Laplace transform — definition, region of convergence

1. The Definition (built from scratch)

WHY the kernel $e^{-st}$ ? We want a "weighting factor" that kills the function's growth at large $t$ so the infinite integral can settle to a finite number. The exponential $e^{-st}$ decays (when $\operatorname{Re}(s)$ is large enough), squashing $f(t)$ down. It also has the magic property $\frac{d}{dt}e^{-st} = -s\,e^{-st}$ , which is exactly what makes derivatives turn into multiplication by $s$ .

WHAT we get: a function $F(s)$ that lives in " $s$ -space." Information about $f$ is repackaged, not lost.

HOW we compute it: plug $f(t)$ into the integral and evaluate. Let's earn one.

Worked example Transform of

$f(t)=1$ $F(s)=\int_0^\infty e^{-st}\cdot 1 \, dt = \left[\frac{-1}{s}e^{-st}\right]_0^\infty$

Why this step? $\int e^{-st}dt = -\frac{1}{s}e^{-st}$ (basic exponential integral).
At the upper limit $t\to\infty$ : $e^{-st}\to 0$ only if $\operatorname{Re}(s)>0$ . (Why? $|e^{-st}|=e^{-\sigma t}\to 0$ iff $\sigma>0$ .)
At $t=0$ : $e^{0}=1$ . $F(s) = 0 - \left(\frac{-1}{s}\right) = \frac{1}{s}, \qquad \operatorname{Re}(s)>0.$ The condition $\operatorname{Re}(s)>0$ is the ROC.

2. Region of Convergence (ROC)

Deriving the bound (WHY it works): $|F(s)| = \left|\int_0^\infty e^{-st}f(t)\,dt\right| \le \int_0^\infty e^{-\sigma t}|f(t)|\,dt \le \int_0^\infty e^{-\sigma t}Me^{at}\,dt = M\int_0^\infty e^{-(\sigma-a)t}\,dt = \frac{M}{\sigma - a},$ finite iff $\sigma - a > 0$ , i.e. $\sigma>a$ . Done — the half-plane drops straight out of the inequality.

Figure — Laplace transform — definition, region of convergence

Worked example Transform of

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

Why integration by parts? Product of $t$ (algebraic) and $e^{-st}$ — classic IBP. Let $u=t,\ dv=e^{-st}dt$ , so $du=dt,\ v=-\frac{1}{s}e^{-st}$ . $= \left[-\frac{t}{s}e^{-st}\right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\,dt.$
The boundary term vanishes for $\operatorname{Re}(s)>0$ (exponential beats $t$ ).
The leftover integral is $\frac{1}{s}\cdot\frac{1}{s}$ . $F(s)=\frac{1}{s^2},\qquad \operatorname{Re}(s)>0.$

3. Common Mistakes (Steel-manned)

4. Active Recall

Recall Quick self-test (cover the answers)

Q: What is the kernel and why that one? → $e^{-st}$ ; it decays to tame growth and converts $d/dt$ into $\times s$ .
Q: ROC of $\mathcal{L}\{e^{3t}\}$ ? → $\operatorname{Re}(s)>3$ .
Q: What two conditions guarantee existence? → piecewise continuous + exponential order.
Q: Why does $e^{t^2}$ have no transform? → grows faster than any $e^{at}$ ; integral diverges for all $s$ .
Q: Shape of an ROC? → a right half-plane $\operatorname{Re}(s) > a$ .

Recall Feynman: explain it to a 12-year-old

Imagine a noisy classroom (a wiggly signal in time). The Laplace transform is like a special pair of headphones that, instead of listening normally, multiplies every sound by a "fade-out" filter that gets quieter as time goes on. If the noise isn't crazy loud (doesn't grow too fast), the headphones add up all the faded sounds into one single number for each "fade speed" $s$ . The list of fade speeds that give a sensible (finite) number is the Region of Convergence — basically "fade fast enough and it always works; the noisier the original, the faster you must fade."

5. The 80/20 (what to truly own)

$F(s)=\int_0^\infty e^{-st}f(t)\,dt$ — and why the kernel.
ROC is the half-plane $\operatorname{Re}(s)>a$ where $a$ = growth rate.
Existence = piecewise continuous + exponential order.
Three anchor results: $\mathcal{L}\{1\}=\frac1s$ , $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ , $\mathcal{L}\{t\}=\frac{1}{s^2}$ .

