Worked examples — Transforms of standard functions — proofs

Everything below leans on the one machine: $\mathcal{L}\{f(t)\}={\int }_0^{\infty }{e}^{-st}f(t)dt,$ and the dictionary built from it. If any symbol here feels unfamiliar, the Laplace Transform — Definition and Existence and Linearity and First Shifting Theorem notes build them from zero.

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The scenario matrix

Before working anything, let us list every kind of cell these transforms live in. A worked example is only useful if it lands in a named cell — so each example below is tagged Cell Xn.

Cell	Situation class	What could go wrong / what to watch
A	Constant / pure power $t^n$	forgetting the $n!$; wrong power of $s$
B	Exponential $e^{at}$, positive $a$	convergence needs $s>a$ (strip shifts)
C	Exponential $e^{at}$, negative $a$	pole at $s=-
D	Trig $\sin /\cos$ (oscillating)	denominator $s^2+a^2$, plus sign
E	Hyperbolic $\sinh /\cosh$ (growing)	denominator $s^2-a^2$, minus, need $s>|a|$
F	Shift $e^{at}f(t)$ (First Shifting)	replace $s\to s-a$ everywhere
G	Zero / degenerate input ($a=0$, $n=0$)	formulas must collapse to simpler ones
H	Limiting behaviour ($s\to \infty$, $a\to 0$)	sanity checks, not new results
I	Linear combination (sum of several)	linearity, term by term
J	Word problem (physics / ODE)	set up $f(t)$, keep units, interpret

The ten examples below cover all ten cells.

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Example 1 — Pure power (Cell A)

Steps.

1. Apply $\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$ with $n=3$. Why this step? This is the direct dictionary entry — no need to redo the integral. $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4},s>0.$
2. For the constant, pull the number out (linearity) and use $\mathcal{L}\{1\}=1/s$. Why this step? $\mathcal{L}$ is linear: constants slide through the integral. $\mathcal{L}\{5\}=5\cdot \frac{1}{s}=\frac{5}{s},s>0.$

Verify: check the $n!$ isn't dropped — $\mathcal{L}\{t^1\}$ should be $1!/s^2=1/s^2$. It does. And $t^3$ growing means we still only need $s>0$ (the weight $e^{-st}$ beats any polynomial). ✓

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Example 2 — Positive exponential (Cell B)

Steps.

1. Use the dictionary entry $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ with $a=3$. Why this step? It is exactly the $\mathcal{L}\{1\}$ proof but with $s\to s-a$. $\mathcal{L}\{e^{3t}\}=\frac{1}{s-3}.$
2. Convergence: we needed $s-a>0$, i.e. $s>3$. Why this step? Only then does $e^{-(s-a)t}\to 0$ as $t\to \infty$.

Verify: the pole sits at $s=3$, exactly the abscissa of convergence, matching "the fastest growth of $f$ sets the threshold." ✓

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Example 3 — Negative exponential (Cell C)

Steps.

1. Put $a=-2$ into $\frac{1}{s-a}$. Why this step? The formula is signed — $a$ can be negative, no special case needed. $\mathcal{L}\{e^{-2t}\}=\frac{1}{s+2},s>-2.$

Verify: the strip $s>-2$ is looser than Cell B's $s>3$ because a decaying input is "easy to weigh." The two exponential cells B and C together cover both signs of $a$. ✓

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Example 4 — Oscillating (Cell D)

Steps.

1. Apply $\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2}$, $a=4$. Why this step? Direct dictionary entry (proved via $e^{iat}$). $\mathcal{L}\{\sin 4t\}=\frac{4}{s^2+16},s>0.$
2. Apply $\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2}$, $a=4$. $\mathcal{L}\{\cos 4t\}=\frac{s}{s^2+16},s>0.$

Look at the figure: the blue oscillating $\sin 4t$ times the orange decaying weight $e^{-st}$ gives the green product, whose net signed area is exactly the transform. The plus/minus lobes nearly cancel — that is why the answer is a modest fraction, not something huge.

Verify: at $s\to \infty$ both transforms $\to 0$ (weight crushes everything). Setting $a=4$, $s=3$: $\mathcal{L}\{\sin 4t\}=4/(9+16)=4/25=0.16$. ✓

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Example 5 — Growing hyperbolic (Cell E)

Steps.

1. $\mathcal{L}\{\cosh at\}=\frac{s}{s^2-a^2}$, $a=2$. Why this step? Built from $\cosh at=\frac{1}{2}(e^{at}+e^{-at})$ plus linearity. $\mathcal{L}\{\cosh 2t\}=\frac{s}{s^2-4},s>2.$
2. $\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2}$, $a=2$. $\mathcal{L}\{\sinh 2t\}=\frac{2}{s^2-4},s>2.$

The figure contrasts the plus (trig) and minus (hyperbolic) denominators as functions of $s$: the hyperbolic curve has a vertical wall (a real pole) at $s=a$, which is exactly the boundary $s>|a|$ of its strip. The trig curve is smooth — no real pole, so $s>0$ suffices.

Verify: at $s=3,a=2$: $\mathcal{L}\{\cosh 2t\}=3/(9-4)=3/5=0.6$ and $\mathcal{L}\{\sinh 2t\}=2/5=0.4$. Note $s=3>2$ keeps us legal. ✓

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Example 6 — First Shifting Theorem (Cell F)

Steps.

