now use the exponential shift property. Since $e^{-2t}$ means $a=-2$, replace $s$ by $s+2$:

$$\mathcal{L}\{e^{-2t}\cos(3t)\} = \frac{s+2}{(s+2)^2 + 9}$$

For this transform, the condition becomes $\operatorname{Re}(s) > -2$.

That is the whole calculation. Once you know the base pair and the shift rule, there is no need to go back to the integral.

Common Mistakes With A Laplace Transform Table

1. Mixing up the sign in the shift rule. For $e^{at}f(t)$ the result is $F(s-a)$, so for $e^{-2t}f(t)$ you get $F(s+2)$.
2. Ignoring convergence conditions. For example, for real $a$, $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ needs $\operatorname{Re}(s) > a$.
3. Forgetting the initial value in the derivative formula. $\mathcal{L}\{f'(t)\}$ is not just $sF(s)$.
4. Using a table entry that almost matches but not exactly. A small change in sign or shift can change the answer completely.
5. Mixing one-sided and two-sided Laplace transforms. Most introductory tables use the one-sided version starting at $t=0$.

When A Laplace Transform Table Is Useful

Laplace tables are most useful when the problem is posed for $t \ge 0$ and initial conditions matter.

• In differential equations, they turn derivatives into algebraic terms and make initial value problems easier to solve.
• In circuits and control, they help analyze inputs, outputs, and transfer functions.
• In signals and systems, they describe decay, oscillation, and system response in a compact form.

The convergence condition still matters here. If the transform does not converge in the region you need, the table entry alone is not enough.

Inverse Laplace Transform: Read The Table Backward

The same table is used for inverse Laplace transforms. If you see

$$\frac{s+2}{(s+2)^2 + 9}$$

you can recognize it as the shifted cosine pattern and read it backward as

$$e^{-2t}\cos(3t)$$

That is often the fastest route in solved examples: identify the pattern first, then justify it with the table and the shift rule.

Try A Similar Problem

Try finding the transform of

$$e^{4t}\sin(2t)$$

Start from the sine row in the table, then apply the shift carefully. If you want one more step after that, try your own version with $e^{-t}\sin(5t)$ and compare how the sign changes the shift.