The Laplace Transform

A machine that swaps calculus for algebra

Imagine a machine with a slot on the left and a slot on the right. You feed a function of time — a voltage f(t) rising and falling, a spring settling, a circuit ringing — into the left slot. Out of the right slot slides a completely different-looking function, one of a new variable s. The two are the same information in disguise, but here is the magic: the operations that are hard on the left — differentiating, solving differential equations, integrating awkward products — become the operations that are easy on the right, mere multiplication and algebra. This machine is the Laplace transform, and it is the engineer's single favourite trick.

Its whole reason for existing, the punchline we build toward, is this: the transform turns a differential equation into an ordinary algebra problem. You transform the equation, solve for the unknown with schoolbook algebra, and transform back. Calculus goes in; algebra comes out. But before we can wield it, we have to build it — and it is built, brick by brick, out of a single improper integral.

The definition: multiply by a decaying weight, then integrate

The Laplace transform of a function f(t) defined for t \ge 0 is written \mathcal{L}\{f\} or F(s), and it is defined by the improper integral

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

Read the recipe literally. Take your function f(t), multiply it by the decaying weight e^{-st} — an exponential that starts at 1 when t = 0 and dies away toward the right — and add up the whole product from t = 0 out to infinity. The result no longer depends on t at all (we integrated it away); it depends only on the parameter s hidden in the weight. Different values of s weight the near and far parts of f differently, so F(s) is a whole curve of readings, one for each s.

That damping weight is the point. However fast f(t) grows, a large enough s makes e^{-st} crush the product back down to something with finite area — which is exactly why the improper integral has a chance of converging.

The transform respects sums and scalar multiples, because integration does. For any constants a, b and functions f, g:

\mathcal{L}\{a f + b g\} = a\,\mathcal{L}\{f\} + b\,\mathcal{L}\{g\};
so you can transform a complicated function term by term — split it into pieces you already know, transform each, and add the results back up.

Linearity is what makes a short table of standard transforms so powerful: almost every function you meet is a sum of scaled copies of the handful in the table, so you rarely integrate from scratch twice.

Worked example 1 — the exponential, \mathcal{L}\{e^{at}\}

The exponential is the cleanest transform to derive, and it exposes the convergence condition in full view. We want \mathcal{L}\{e^{at}\} for a constant a.

Step 1 — write the definition and combine the exponentials. Substitute f(t) = e^{at} and merge the two powers of e:

\mathcal{L}\{e^{at}\} = \int_0^{\infty} e^{-st}\, e^{at}\, dt = \int_0^{\infty} e^{-(s-a)t}\, dt.

Step 2 — replace \infty by a cut-off and integrate. This is a plain exponential integral; write k = s - a for a moment:

\int_0^{T} e^{-(s-a)t}\, dt = \left[\frac{e^{-(s-a)t}}{-(s-a)}\right]_0^{T} = \frac{1 - e^{-(s-a)T}}{s-a}.

Step 3 — take the limit, and watch the sign of s - a. Everything hinges on e^{-(s-a)T} as T \to \infty. If s > a the exponent is negative, so the term dies to 0 and the integral converges:

\mathcal{L}\{e^{at}\} = \lim_{T \to \infty} \frac{1 - e^{-(s-a)T}}{s-a} = \frac{1}{s-a}, \qquad s > a.

If instead s \le a, the weight e^{-st} is too weak to tame the growth of e^{at}, the exponential term blows up, and the integral diverges. So the transform exists only on the half-line s > a — its region of convergence. Setting a = 0 recovers a fact we will lean on constantly: \mathcal{L}\{1\} = 1/s for s > 0.

Worked example 2 — \mathcal{L}\{1\} and \mathcal{L}\{t\} by parts

Now the constant and the ramp, which together seed the whole powers-of-t family. First f(t) = 1, straight from the definition:

\mathcal{L}\{1\} = \int_0^{\infty} e^{-st}\, dt = \lim_{T\to\infty}\left[\frac{e^{-st}}{-s}\right]_0^{T} = \lim_{T\to\infty}\frac{1 - e^{-sT}}{s} = \frac{1}{s}, \qquad s > 0.

Next f(t) = t. This one needs integration by parts, with u = t and dv = e^{-st}\,dt, so du = dt and v = -\tfrac{1}{s}e^{-st}.

Step 1 — apply the by-parts formula.

\mathcal{L}\{t\} = \int_0^{\infty} t\,e^{-st}\, dt = \underbrace{\left[-\frac{t}{s}e^{-st}\right]_0^{\infty}}_{\text{boundary term}} + \frac{1}{s}\int_0^{\infty} e^{-st}\, dt.

Step 2 — kill the boundary term. At t = 0 it is 0; as t \to \infty the exponential e^{-st} beats the linear t to zero (for s > 0), so the boundary term vanishes at both ends.

Step 3 — recognise the leftover integral. What remains is \tfrac{1}{s}\,\mathcal{L}\{1\}, which we just computed:

\mathcal{L}\{t\} = \frac{1}{s}\cdot\frac{1}{s} = \frac{1}{s^2}, \qquad s > 0.

The same by-parts move, applied to t^2, drops it onto t; applied to t^3 it drops onto t^2; and so, one rung at a time, each power hands you down to the one below. The pattern that falls out is the general power rule \mathcal{L}\{t^n\} = n!/s^{\,n+1} — and you can see the factorial being born, one factor of 1/s and one integer per step down the ladder.

The transform table — the working reference

Here is the core dictionary. Every entry is derived exactly as above — definition, cut-off, limit — and each carries a region of convergence, the range of s for which the improper integral actually converges. Together with linearity, this handful covers a large fraction of everything a first course throws at you.

f(t)	F(s) = \mathcal{L}\{f\}(s)	Converges for
1	\dfrac{1}{s}	s > 0
t	\dfrac{1}{s^2}	s > 0
t^n (n a non-negative integer)	\dfrac{n!}{s^{\,n+1}}	s > 0
e^{at}	\dfrac{1}{s-a}	s > a
\sin(\omega t)	\dfrac{\omega}{s^2 + \omega^2}	s > 0
\cos(\omega t)	\dfrac{s}{s^2 + \omega^2}	s > 0

Linearity lets you combine rows on sight. For instance \mathcal{L}\{3 + 2t - 5e^{4t}\} = \dfrac{3}{s} + \dfrac{2}{s^2} - \dfrac{5}{s-4}, valid wherever all three pieces converge — here s > 4, the most restrictive of the three.

You could integrate \int_0^\infty e^{-st}\sin(\omega t)\,dt by parts twice and solve the resulting equation for the integral — the classic "the integral reappears, move it to the other side" trick. But there is a slicker route through the exponential entry we already proved. Euler's formula says e^{i\omega t} = \cos(\omega t) + i\sin(\omega t), so by linearity

\mathcal{L}\{e^{i\omega t}\} = \frac{1}{s - i\omega} = \frac{s + i\omega}{s^2 + \omega^2} = \underbrace{\frac{s}{s^2+\omega^2}}_{\mathcal{L}\{\cos\omega t\}} + i\,\underbrace{\frac{\omega}{s^2+\omega^2}}_{\mathcal{L}\{\sin\omega t\}}.

Read off the real and imaginary parts and both trig transforms drop out at once, sharing a denominator s^2 + \omega^2 — the fingerprint of oscillation that will reappear every time an ODE describes something that vibrates.

See the transform being computed

The picture below makes the definition concrete for f(t) = t. In faint grey is f(t) = t, climbing forever. The falling curve is the weight e^{-st}. Their product t\,e^{-st} is the bold curve, and the shaded area beneath it — added up all the way to infinity — is precisely the transform value F(s) = \mathcal{L}\{t\}(s) = 1/s^2.

Slide s and watch two things at once. As s grows, the weight e^{-st} collapses toward the axis faster, the product bump shrinks and slides left, and the shaded area drops — exactly as 1/s^2 predicts (double s and the area quarters). As s shrinks toward 0 the weight barely decays, the product stays fat far out to the right, and the area balloons toward infinity — the edge of the region of convergence coming into view.

Why it matters: derivatives become multiplication by s

Here is the property that justifies the whole enterprise. One more integration by parts, applied to f'(t), gives the derivative rule:

\mathcal{L}\{f'\}(s) = s\,F(s) - f(0).

Look at what happened: differentiating f in time became multiplying F by s (and subtracting a starting value). Apply it twice and a second derivative becomes an s^2. So a differential equation full of y', y'' — calculus — transforms into a polynomial equation in s — algebra. You solve that algebraically for Y(s), then read the answer back off the table in reverse. That reversal, and a full worked ODE, is the next lesson. For now, hold onto the headline: Laplace trades derivatives for factors of s, and so trades calculus for algebra.

Three things trip up almost everyone meeting the transform for the first time.

The Laplace integral is improper, and only converges for some s. \mathcal{L}\{e^{at}\} = 1/(s-a) is only valid for s > a; feed in s \le a and the integral diverges — the formula 1/(s-a) is a value the transform approaches on its region of convergence, not a magic identity that holds everywhere. Every table entry silently carries its own range of s.
\mathcal{L} starts the clock at t = 0. The lower limit of the integral is 0, not -\infty. The transform only ever sees f(t) for t \ge 0; whatever the function did for negative time is invisible to it. That is why Laplace suits initial-value problems — systems switched on at t = 0 — and why the starting value f(0) shows up explicitly in the derivative rule.
F(s) is a function of s, not of t. The variable t is integrated away and gone; an answer with a leftover t in it is a sign the integral was never finished. Time goes in; the s-domain comes out.

Pierre-Simon Laplace (1749–1827) was a French mathematician and astronomer so formidable that contemporaries called him "the Newton of France". The transform grew out of his work on probability and on "generating functions" — clever weighted sums that pack a whole sequence or function into one tidy object you can manipulate algebraically. The continuous integral version, \int_0^\infty e^{-st}f(t)\,dt, is exactly that idea for functions of a continuous time.

Its rise to fame came a century later through the English engineer Oliver Heaviside, who used transform-like "operational" methods to solve the electrical circuit equations of the telegraph — long before anyone had made the method rigorous. Mathematicians grumbled that his manipulations were unjustified; Heaviside shot back, "Shall I refuse my dinner because I do not fully understand the process of digestion?" The transform was later put on firm footing, and today it is bedrock in control theory, signal processing, and electrical engineering — a machine that really does swap calculus for algebra, exactly as advertised.