1. Get the base transform: $\mathcal{L}\{\cos 3t\}=\frac{s}{s^2+9}=F(s)$. Why this step? Shifting acts on an already-known transform.
2. Replace every $s$ by $s+1$. Why this step? The shift theorem: $\mathcal{L}\{e^{-t}\cos 3t\}=F(s+1)$. $\mathcal{L}\{e^{-t}\cos 3t\}=\frac{(s+1)}{(s+1{)}^2+9},s>-1.$

Verify: strip moves from $s>0$ to $s>-1$ (shifted left by $1$, matching $a=-1$). If we instead had $e^{+t}\cos 3t$, the strip would be $s>1$. ✓

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Example 7 — Degenerate / zero inputs (Cell G)

Steps.

1. Cosine at $a=0$: $\frac{s}{s^2+a^2}{|}_{a=0}=\frac{s}{s^2}=\frac{1}{s}$. Why this step? Set $a=0$ directly and simplify — this is the degenerate limit. $\mathcal{L}\{\cos 0\}=\frac{1}{s}=\mathcal{L}\{1\}.✓$
2. Sine at $a=0$: $\frac{a}{s^2+a^2}{|}_{a=0}=\frac{0}{s^2}=0$. $\mathcal{L}\{\sin 0\}=0=\mathcal{L}\{0\}.✓$
3. Power at $n=0$: $\frac{n!}{s^{n+1}}{|}_{n=0}=\frac{0!}{s^1}=\frac{1}{s}$ (recall $0!=1$). $\mathcal{L}\{t^0\}=\frac{1}{s}.✓$

Verify: all three degenerate cases fold onto the single anchor $\mathcal{L}\{1\}=1/s$ (or $0$). Consistency confirmed. ✓

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Example 8 — Limiting behaviour as a sanity engine (Cell H)

Steps.

1. $s\to \infty$: $\frac{a}{s^2+a^2}\to 0$. Why this step? The denominator grows without bound — every transform must decay in $s$.
2. $a\to 0$: $\frac{a}{s^2+a^2}\to \frac{0}{s^2}=0$, agreeing with Cell G.
3. Initial-value shape: $s\cdot \frac{s}{s^2+a^2}=\frac{s^2}{s^2+a^2}\stackrel{s\to \infty }{\to }1=\cos 0$. Why this step? ${\lim }_{s\to \infty }sF(s)=f(0^{+})$ is a free correctness check.

Verify: with $a=5$: ${\lim }_{s\to \infty }\frac{s^2}{s^2+25}=1$, and $\cos (0)=1$. Match. ✓

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Example 9 — Linear combination (Cell I)

Steps.

1. $\mathcal{L}\{3\}=\frac{3}{s}$; $\mathcal{L}\{-2e^{4t}\}=-\frac{2}{s-4}$. Why this step? Linearity lets us handle each term separately.
2. $\mathcal{L}\{t^2\}=\frac{2!}{s^3}=\frac{2}{s^3}$; $\mathcal{L}\{-5\sin 3t\}=-\frac{15}{s^2+9}$.
3. Add and record the strip. $\mathcal{L}\{\cdots \}=\frac{3}{s}-\frac{2}{s-4}+\frac{2}{s^3}-\frac{15}{s^2+9},s>4.$ Why $s>4$? Every term must converge simultaneously; the growing $e^{4t}$ demands the strictest bound.

Verify: each numerator carries the right factorial/coefficient; the strip is the max of $\{0,4,0,0\}=4$. ✓

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Example 10 — Word problem: a damped spring (Cell J)

Steps.

1. Base transform: $\mathcal{L}\{\sin 2t\}=\frac{2}{s^2+4}=F(s)$. Why this step? Peel off the exponential; transform the plain sine first.
2. Apply First Shifting with $a=-1$: replace $s\to s+1$. Why this step? $\mathcal{L}\{e^{-t}\sin 2t\}=F(s+1)$. $X(s)=\frac{2}{(s+1{)}^2+4},s>-1.$
3. Initial-value check: ${\lim }_{s\to \infty }sX(s)={\lim }_{s\to \infty }\frac{2s}{(s+1{)}^2+4}=0=x(0)$. Why this step? ${\lim }_{s\to \infty }sX(s)=x(0^{+})$; a released-from-rest sine should give $0$.

Verify (units + number): $X(s)$ has units of metre·seconds (a displacement integrated over time), consistent with $\int x{e}^{-st}dt$. At $s=1$: $X(1)=\frac{2}{(2{)}^2+4}=\frac{2}{8}=0.25$. And $\lim sX(s)=0$ matches $x(0)=e^0\sin 0=0$. ✓ This $X(s)$ is exactly the shape you would invert with Inverse Laplace Transform — Partial Fractions to recover the motion.

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Coverage check

Recall Did we hit every cell?

A (Ex 1) · B (Ex 2) · C (Ex 3) · D (Ex 4) · E (Ex 5) · F (Ex 6) · G (Ex 7) · H (Ex 8) · I (Ex 9) · J (Ex 10). Every sign of $a$, both zero cases, the limiting checks, a linear pile-up, and a physics setup — all covered. Which cell forced the tightest convergence strip in Example 9? ::: Cell B's $e^{4t}$ term, giving $s>4$. Why is the hyperbolic strip $s>|a|$ and not just $s>a$? ::: Both $e^{at}$ and $e^{-at}$ pieces must converge, so $s>a$ and $s>-a$.

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Connections

• 4.6.26 Transforms of standard functions — proofs (Hinglish) — the parent proofs
• Laplace Transform — Definition and Existence
• Linearity and First Shifting Theorem — used in Cells F, I, J
• Transform of Derivatives — solving ODEs — the payoff of this dictionary
• Inverse Laplace Transform — Partial Fractions — undoing Example 10
• Integration by Parts · Euler's Formula $e^{i\theta}$ · Improper Integrals

Concept Map