Laplace transform — definition, region of convergence: Worked examples

Before anything, one reminder we lean on constantly. The variable $s$ is a complex number, written $s = σ + iω$ , where $σ = Re (s)$ is its "real part" (a plain number on the horizontal axis) and $ω$ is its "imaginary part." The only thing that decides whether the infinite integral settles down is $σ$ , because the size of the kernel is $∣ e^{- s t} ∣ = ∣ e^{- (σ + iω) t} ∣ = e^{- σ t} \cdot = 1 ∣ e^{- iω t} ∣ = e^{- σ t} .$ The $e^{- iω t}$ piece just spins on a circle of radius $1$ — it never changes size. So convergence is a race between $e^{- σ t}$ shrinking and $f (t)$ growing. Hold that picture.

One more tool we quote several times, so let's pin its exact meaning here (the parent defined it too):

The scenario matrix

Every Laplace problem this topic can throw at you falls into one of these cells. Each worked example below is tagged with the cell(s) it covers. (Reminder: ROC = Region Of Convergence.)

#	Case class	What's special	Example
A	Positive growth rate $f = e^{a t}, a > 0$	ROC boundary is at $s > a > 0$	Ex 1
B	Negative growth rate $f = e^{a t}, a < 0$	ROC extends left of $0$ : $s > a$	Ex 2
C	Zero growth (bounded / constant)	$a = 0$ , ROC is $s > 0$	Ex 3
D	Oscillation (sine / cosine)	complex $a = \pm iω$ , real answer	Ex 4
E	Polynomial × exponential	needs shifting or repeated IBP	Ex 5
F	Degenerate: no transform exists	grows faster than any $e^{a t}$	Ex 6
G	Limiting case ( $a \to 0$ recovers a simpler formula)	continuity check	Ex 7
H	Piecewise / shifted "switch-on"	integral splits at a jump	Ex 8
I	Word problem (radioactive decay signal)	build $f (t)$ from a story	Ex 9
J	Exam twist (same $F (s)$ , which ROC?)	ROC decides the answer	Ex 10

Intuition Edge case worth its own paragraph: what happens exactly ON the boundary

$σ = a$ ? Every right-sided ROC below is written with a strict inequality $Re (s) > a$ . Why not $\geq$ ? On the boundary line $σ = a$ the race is a tie: the kernel's decay $e^{- σ t}$ exactly cancels the function's growth $e^{a t}$ , leaving something that does not shrink. For $f = e^{a t}$ at $σ = a$ the integrand is literally $e^{0} = 1$ , and $\int_{0}^{\infty} 1 d t$ diverges — so the boundary is excluded. For a bounded oscillation like $cos ω t$ , exactly on $σ = 0$ the integrand $cos ω t$ never decays either; it oscillates forever and the integral fails to converge absolutely (at best it wobbles). So the honest statement is: the transform converges (absolutely) on the open half-plane $Re (s) > a$ , and generally not on the boundary line itself. Only special finite-energy cases can converge conditionally there — we do not rely on that. Keep the strict $>$ .

Ex 1 — Cell A: positive growth rate

Worked example

$L {e^{2 t}}$ and its ROC Forecast (guess first): the function blows up like $e^{2 t}$ . For the kernel $e^{- σ t}$ to win the race, we'll need $σ$ bigger than $2$ . Guess: $F (s) = \frac{1}{s - 2}$ , ROC $Re (s) > 2$ .

1. Write the integral and merge the exponents. $F (s) = \int_{0}^{\infty} e^{- s t} e^{2 t} d t = \int_{0}^{\infty} e^{- (s - 2) t} d t .$ Why this step? Two exponentials multiply by adding exponents; merging turns a product into a single decaying exponential we already know how to integrate.

2. Integrate. $= [\frac{- 1}{s - 2} e^{- (s - 2) t}]_{0}^{\infty} .$ Why this step? $\int e^{k t} d t = \frac{1}{k} e^{k t}$ ; here $k = - (s - 2)$ .

3. Evaluate both limits.

Upper limit $t \to \infty$ : the size is $e^{- (σ - 2) t} \to 0$ only if $σ - 2 > 0$ , i.e. $Re (s) > 2$ .
Lower limit $t = 0$ : $e^{0} = 1$ . $F (s) = 0 - (\frac{- 1}{s - 2}) = \frac{1}{s - 2}, Re (s) > 2.$

Verify: growth rate $a = 2$ appears exactly as the ROC boundary — matches the parent's rule "ROC = right half-plane past the growth rate." Forecast confirmed. ✓

Ex 2 — Cell B: negative growth rate

Worked example

$L {e^{- 3 t}}$ and its ROC Forecast: now the function decays. The kernel barely needs to help. We should be able to allow $σ$ down to $- 3$ . Guess: $F (s) = \frac{1}{s + 3}$ , ROC $Re (s) > - 3$ .

1. Merge exponents: $F (s) = \int_{0}^{\infty} e^{- s t} e^{- 3 t} d t = \int_{0}^{\infty} e^{- (s + 3) t} d t .$ Why this step? Same "add exponents" move; here the two decays reinforce each other.

2. Integrate and evaluate. $= [\frac{- 1}{s + 3} e^{- (s + 3) t}]_{0}^{\infty} .$ Why this step? We use the standard exponential integral $\int e^{k t} d t = \frac{1}{k} e^{k t}$ with $k = - (s + 3)$ , so the antiderivative is $\frac{- 1}{s + 3} e^{- (s + 3) t}$ . Then the upper limit $\to 0$ iff $Re (s + 3) > 0 ⟺ Re (s) > - 3$ , and the lower limit at $t = 0$ gives $\frac{- 1}{s + 3}$ .

3. Collect: $F (s) = \frac{1}{s + 3}, Re (s) > - 3.$

Verify: the abscissa of convergence is $a = - 3$ , which is left of the imaginary axis. This is the key contrast with Ex 1: the ROC boundary follows the sign of the growth rate. ✓

Figure — Laplace transform — definition, region of convergence

The figure above shows both ROCs as shaded half-planes on the $s$ -plane. Notice the shaded region for $e^{2 t}$ (magenta boundary at $σ = 2$ ) sits entirely to the right, while $e^{- 3 t}$ (violet boundary at $σ = - 3$ ) reaches far left. Both are half-planes opening rightward — that shape never changes.

Ex 3 — Cell C: zero growth (constant)

Worked example

$L {5}$ Forecast: a constant doesn't grow at all ( $a = 0$ ), so ROC should be $Re (s) > 0$ . Since the parent showed $L {1} = \frac{1}{s}$ , linearity should give $\frac{5}{s}$ .

1. Pull the constant out: $F (s) = \int_{0}^{\infty} e^{- s t} \cdot 5 d t = 5 \int_{0}^{\infty} e^{- s t} d t .$ Why this step? The integral is linear — a constant multiplier passes straight through, so we reduce to the known $L {1}$ .

2. Use $L {1} = \frac{1}{s}$ (valid for $Re (s) > 0$ ): $F (s) = 5 \cdot \frac{1}{s} = \frac{5}{s}, Re (s) > 0.$ Why this step? Rather than integrate again from scratch, we plug in the already-earned anchor result $L {1} = \frac{1}{s}$ (proved in the parent note); reusing a known transform is faster and its ROC $Re (s) > 0$ carries over unchanged because multiplying by the constant $5$ cannot affect where the integral converges.

Verify: plug $s = 1$ : $F (1) = 5$ . Directly, $\int_{0}^{\infty} 5 e^{- t} d t = 5 \cdot [- e^{- t}]_{0}^{\infty} = 5 (0 - (- 1)) = 5$ . ✓

Ex 4 — Cell D: oscillation

Worked example

$L {cos (2 t)}$ Forecast: cosine is bounded, so it's exponential order $a = 0$ → ROC $Re (s) > 0$ . Bounded oscillation → answer should be real and involve $s^{2} + 4$ . Guess $\frac{s}{s ^{2} + 4}$ .

1. Rewrite cosine using exponentials (Euler): $cos 2 t = \frac{e ^{i 2 t} + e ^{- i 2 t}}{2} .$ Why this step? We already own $L {e^{a t}} = \frac{1}{s - a}$ . Turning cosine into exponentials lets us reuse that instead of integrating by parts twice. The tool is chosen because it converts oscillation into two things we've already solved.

2. Apply the exponential result with $a = + i 2$ and $a = - i 2$ : $F (s) = \frac{1}{2} (\frac{1}{s - i 2} + \frac{1}{s + i 2}) .$ Why this step? Linearity again; each piece is a known transform.

3. Add the fractions over the common denominator $(s - i 2) (s + i 2) = s^{2} - (i 2)^{2} = s^{2} + 4$ : $F (s) = \frac{1}{2} \cdot \frac{( s + i 2 ) + ( s - i 2 )}{s ^{2} + 4} = \frac{1}{2} \cdot \frac{2 s}{s ^{2} + 4} = \frac{s}{s ^{2} + 4}, Re (s) > 0.$ Why the ROC? Each exponential $e^{\pm i 2 t}$ has size $1$ (pure oscillation, $a = 0$ ), so both need $Re (s) > 0$ ; the joint region is $Re (s) > 0$ . (Exactly on the boundary $Re (s) = 0$ the integrand never decays — the boundary edge case flagged above.)

Verify: at $s = 2$ , $F (2) = \frac{2}{4 + 4} = \frac{1}{4}$ . And $\frac{s}{s ^{2} + 4}$ is real for real $s$ — matches "real function → real transform." ✓

Ex 5 — Cell E: polynomial × exponential

Worked example

$L {t e^{- t}}$ Forecast: we know $L {t} = \frac{1}{s ^{2}}$ (ROC $Re (s) > 0$ ). Multiplying $f (t)$ by $e^{- t}$ should shift $s \to s + 1$ (a decay makes convergence easier, pushing the ROC left). Guess $\frac{1}{( s + 1 ) ^{2}}$ , ROC $Re (s) > - 1$ .

1. Set up and merge the exponential: $F (s) = \int_{0}^{\infty} t e^{- t} e^{- s t} d t = \int_{0}^{\infty} t e^{- (s + 1) t} d t .$ Why this step? Grouping the two exponentials shows that the whole problem is just " $L {t}$ but with $s$ replaced by $s + 1$ " — the first shifting property in action.

2. Recognise the pattern. Since $\int_{0}^{\infty} t e^{- k t} d t = \frac{1}{k ^{2}}$ (from $L {t} = \frac{1}{s ^{2}}$ with $k$ in place of $s$ ), set $k = s + 1$ : $F (s) = \frac{1}{( s + 1 ) ^{2}} .$ Why this works? The integral only "sees" the combined constant $k = s + 1$ ; the derivation of $\frac{1}{s ^{2}}$ never assumed $s$ was special, so it holds for any $k$ with $Re (k) > 0$ .

3. ROC: need $Re (k) = Re (s) + 1 > 0$ , i.e. $Re (s) > - 1$ . $F (s) = \frac{1}{( s + 1 ) ^{2}}, Re (s) > - 1.$

Verify: at $s = 1$ : $F (1) = \frac{1}{4} = 0.25$ . Direct numeric integral of $t e^{- 2 t}$ from $0$ to $\infty$ is $\frac{1}{2 ^{2}} = 0.25$ . ✓

Ex 6 — Cell F: degenerate, NO transform

Worked example Does

$L {e^{t^{2}}}$ exist? Forecast: the parent warned this one grows faster than any $e^{a t}$ . Guess: no transform for any $s$ .

1. Look at the integrand's size for any real $σ$ : $∣ e^{- s t} e^{t^{2}} ∣ = e^{- σ t} e^{t^{2}} = e^{t^{2} - σ t} .$ Why this step? Convergence is decided by whether the magnitude eventually shrinks. So we examine the exponent $t^{2} - σ t$ .

2. Ask: does $t^{2} - σ t \to - \infty$ (needed for decay) or $+ \infty$ (blow-up)? Factor: $t^{2} - σ t = t (t - σ)$ . For any fixed $σ$ , once $t > σ$ both factors are positive, so $t (t - σ) \to + \infty$ . Why this step? This shows the exponent grows without bound no matter how large we pick $σ$ — the $t^{2}$ term always eventually dominates the linear $σ t$ .

3. Therefore $e^{t^{2} - σ t} \to \infty$ , the integrand grows, and the integral diverges for every $s$ .

Verify (against the definition of exponential order above): recall we need a size bound $M > 0$ and a start time $T$ with $e^{t^{2}} \leq M e^{a t}$ for all $t \geq T$ . Rearranging, this demands $\frac{e ^{t^{2}}}{e ^{a t}} = e^{t^{2} - a t} \leq M$ — i.e. the ratio stays below the fixed ceiling $M$ . But $e^{t^{2} - a t} \to \infty$ , so it eventually punches through any ceiling $M$ you name, for any growth rate $a$ . No valid $(M, a)$ pair exists, so $e^{t^{2}}$ is not of exponential order and the existence theorem's hypothesis fails — correctly predicting no transform. ✓ (Nothing to check numerically — that is the answer.)

Ex 7 — Cell G: limiting case (continuity)

Worked example Recover

$L {1}$ from $L {e^{a t}}$ as $a \to 0$ Forecast: setting $a = 0$ turns $e^{a t}$ into the constant $1$ . So $\frac{1}{s - a}$ should smoothly become $\frac{1}{s}$ . Guess: yes, no surprises.

1. Start from the general result $L {e^{a t}} = \frac{1}{s - a}$ , valid for $Re (s) > a$ . Why this step? We test the machine's consistency: a special case ( $a = 0$ ) should agree with the directly-computed answer $L {1} = \frac{1}{s}$ .

2. Take the limit: $lim_{a \to 0} \frac{1}{s - a} = \frac{1}{s - 0} = \frac{1}{s} .$ The ROC boundary $Re (s) > a$ slides to $Re (s) > 0$ . Why this step? Since $\frac{1}{s - a}$ is a continuous function of the parameter $a$ (as long as $s \neq = a$ ), letting $a \to 0$ is legitimate and simply substitutes $a = 0$ ; this is exactly how we confirm the two formulas agree at the shared value $a = 0$ instead of merely asserting it.

Verify: at $s = 3$ : $\frac{1}{3 - a} \to \frac{1}{3}$ as $a \to 0$ , and $L {1} (3) = \frac{1}{3}$ . Perfect match — the formula family is continuous in $a$ . ✓

Ex 8 — Cell H: piecewise / switch-on

Worked example Signal switched on at

$t = 0$ , off at $t = 1$ : $f (t) = {10 0 \leq t < 1 t \geq 1$ Forecast: finite-duration signal — the integral only runs over $[0, 1)$ , a finite interval, so it converges for all $s$ (no infinite tail to fight, so no growth condition at all). Answer should be $\frac{1 - e ^{- s}}{s}$ .

1. Because $f = 0$ after $t = 1$ , split the infinite integral and drop the empty tail: $F (s) = \int_{0}^{1} e^{- s t} \cdot 1 d t + \int_{1}^{\infty} e^{- s t} \cdot 0 d t = \int_{0}^{1} e^{- s t} d t .$ Why this step? A piecewise function must be integrated piece by piece; the second piece contributes nothing.

2. Integrate over the finite interval: $= [\frac{- 1}{s} e^{- s t}]_{0}^{1} = \frac{- 1}{s} e^{- s} - (\frac{- 1}{s}) = \frac{1 - e ^{- s}}{s} .$ Why this step? We again use $\int e^{- s t} d t = \frac{- 1}{s} e^{- s t}$ , but now over a finite interval, so we simply substitute the two endpoints $t = 1$ and $t = 0$ — there is no limit-to-infinity to worry about, hence no convergence condition arises here.

ROC — the whole plane. Because both endpoints are finite, the integral $\int_{0}^{1} e^{- s t} d t$ is a plain integral of a bounded, continuous function over a bounded interval — it produces a finite number for every complex $s$ , with no exception. In particular at $s = 0$ : the closed form $\frac{1 - e ^{- s}}{s}$ looks like $\frac{0}{0}$ , but that is only an artefact of the algebra (a removable singularity fixed by analytic continuation); the actual integral there is $\int_{0}^{1} 1 d t = 1$ , perfectly finite.

Handling the removable singularity at $s = 0$ properly. The closed form $\frac{1 - e ^{- s}}{s}$ is undefined as written at $s = 0$ (division by zero), yet we just saw the transform genuinely equals $1$ there. Both facts are true and compatible: $s = 0$ is a removable singularity — a hole in the formula that the underlying function fills in smoothly. We confirm the fill-in value by a limit (using $e^{- s} \approx 1 - s$ for tiny $s$ ): $lim_{s \to 0} \frac{1 - e ^{- s}}{s} = lim_{s \to 0} \frac{1 - ( 1 - s + \dots )}{s} = lim_{s \to 0} \frac{s - \dots}{s} = 1,$ which matches the direct integral $\int_{0}^{1} 1 d t = 1$ . So we extend the formula by continuity, defining $F (0) := 1$ . The honest closed form is therefore a two-line definition: $F (s) = \frac{1 - e ^{- s}}{s} (s \neq = 0), F (0) = 1; ROC = all of the s -plane .$

Verify: at $s = 1$ : $F (1) = \frac{1 - e ^{- 1}}{1} = 1 - 0.367879 \dots = 0.632120 \dots$ . Direct: $\int_{0}^{1} e^{- t} d t = 1 - e^{- 1} = 0.632120 \dots$ . And the removable-singularity fill-in checks out: $lim_{s \to 0} \frac{1 - e ^{- s}}{s} = 1 = \int_{0}^{1} 1 d t$ . ✓

Ex 9 — Cell I: word problem

Worked example A radioactive sample emits a decaying signal

$f (t) = A e^{- λ t}$ (amplitude $A = 4$ counts/s, decay constant $λ = 0.5 s^{- 1}$ ). Find its Laplace transform and ROC. Forecast: decaying exponential → Cell B pattern. Expect $\frac{4}{s + 0.5}$ , ROC $Re (s) > - 0.5$ .

1. Translate the story into a function: $f (t) = 4 e^{- 0.5 t}$ for $t \geq 0$ . Why this step? "Decays with constant $λ$ " means multiply by $e^{- λ t}$ ; the amplitude $A$ is the starting value at $t = 0$ .

2. Use linearity + the $e^{a t}$ result with $a = - 0.5$ : $F (s) = 4 \cdot \frac{1}{s - ( - 0.5 )} = \frac{4}{s + 0.5}, Re (s) > - 0.5.$ Why this tool? We already solved $L {e^{a t}}$ ; the word problem is that result dressed in physical units. Reusing it (with $a = - λ = - 0.5$ ) beats re-integrating, and the constant $A = 4$ passes through by linearity without touching the ROC.

3. Units sanity: $f$ has units of counts/s; the transform variable $s$ has units of $s^{- 1}$ , so $F (s) = \int e^{- s t} f d t$ has units $(counts/s) \cdot s = counts$ . And $\frac{4}{s + 0.5}$ : numerator counts/s over denominator $s^{- 1}$ = counts. ✓

Verify: at $s = 1.5 s^{- 1}$ : $F = \frac{4}{1.5 + 0.5} = \frac{4}{2} = 2$ . ✓

Ex 10 — Cell J: exam twist (ROC decides everything)

Worked example Two problems produce the

same algebra $F (s) = \frac{1}{s - 1}$ . In problem $P$ we're told the signal is causal ( $f = 0$ for $t < 0$ ); its ROC is $Re (s) > 1$ . A tempting "trap answer" claims the ROC could instead be $Re (s) < 1$ . Which is correct for our right-sided definition, and why? Forecast: our whole course uses $\int_{0}^{\infty}$ (right-sided). So the ROC must be a right half-plane. Guess: $Re (s) > 1$ is the only valid one here.

1. Recall the shape rule. Every one-sided (right-sided) Laplace transform converges on a right half-plane $Re (s) > a$ , because the only convergence battle happens at the $\infty$ end, controlled by $e^{- σ t}$ . Why this step? The ROC shape is forced by which end of the integral can diverge; a $\int_{0}^{\infty}$ integral can only be tamed by making $σ$ large (rightward).

2. The abscissa $a$ is the growth rate. Here $\frac{1}{s - 1}$ is exactly the transform of $e^{1 \cdot t} = e^{t}$ (from $L {e^{a t}} = \frac{1}{s - a}$ with $a = 1$ ), and $e^{t}$ has growth rate $1$ . So $a = 1$ and the ROC is $Re (s) > 1$ . Why this step? The ROC boundary is always the growth rate $a$ of the underlying time-signal (proved in the parent: the bound $\frac{M}{σ - a}$ is finite iff $σ > a$ ). We read $a$ off by asking "which $e^{a t}$ produced this $\frac{1}{s - a}$ ?" — here the denominator $s - 1$ answers $a = 1$ .

3. Resolve the trap. The alternative $Re (s) < 1$ is a left half-plane. For our $\int_{0}^{\infty}$ definition that shape is impossible — a right-sided integral can never converge only to the left, because increasing $σ$ (moving right) only ever helps convergence, never hurts it. The left half-plane $Re (s) < 1$ actually belongs to a different, anticausal signal, namely $f (t) = - e^{t}$ for $t < 0$ and $0$ for $t \geq 0$ (a two-sided transform, outside our course's $\int_{0}^{\infty}$ definition). Same algebra $\frac{1}{s - 1}$ , different function, distinguished only by the ROC.

Conclusion. For the causal signal of problem $P$ : $F (s) = \frac{1}{s - 1}$ with ROC $Re (s) > 1$ . The trap answer $Re (s) < 1$ is rejected because it contradicts the right-half-plane shape our one-sided definition forces. This is precisely why the parent insists you always state the half-plane — the algebra alone is ambiguous; the ROC is the tie-breaker.

Verify: for our causal $e^{t}$ , take $s = 2$ (inside $Re (s) > 1$ ): $\int_{0}^{\infty} e^{- 2 t} e^{t} d t = \int_{0}^{\infty} e^{- t} d t = 1 = \frac{1}{2 - 1}$ . ✓ Consistent only in the right half-plane; picking $s$ with $Re (s) < 1$ would make the original integral diverge.

Wrap-up recall

Recall Match each cell to its lesson

Cell A (positive growth) ::: ROC starts to the right of $0$ : $Re (s) > a > 0$ . Cell B (negative growth) ::: ROC extends left of $0$ : $Re (s) > a$ with $a < 0$ . Cell C (constant) ::: $a = 0$ , ROC $Re (s) > 0$ , transform is $\frac{c}{s}$ . Cell D (oscillation) ::: use Euler to reuse $\frac{1}{s - a}$ ; real answer, ROC $Re (s) > 0$ . Cell E (poly × exp) ::: shifting $s \to s + 1$ gives $\frac{1}{( s + 1 ) ^{2}}$ . Cell F (super-fast growth) ::: not exponential order → NO transform for any $s$ . Cell G (limit $a \to 0$ ) ::: $\frac{1}{s - a} \to \frac{1}{s}$ ; formulas are continuous. Cell H (finite duration) ::: split at the jump; converges for the ==entire $s$ -plane== (including $s = 0$ via the removable-singularity fill-in $F (0) = 1$ ). Cell I (word problem) ::: build $f (t) = A e^{- λ t}$ from the story, then reuse $\frac{1}{s - a}$ with $a = - λ$ . Cell J (ambiguous algebra) ::: the ROC is what distinguishes two signals with the same $F (s)$ . On the boundary $σ = a$ ::: the race is a tie; generally diverges, so ROC is the strict-open half-plane.

Connections

Laplace Transform of Derivatives — Ex 5's shifting trick reappears when derivatives hit.
Solving ODEs with Laplace Transforms — these building-block transforms are the toolkit.
Inverse Laplace Transform — Ex 10 is exactly why inversion needs the ROC.
Improper Integrals — the convergence tests in Ex 1, 2, 6.
Exponential Order and Growth Rates — Cell F's failure explained in full.
Fourier Transform — the boundary line $Re (s) = 0$ touched in Ex 4.

#flashcard