Connections

Inverse Laplace Transform — going back from $F(s)$ to $f(t)$ ; ROC fixes uniqueness.
Laplace Transform of Derivatives — the $\times s$ property used to solve ODEs.
Solving ODEs with Laplace Transforms — the payoff application.
Improper Integrals — convergence machinery behind the ROC.
Fourier Transform — special case along $\operatorname{Re}(s)=0$ .
Exponential Order and Growth Rates

Define the Laplace transform of

f(t)

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

, for

s

where the integral converges.

Why is the kernel chosen as

e^{-st}

It decays to tame the function's growth (so the infinite integral converges) and satisfies

\frac{d}{dt}e^{-st}=-s\,e^{-st}

, turning differentiation into multiplication by

s

What is the Region of Convergence (ROC)?

The set of

s

for which the defining integral converges; for one-sided transforms it is a right half-plane

\operatorname{Re}(s)>a

What is the abscissa of convergence?

The boundary value

a

such that

F(s)

converges for

\operatorname{Re}(s)>a

; equals the function's exponential growth rate.

State the two sufficient conditions for existence of

\mathcal{L}\{f\}

f

is piecewise continuous on

[0,\infty)

and of exponential order

a

(

|f(t)|\le Me^{at}

Compute

\mathcal{L}\{1\}

and its ROC.

\frac{1}{s}

, with

\operatorname{Re}(s)>0

Compute

\mathcal{L}\{e^{at}\}

and its ROC.

\frac{1}{s-a}

, with

\operatorname{Re}(s)>a

Compute

\mathcal{L}\{t\}

and its ROC.

\frac{1}{s^2}

, with

\operatorname{Re}(s)>0

Compute

\mathcal{L}\{\sin\omega t\}

and its ROC.

\frac{\omega}{s^2+\omega^2}

, with

\operatorname{Re}(s)>0

Why does

f(t)=e^{t^2}

have no Laplace transform?

It grows faster than any

e^{at}

, so

e^{-st}e^{t^2}\to\infty

for every

s

and the integral diverges.

The convergence condition is on which part of

s

\operatorname{Re}(s)=\sigma

, since

|e^{-st}|=e^{-\sigma t}

Bound showing convergence for exponential-order

f

|F(s)|\le \frac{M}{\sigma-a}

for

\sigma>a

, hence finite (converges).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Laplace transform basically ek machine hai jo time ke function $f(t)$ ko ek naye variable $s$ ke function $F(s)$ mein convert kar deta hai. Formula hai $F(s)=\int_0^\infty e^{-st}f(t)\,dt$ . Yeh $e^{-st}$ jo kernel hai, woh ek "fade-out" filter ki tarah kaam karta hai — jaise time badhta hai, yeh function ko dabaata jaata hai, taaki infinite integral ka answer finite mile. Bonus yeh ki differentiation algebra ban jaati hai (multiply by $s$ ), isliye ODEs solve karna easy ho jaata hai.

Ab Region of Convergence (ROC) ka funda: integral hamesha converge nahi karta. Agar $f(t)$ tezi se grow karta hai jaise $e^{at}$ , toh integrand $e^{(a-\sigma)t}$ jaisa behave karta hai, jahan $\sigma=\operatorname{Re}(s)$ . Yeh tabhi decay karega jab $\sigma>a$ . Matlab ROC hamesha ek right half-plane hota hai: $\operatorname{Re}(s)>a$ . Yeh $a$ ko abscissa of convergence kehte hain, aur yeh function ki growth rate hai.

Existence ke liye do conditions: $f$ piecewise continuous ho aur exponential order ka ho (yaani $|f(t)|\le Me^{at}$ ). Agar function bahut hi zyada tezi se grow kare — jaise $e^{t^2}$ — toh koi bhi $s$ kaam nahi karega, integral har jagah diverge, aur uska Laplace transform exist hi nahi karta. Isliye answer likhte waqt ROC zaroor likho, sirf $\frac{1}{s-a}$ likhna adhoora hai.

Yaad rakhne ka tarika: "fade fast enough, right half-plane mein sab set." Teen results muh-zabaani aane chahiye: $\mathcal{L}\{1\}=\frac1s$ , $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ , aur $\mathcal{L}\{t\}=\frac{1}{s^2}$ — sab ke saath $\operatorname{Re}(s)$ wali condition. Bas itna pakka, toh chapter ka 80% clear.

Laplace transform — definition, region of convergence

1. The Definition (built from scratch)

2. Region of Convergence (ROC)

3. Common Mistakes (Steel-manned)

4. Active Recall

5. The 80/20 (what to truly own)

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections