Laplace transform

Given a complex-valued function $f (t)$ with $t \in R$ , the Laplace transform is a function $F : Ω \subseteq C \to C$ such that every $F (s)$ represents the weight of the function $e^{s t}$ in the decomposition

f (t) = \frac{1}{2 π i} (\dots + F (c + 0.1 i) e^{(c + 0.1 i) t} + F (c + 0.2 i) e^{(c + 0.2 i) t} +

+ F (s) e^{s t} + \dots) =

= \frac{1}{2 π i} lim_{T \to \infty} \int_{c - i T}^{c + i T} e^{s t} F (s) d s

for any $c$ inside the Region of Convergence (ROC) — the set of $s \in C$ for which the forward transform converges.

Bilateral vs. unilateral. The formulas above describe the bilateral Laplace transform, where $F (s) = \int_{- \infty}^{\infty} f (t) e^{- s t} d t$ and the ROC is a vertical strip $σ_{1} < Re (s) < σ_{2}$ . The more common unilateral transform integrates from $0$ to $\infty$ , with ROC a right half-plane $Re (s) > σ_{c}$ .

A particular case would be the Fourier transform, whose picture corresponds to the case $c = 0$ (provided $c = 0$ lies in the ROC).

Both transformations have to do with the fact that some spaces of functions are Hilbert spaces, so we have a "basis". The original function can be thought as expressed in the Dirac delta basis, and the Fourier transform and Laplace transform are nothing but the expression of the same function in a different basis. See Fourier, Laplace and Dirichlet relationship.

What basis?, the natural for the generator of the translations.

Why a 1D Line Integral instead of $\int_{C} F (s) e^{s t} d s$ ?

Dimensionality: Since $f (t)$ is a 1D function of time ( $t \in R$ ), it requires a 1D continuum of basis elements to represent it.
Linear Independence: Fixing $Re (s) = c$ yields $s = c + i ω$ , giving a single degree of freedom $ω \in R$ . The family ${e^{(c + i ω) t}}_{ω \in R}$ forms an orthogonal basis for $e^{- c t} f (t)$ .
Overcompleteness: The set of all exponentials ${e^{s t}}_{s \in C}$ across the entire 2D complex plane is massively redundant (overcomplete). Integrating over all of $C$ would sum infinitely many duplicate reconstructions of $f (t)$ together, causing the surface integral to diverge.
Analyticity: By Cauchy's Integral Theorem, every vertical line $c$ within the ROC carries 100% of the information needed to reconstruct $f (t)$ . Shifting $c$ changes the damping factor $e^{- c t}$ , but yields the exact same signal.

One Laplace transform or several ones?

It may seems that we have a Laplace transform for each $c$ . But we have only one complex variable function $F (s)$ , with $s \in C$ . When you calculate the (bilateral) Laplace transform of $f (t)$ , $c$ does not exist in the formula:

L {f (t)} (s) = F (s) = \int_{- \infty}^{\infty} f (t) e^{- s t} d t

(The unilateral version integrates from $0$ to $\infty$ and is often denoted $L {f (t)} (s) = \int_{0}^{\infty} f (t) e^{- s t} d t$ .)
We can think that since

f (t) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} F (s) e^{s t} d s

we have, indeed, different real-valued functions, a 1D restriction of $F (s)$ along the line $Re (s) = c$ :

F_{c} (ω) := F (c + i ω) $ $ B u t t h e n $ F_{c} (ω) $ i s l i t e r a l l y t h e 1 D F o u r i e r T r a n s f o r m o f a d i f f e r e n t f u n c t i o n : $ e^{- c t} f (t) $ . I n e f f e c t, $ $ F_{c} (ω) = F {e^{- c t} f (t)} (ω)

since

e^{- c t} f (t) = e^{- c t} \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} F (s) e^{s t} d s = e^{- c t} \frac{1}{2 π i} \int_{- \infty}^{\infty} F_{c} (ω) e^{(c + i ω) t} d ω =

\frac{1}{2 π i} \int_{- \infty}^{\infty} F_{c} (ω) e^{i ω t} d ω

Each "different" Laplace transform is, indeed, a Fourier transform of a "modified" function. And all these 1D slices $F_{c} (ω)$ are rigidly bound to one another:

Identity Theorem: If you know the 1D slice $F_{c_{1}} (ω)$ along just one vertical line, you can use analytic continuation to uniquely determine $F_{c} (ω)$ along every other valid line $c$ within the ROC.
- For the bilateral transform (ROC = strip), knowing $F (s)$ on one vertical line determines it everywhere in the strip — but not beyond the boundaries $σ_{1}, σ_{2}$ .
- For the unilateral transform (ROC = half-plane), the ROC is $Re (s) > σ_{c}$ , so analytic continuation extends $F (s)$ to the entire right half-plane — a much stronger statement: one vertical line determines everything.
None of these 1D slices can vary independently. Once you choose $f (t)$ , the entire 2D landscape $F (s)$ —and thus all of its 1D slices $F_{c} (ω)$ simultaneously—is locked into place.

My interpretation

We can think of Laplace transform $F (s)$ of $f (t)$ as a kind of recipe to construct $f (t)$ from exponential functions ${e^{s t} : s \in C}$ . But we don't need the whole $F$ , we can fix a vertical line in $C$ , $L_{c} = {c + i ω : ω \in R}, c \in R$ . Consider the $s$ -plane as a collection of all the exponential functions. We can reconstruct $f$ from the exponentials living in any of those lines

f (t) = \frac{1}{2 π i} \int_{c_{1} - i \infty}^{c_{1} + i \infty} F (s) e^{s t} d s = \frac{1}{2 π i} \int_{c_{2} - i \infty}^{c_{2} + i \infty} F (s) e^{s t} d s = \dots

Example

This figure illustrates how a single complex landscape $F (s) = \frac{1}{s - i}$ unifies an infinite family of 1D Fourier transforms for the causal signal $f (t) = e^{i t} u (t)$ . (Here $u (t)$ is the Heaviside step function, for issues of convergence)
Pasted image 20260803163409.png
For any vertical line $s = c + i ω$ within the Region of Convergence ( $c > 0$ ), the 1D cross-section $F_{c} (ω)$ is literally the continuous Fourier transform of the exponentially damped signal $e^{- c t} f (t)$ . As $c$ decreases from $0.50$ down to $0.05$ , the damping weakens: the smooth slices of the real part $Re [F_{c} (ω)]$ grow taller (height $1 / c$ ) and narrower while maintaining a constant area of $π$ . On the boundary $c = 0$ , the damping vanishes entirely, and this real part collapses into the Dirac delta distribution $π δ (ω - 1)$ —matching the real part of the Fourier transform of the causal tone $e^{i t} u (t)$ . Because all these 1D slices are rigidly bound together across the 2D surface $F (s)$ , reconstructing the original signal $f (t)$ via the inverse Laplace transform along any vertical path $L_{c}$ with $c > 0$ yields the exact same underlying function.

Poles and the shape of the original function

Reading $f (t)$ from the poles of $F (s)$ is reading the weights of the exponential basis ${e^{s t}}$ : a simple pole at $s = a$ contributes one exponential,

L^{- 1} {\frac{A}{s - a}} = A e^{a t} .

Two cases matter:

Purely imaginary pole ( $a = i ω$ ): then $e^{i ω t} = \cos (ω t) + i \sin (ω t)$ , a pure oscillation. Functions whose transform has poles only on the imaginary axis are the almost-periodic ones — the building blocks of Fourier decompositions.
Real pole ( $a \in R$ ): then $e^{a t}$ is a real exponential (growth if $a > 0$ , decay if $a < 0$ ), e.g. $\frac{1}{s - 1} \leftrightarrow e^{t}$ . The exponential is itself a Taylor series, $e^{t} = \sum_{n = 0}^{\infty} \frac{t^{n}}{n!}$ : a simple real pole encodes the Taylor coefficients, not a polynomial term.

Polynomial terms. A bare $t^{n}$ comes only from multiple poles at the origin: a pole of order $n + 1$ at $s = 0$ gives

L^{- 1} {\frac{n!}{s^{n + 1}}} = t^{n},

and more generally a pole of the same multiplicity at $s = a$ gives

L^{- 1} {\frac{n!}{(s - a)^{n + 1}}} = t^{n} e^{a t} .

So: oscillatory modes ↔ imaginary poles, exponential growth/decay ↔ real poles, and pure polynomials require the special position $s = 0$ (any other pole position injects the extra factor $e^{a t}$ ).

Example. The transform of $f (t) = \sin (t) + t^{2} + 5 t + 6$ isx

F (s) = \frac{1}{s^{2} + 1} + \frac{2}{s^{3}} + \frac{5}{s^{2}} + \frac{6}{s},

and its poles read exactly as predicted: