Laplace Transform Table: Common Pairs, Properties and LaTeX Source

The Laplace transform turns a function of time f(t) into a function of a complex variable F(s), and in doing so turns derivatives into multiplication by s. That is why it solves linear differential equations, circuits and control systems by algebra. This page is the table you keep open while doing that: 26 transform pairs with their regions of convergence, every property and theorem, two worked inverse transforms, and the LaTeX source for the whole table so you can paste it into your own notes.

Definition and Region of Convergence

For a function f(t) defined for t ≥ 0, the (one-sided) Laplace transform is

F(s) = ℒ{f(t)} = ∫₀^∞ e^−st f(t) dt

The integral converges only when s is far enough to the right in the complex plane: there is a real number a, the abscissa of convergence, such that the integral exists for every s with Re s > a and (for most functions) diverges to its left. That half-plane is the region of convergence, and it is the third column of the table below. The transform is written ℒ{f}, ℒ[f] or simply F(s); the inverse transform ℒ⁻¹{F} recovers f(t) for t ≥ 0.

Throughout the table, n is a non-negative integer, a, b and ω are constants, u(t − a) is the unit step (Heaviside function, also written H or θ) that jumps from 0 to 1 at t = a, and δ is the Dirac delta.

Table of Laplace Transform Pairs

The pairs are grouped as they are usually taught: powers of t, exponentials, sines and cosines, damped and growing oscillations, then the step and impulse functions that handle switching. Read it left to right for a forward transform and right to left for an inverse one.

Notes: Γ is the gamma function, which reduces row 4 to row 3 when p is an integer because Γ(n + 1) = n!. Rows 5 and 6 are the p = ½ and p = −½ cases of row 4, using Γ(½) = √π. In row 25, γ ≈ 0.5772 is the Euler–Mascheroni constant. Rows 13 and 14 are rows 11 and 12 with ω = ia, which is why the sign in the denominator flips. Row 10 is the partial-fraction pair you meet most often when inverting a transform with two real poles.

Properties of the Laplace Transform

The properties are what make the table small enough to fit on a page: any function you can build from the pairs above by shifting, scaling, differentiating or multiplying by t has a transform you can write down without integrating. Here F = ℒ{f} and G = ℒ{g}.

Two cautions. The derivative rules assume f is continuous on [0, ∞) with a piecewise-continuous derivative; a jump at t = 0 means f(0) must be read as f(0⁺). The final value theorem only holds when the limit exists, which requires every pole of s F(s) to lie in the open left half-plane: applying it to F(s) = ω/(s² + ω²) gives 0, but sin(ωt) has no limit.

Inverse Transforms with Partial Fractions

Going backwards is where the table earns its keep. A transform that comes out of a differential equation is almost always a rational function, and the strategy is the same every time: factor the denominator, split into partial fractions, then match each term to a row of the table, pulling constants out by linearity.

Example 1: two real poles

Find f(t) when F(s) = (s + 3) / (s² + 3s + 2).

1. Factor: s² + 3s + 2 = (s + 1)(s + 2), so F(s) = A/(s + 1) + B/(s + 2).
2. Cover-up method: A = (s + 3)/(s + 2) at s = −1, which is 2; B = (s + 3)/(s + 1) at s = −2, which is −1.
3. Row 7 with a = −1 and a = −2: f(t) = 2e^−t − e^−2t for t ≥ 0.

Check with the initial value theorem: s F(s) → 1 as s → ∞, and f(0) = 2 − 1 = 1.

Example 2: complex poles and an initial value problem

Solve y″ + 2y′ + 5y = 0 with y(0) = 1 and y′(0) = 0.

1. Transform both sides with the derivative rules, writing Y = ℒ{y}: (s²Y − s) + 2(sY − 1) + 5Y = 0, so Y(s) = (s + 2) / (s² + 2s + 5).
2. The denominator has complex roots, so complete the square instead of factoring: s² + 2s + 5 = (s + 1)² + 2².
3. Split the numerator to match rows 15 and 16 with a = −1 and ω = 2: Y(s) = (s + 1) / ((s + 1)² + 4) + ½ · 2 / ((s + 1)² + 4).
4. Read off: y(t) = e^−t cos 2t + ½ e^−t sin 2t. It satisfies y(0) = 1, and differentiating gives y′(0) = −1 + 1 = 0.

A delayed input works the same way with row 24 in reverse: a factor e^−as in F(s) means "invert the rest, then replace t by t − a and multiply by the stepu(t − a)". For instance e^−3s / (s² + 4) inverts to ½ u(t − 3) sin 2(t − 3). Repeated poles use rows 8 and 9, and a term A/(s − a)² is A t e^at.

Typesetting the Laplace Transform Table in LaTeX

Everything above was produced from the LaTeX below, which compiles with pdfLaTeX and only standard packages: amsmath for \dfrac and \text, booktabs for the rules, array for math-mode columns and longtable because 26 rows of fractions will not fit on one page. Start with the operator symbol.

The ℒ symbol and a definition

\mathcal{L} gives the calligraphic ℒ. Define it once as \Lap so you can change it everywhere later (many textbooks prefer the script \mathscr{L} from mathrsfs). The braces around the argument must be escaped as \{ and \}, otherwise LaTeX treats them as grouping and prints nothing. \operatorname{Re} gives an upright "Re"; the built-in \Re prints a Fraktur ℜ that most readers find harder to parse.

\documentclass{article}
\usepackage{amsmath,amssymb}

\newcommand{\Lap}{\mathcal{L}}          % the transform operator
\newcommand{\iLap}{\mathcal{L}^{-1}}    % its inverse
\DeclareMathOperator{\re}{Re}           % upright Re, for the region of convergence

\begin{document}

The Laplace transform of $f(t)$ is
\[
  F(s) = \Lap\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t)\,\mathrm{d}t,
  \qquad \re s > a,
\]
and $f(t) = \iLap\{F(s)\}$ for $t \ge 0$. For example,
$\Lap\{e^{at}\} = \dfrac{1}{s-a}$ for $\re s > \re a$.

\end{document}

The pairs table with booktabs and longtable

The array package lets a column type open and close math mode around every cell, so \newcolumntype{M}{>{$}l<{$}} means you type \dfrac{1}{s} rather than $\dfrac{1}{s}$ in each cell. Header text in such a column must be wrapped in \multicolumn{1}{l}{...} to switch it back to text mode. \arraystretch opens the rows up so stacked fractions do not touch; \endhead repeats the header on every page and \endlastfoot puts the bottom rule only after the final row.

% Preamble: \usepackage{booktabs,array,longtable} and
% \newcolumntype{M}{>{$}l<{$}}  (see the complete document below)

\renewcommand{\arraystretch}{1.9}
\begin{longtable}[l]{M M M}
  \toprule
  \multicolumn{1}{l}{$f(t)$} &
  \multicolumn{1}{l}{$F(s) = \Lap\{f(t)\}$} &
  \multicolumn{1}{l}{Converges for} \\
  \midrule
  \endhead
  \bottomrule
  \endlastfoot
  1 & \dfrac{1}{s} & \re s > 0 \\
  t & \dfrac{1}{s^{2}} & \re s > 0 \\
  t^{n}\quad (n = 0, 1, 2, \dots) & \dfrac{n!}{s^{n+1}} & \re s > 0 \\
  t^{p}\quad (p > -1) & \dfrac{\Gamma(p+1)}{s^{p+1}} & \re s > 0 \\
  \sqrt{t} & \dfrac{\sqrt{\pi}}{2 s^{3/2}} & \re s > 0 \\
  \dfrac{1}{\sqrt{t}} & \sqrt{\dfrac{\pi}{s}} & \re s > 0 \\
  e^{at} & \dfrac{1}{s-a} & \re s > \re a \\
  t\,e^{at} & \dfrac{1}{(s-a)^{2}} & \re s > \re a \\
  t^{n} e^{at} & \dfrac{n!}{(s-a)^{n+1}} & \re s > \re a \\
  \dfrac{e^{at} - e^{bt}}{a - b} & \dfrac{1}{(s-a)(s-b)} & \re s > \max(\re a, \re b) \\
  \sin(\omega t) & \dfrac{\omega}{s^{2} + \omega^{2}} & \re s > 0 \\
  \cos(\omega t) & \dfrac{s}{s^{2} + \omega^{2}} & \re s > 0 \\
  \sinh(at) & \dfrac{a}{s^{2} - a^{2}} & \re s > |a| \\
  \cosh(at) & \dfrac{s}{s^{2} - a^{2}} & \re s > |a| \\
  e^{at} \sin(\omega t) & \dfrac{\omega}{(s-a)^{2} + \omega^{2}} & \re s > \re a \\
  e^{at} \cos(\omega t) & \dfrac{s-a}{(s-a)^{2} + \omega^{2}} & \re s > \re a \\
  t \sin(\omega t) & \dfrac{2 \omega s}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  t \cos(\omega t) & \dfrac{s^{2} - \omega^{2}}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  \sin(\omega t) - \omega t \cos(\omega t) & \dfrac{2 \omega^{3}}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  \dfrac{\sin(\omega t)}{t} & \arctan\dfrac{\omega}{s} & \re s > 0 \\
  \delta(t) & 1 & \text{all } s \\
  \delta(t - a)\quad (a \ge 0) & e^{-as} & \text{all } s \\
  u(t - a)\quad (a \ge 0) & \dfrac{e^{-as}}{s} & \re s > 0 \\
  u(t - a)\, f(t - a) & e^{-as} F(s) & \text{as for } f \\
  \ln t & -\dfrac{\gamma + \ln s}{s} & \re s > 0 \\
  \operatorname{erf}\bigl(\sqrt{t}\bigr) & \dfrac{1}{s \sqrt{s + 1}} & \re s > 0 \\
\end{longtable}

If you only need a dozen rows, swap longtable for a plain tabular and delete the \endhead and \endlastfoot lines. Our LaTeX tables guide covers both environments in detail, including captions and column widths.

The properties table

The first column is ordinary text, so its type is l, and the other two use the M type again. Integrals and limits inside a table are set in text style by default; \displaystyle restores the full-size operator with the limits in the right place.

\renewcommand{\arraystretch}{1.9}
\begin{tabular}{l M M}
  \toprule
  Property & f(t) & F(s) \\
  \midrule
  Linearity & a f(t) + b g(t) & a F(s) + b G(s) \\
  First derivative & f'(t) & s F(s) - f(0) \\
  Second derivative & f''(t) & s^{2} F(s) - s f(0) - f'(0) \\
  $n$th derivative & f^{(n)}(t) & s^{n} F(s) - s^{n-1} f(0) - \dots - f^{(n-1)}(0) \\
  Integral & \displaystyle\int_{0}^{t} f(\tau)\,\mathrm{d}\tau & \dfrac{F(s)}{s} \\
  Shift in $s$ & e^{at} f(t) & F(s - a) \\
  Shift in $t$ & u(t - a)\, f(t - a) & e^{-as} F(s) \\
  Scaling & f(at)\quad (a > 0) & \dfrac{1}{a} F\!\left(\dfrac{s}{a}\right) \\
  Multiply by $t$ & t f(t) & -F'(s) \\
  Multiply by $t^{n}$ & t^{n} f(t) & (-1)^{n} F^{(n)}(s) \\
  Divide by $t$ & \dfrac{f(t)}{t} & \displaystyle\int_{s}^{\infty} F(\sigma)\,\mathrm{d}\sigma \\
  Convolution & (f * g)(t) = \displaystyle\int_{0}^{t} f(\tau) g(t - \tau)\,\mathrm{d}\tau & F(s)\, G(s) \\
  Periodic, period $T$ & f(t + T) = f(t) & \dfrac{1}{1 - e^{-sT}} \displaystyle\int_{0}^{T} e^{-st} f(t)\,\mathrm{d}t \\
  Initial value & f(0^{+}) & \displaystyle\lim_{s \to \infty} s F(s) \\
  Final value & \displaystyle\lim_{t \to \infty} f(t) & \displaystyle\lim_{s \to 0} s F(s) \\
  \bottomrule
\end{tabular}

A worked inverse transform with align*

For the derivations, align* lines up the equals signs and \left. ... \right|_{s=-1} produces the evaluation bar used in the cover-up method. Inverse transforms use \iLap, defined in the preamble as \mathcal{L}^{-1}.

\begin{align*}
  F(s) &= \frac{s + 3}{s^{2} + 3s + 2}
        = \frac{s + 3}{(s + 1)(s + 2)}
        = \frac{A}{s + 1} + \frac{B}{s + 2} \\
  A &= \left.\frac{s + 3}{s + 2}\right|_{s = -1} = 2, \qquad
  B = \left.\frac{s + 3}{s + 1}\right|_{s = -2} = -1 \\
  f(t) &= \iLap\{F(s)\} = 2e^{-t} - e^{-2t}, \qquad t \ge 0.
\end{align*}

Complete document

Here is everything in one file. Compile it twice so longtable can settle the column widths; the output is three pages with the pairs table breaking cleanly across the first two.

\documentclass[11pt]{article}
\usepackage{amsmath,amssymb}
\usepackage{booktabs,array,longtable}
\usepackage[margin=2.5cm]{geometry}

% Shorthand for the transform operators
\newcommand{\Lap}{\mathcal{L}}
\newcommand{\iLap}{\mathcal{L}^{-1}}
\DeclareMathOperator{\re}{Re}

% Math-mode columns: no need to type $...$ in every cell
\newcolumntype{M}{>{$}l<{$}}

\begin{document}

\section*{Laplace transform table}

The Laplace transform of $f(t)$ is
\[
  F(s) = \Lap\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t)\,\mathrm{d}t,
\]
defined for every complex $s$ with $\re s$ larger than the abscissa of
convergence of $f$.

\subsection*{Transform pairs}

\renewcommand{\arraystretch}{1.9}
\begin{longtable}[l]{M M M}
  \toprule
  \multicolumn{1}{l}{$f(t)$} &
  \multicolumn{1}{l}{$F(s) = \Lap\{f(t)\}$} &
  \multicolumn{1}{l}{Converges for} \\
  \midrule
  \endhead
  \bottomrule
  \endlastfoot
  1 & \dfrac{1}{s} & \re s > 0 \\
  t & \dfrac{1}{s^{2}} & \re s > 0 \\
  t^{n}\quad (n = 0, 1, 2, \dots) & \dfrac{n!}{s^{n+1}} & \re s > 0 \\
  t^{p}\quad (p > -1) & \dfrac{\Gamma(p+1)}{s^{p+1}} & \re s > 0 \\
  \sqrt{t} & \dfrac{\sqrt{\pi}}{2 s^{3/2}} & \re s > 0 \\
  \dfrac{1}{\sqrt{t}} & \sqrt{\dfrac{\pi}{s}} & \re s > 0 \\
  e^{at} & \dfrac{1}{s-a} & \re s > \re a \\
  t\,e^{at} & \dfrac{1}{(s-a)^{2}} & \re s > \re a \\
  t^{n} e^{at} & \dfrac{n!}{(s-a)^{n+1}} & \re s > \re a \\
  \dfrac{e^{at} - e^{bt}}{a - b} & \dfrac{1}{(s-a)(s-b)} & \re s > \max(\re a, \re b) \\
  \sin(\omega t) & \dfrac{\omega}{s^{2} + \omega^{2}} & \re s > 0 \\
  \cos(\omega t) & \dfrac{s}{s^{2} + \omega^{2}} & \re s > 0 \\
  \sinh(at) & \dfrac{a}{s^{2} - a^{2}} & \re s > |a| \\
  \cosh(at) & \dfrac{s}{s^{2} - a^{2}} & \re s > |a| \\
  e^{at} \sin(\omega t) & \dfrac{\omega}{(s-a)^{2} + \omega^{2}} & \re s > \re a \\
  e^{at} \cos(\omega t) & \dfrac{s-a}{(s-a)^{2} + \omega^{2}} & \re s > \re a \\
  t \sin(\omega t) & \dfrac{2 \omega s}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  t \cos(\omega t) & \dfrac{s^{2} - \omega^{2}}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  \sin(\omega t) - \omega t \cos(\omega t) & \dfrac{2 \omega^{3}}{(s^{2} + \omega^{2})^{2}} & \re s > 0 \\
  \dfrac{\sin(\omega t)}{t} & \arctan\dfrac{\omega}{s} & \re s > 0 \\
  \delta(t) & 1 & \text{all } s \\
  \delta(t - a)\quad (a \ge 0) & e^{-as} & \text{all } s \\
  u(t - a)\quad (a \ge 0) & \dfrac{e^{-as}}{s} & \re s > 0 \\
  u(t - a)\, f(t - a) & e^{-as} F(s) & \text{as for } f \\
  \ln t & -\dfrac{\gamma + \ln s}{s} & \re s > 0 \\
  \operatorname{erf}\bigl(\sqrt{t}\bigr) & \dfrac{1}{s \sqrt{s + 1}} & \re s > 0 \\
\end{longtable}

\subsection*{Properties}

\renewcommand{\arraystretch}{1.9}
\begin{tabular}{l M M}
  \toprule
  Property & f(t) & F(s) \\
  \midrule
  Linearity & a f(t) + b g(t) & a F(s) + b G(s) \\
  First derivative & f'(t) & s F(s) - f(0) \\
  Second derivative & f''(t) & s^{2} F(s) - s f(0) - f'(0) \\
  $n$th derivative & f^{(n)}(t) & s^{n} F(s) - s^{n-1} f(0) - \dots - f^{(n-1)}(0) \\
  Integral & \displaystyle\int_{0}^{t} f(\tau)\,\mathrm{d}\tau & \dfrac{F(s)}{s} \\
  Shift in $s$ & e^{at} f(t) & F(s - a) \\
  Shift in $t$ & u(t - a)\, f(t - a) & e^{-as} F(s) \\
  Scaling & f(at)\quad (a > 0) & \dfrac{1}{a} F\!\left(\dfrac{s}{a}\right) \\
  Multiply by $t$ & t f(t) & -F'(s) \\
  Multiply by $t^{n}$ & t^{n} f(t) & (-1)^{n} F^{(n)}(s) \\
  Divide by $t$ & \dfrac{f(t)}{t} & \displaystyle\int_{s}^{\infty} F(\sigma)\,\mathrm{d}\sigma \\
  Convolution & (f * g)(t) = \displaystyle\int_{0}^{t} f(\tau) g(t - \tau)\,\mathrm{d}\tau & F(s)\, G(s) \\
  Periodic, period $T$ & f(t + T) = f(t) & \dfrac{1}{1 - e^{-sT}} \displaystyle\int_{0}^{T} e^{-st} f(t)\,\mathrm{d}t \\
  Initial value & f(0^{+}) & \displaystyle\lim_{s \to \infty} s F(s) \\
  Final value & \displaystyle\lim_{t \to \infty} f(t) & \displaystyle\lim_{s \to 0} s F(s) \\
  \bottomrule
\end{tabular}

\subsection*{Worked inverse transform}

\begin{align*}
  F(s) &= \frac{s + 3}{s^{2} + 3s + 2}
        = \frac{s + 3}{(s + 1)(s + 2)}
        = \frac{A}{s + 1} + \frac{B}{s + 2} \\
  A &= \left.\frac{s + 3}{s + 2}\right|_{s = -1} = 2, \qquad
  B = \left.\frac{s + 3}{s + 1}\right|_{s = -2} = -1 \\
  f(t) &= \iLap\{F(s)\} = 2e^{-t} - e^{-2t}, \qquad t \ge 0.
\end{align*}

\end{document}

Frequently Asked Questions

What is the Laplace transform of 1?

1/s, valid for Re s > 0. A constant c transforms to c/s by linearity, and the unit step u(t) has the same transform as 1 because the integral only runs from t = 0.

What is the Laplace transform of e^at?

1/(s − a), valid for Re s > Re a. Replacing a by −a gives 1/(s + a) for the decaying exponential e^−at, which is the pair that appears in almost every circuit and differential-equation problem.

What are the Laplace transforms of sin(ωt) and cos(ωt)?

sin(ωt) transforms to ω/(s² + ω²) and cos(ωt) to s/(s² + ω²), both for Re s > 0. The one with s in the numerator is the cosine: cos(0) = 1, so its transform has to behave like 1/s for large s, while sin(0) = 0 makes its transform fall off like 1/s².

What does the "Re s > a" column mean?

It is the region of convergence: the defining integral only converges for s to the right of the abscissa a, and the closed-form F(s) is only guaranteed to equal the integral there. For solving initial value problems you rarely need it, because you read the inverse transform straight off the table.

Is this the one-sided or two-sided Laplace transform?

One-sided (unilateral), integrating from 0 to ∞. That is the transform used for initial value problems and in engineering courses, and the derivative rules with f(0) and f′(0) only make sense for it. The bilateral transform integrates from −∞ to ∞, is closer to the Fourier transform, and has different tables for functions that are nonzero for t < 0.

How do I write the Laplace transform symbol in LaTeX?

\mathcal{L} in math mode gives the calligraphic ℒ; \mathscr{L} from the mathrsfs package gives the script form many textbooks use. Define \newcommand{\Lap}{\mathcal{L}} once in the preamble and write \Lap\{f(t)\}, with the braces escaped so they are printed.

Next Steps

The table is the easy part; these guides cover the LaTeX around it:

• Mathematical Expressions in LaTeX – Fractions, roots, aligned equations and the amsmath environments used above
• LaTeX Tables – booktabs, longtable, column types, captions and multi-page tables
• Integrals, Sums and Limits – Typeset the defining integral and the convolution with proper limits
• The Greek Alphabet – ω, τ, σ, γ and δ with their LaTeX commands
• Equation Numbering and Referencing – Number the transform definition and cite it from the text
• Math Functions in LaTeX – \sin, \cos, \ln, \operatorname and \DeclareMathOperator

Free LaTeX Tools

• LaTeX Preview – Paste the table source and see the PDF without installing TeX
• Math to LaTeX – Turn a handwritten transform pair into LaTeX
• LaTeX Symbols Reference – Look up \mathcal, \infty, \omega and every other symbol