Visual walkthrough — Transforms of standard functions — proofs

Everything below assumes you have met the definition from Laplace Transform — Definition and Existence, but I will re-earn every symbol anyway.

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Step 1 — What "transform" even means (the weighing machine)

WHAT. We have a function $f(t)$. The letter $t$ is time, and it only runs forward: $t\ge 0$. A transform is a machine that takes this whole function and produces one new function of a different variable, which we call $s$.

WHY. Why bother swapping $t$ for $s$? Because in the $s$-world, the hard operation of calculus (differentiation) turns into the easy operation of algebra (multiply by $s$). That superpower is the whole reason the machine exists — see Transform of Derivatives — solving ODEs. But to use it we first need to know what the machine does to simple inputs.

PICTURE. Think of a set of scales. For each moment $t$ we place a weight on the pan. The weight is $e^{-st}$ — it starts heavy near $t=0$ and fades as time goes on. We multiply the function's value at each moment by that fading weight, then add up everything. The total is one number. Change $s$, get a different total. All those totals together form the new function $F(s)$.

$$F(s)=\underset{\begin{array}{c}\text{add up over}\\ \text{all time }t\ge 0\end{array}}{\underbrace{{\int }_0^{\infty }}}\underset{\begin{array}{c}\text{fading}\\ \text{weight}\end{array}}{\underbrace{e^{-st}}}\underset{\begin{array}{c}\text{the value}\\ \text{being weighed}\end{array}}{\underbrace{f(t)}}dt$$

• $s$ is a knob we hold fixed while integrating. $t$ is the variable we sum over and which disappears after the integral.
• ${\int }_0^{\infty }$ is an improper integral — the top limit is infinity, so we secretly mean "let the top climb to $\infty$ and see where the area settles" (details in Improper Integrals).

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Step 2 — Look at the weight $e^{-st}$ before we use it

WHAT. Before feeding anything in, stare at the weight alone: the curve $t\mapsto e^{-st}$.

WHY. Everything about convergence — the "$s>0$" conditions scattered through the dictionary — comes from whether this curve dies fast enough. If we understand the weight, we understand every condition for free.

PICTURE. For $s>0$ the curve starts at height $1$ (because $e^0=1$) and slides down toward $0$. The bigger $s$ is, the faster it dives. For $s=0$ it is a flat line at height $1$ (never dies). For $s<0$ it grows and shoots to infinity — hopeless for adding up a finite area.

• $e^{-st}$: base $e\approx 2.718$, exponent $-st$. Negative exponent ⇒ shrinking, as long as $s>0$.
• The shaded area under this curve is exactly the number $\mathcal{L}\{1\}$ we are about to compute — the flat line $f(t)=1$ just leaves the weight untouched.

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Step 3 — Compute $\mathcal{L}\{1\}$: find the area under the weight

WHAT. Put $f(t)=1$ into the recipe. Then $e^{-st}\cdot 1=e^{-st}$, so we need the area under the curve of Step 2.

$$\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}dt.$$

WHY the antiderivative $e^{-st}/(-s)$? We need a function whose slope is $e^{-st}$. Differentiating $e^{-st}$ pulls down a factor $-s$ (chain rule). To cancel that unwanted $-s$, we divide by it in advance:

$$\frac{d}{dt}\left[\frac{e^{-st}}{-s}\right]=\frac{-s{e}^{-st}}{-s}=e^{-st}.✓$$

So the running area up to a top limit $T$ is

$${\int }_0^T{e}^{-st}dt={\left[\frac{e^{-st}}{-s}\right]}_0^T=\underset{\text{top}}{\underbrace{\frac{e^{-sT}}{-s}}}-\underset{\text{bottom}}{\underbrace{\frac{e^0}{-s}}}.$$

PICTURE. Watch the top edge $T$ slide rightward. The area fills up but never past a ceiling — it saturates. That ceiling is our answer.

• Bottom term ($t=0$): $e^0=1$, giving $-\frac{1}{-s}=\frac{1}{s}$.
• Top term ($t=T\to \infty$): here the condition is born. For $s>0$, $e^{-sT}\to 0$, so this term vanishes. For $s\le 0$ it does not vanish and the area runs off — no transform exists.

$$\boxed{\mathcal{L}\{1\}=0-\left(-\frac{1}{s}\right)=\frac{1}{s},s>0.}$$

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Step 4 — Shift the knob: $\mathcal{L}\{e^{at}\}$ is the same move

WHAT. Feed in a growing (or shrinking) exponential $f(t)=e^{at}$. Inside the integral two exponentials meet and merge:

$$\mathcal{L}\{e^{at}\}={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{-(s-a)t}dt.$$

• Combining exponents: $-st+at=-(s-a)t$. The knob $s$ has simply been replaced by $s-a$.

WHY. We do not redo any work. This integral is identical to Step 3 with the letter $s$ swapped for $s-a$. So copy the answer:

$$\boxed{\mathcal{L}\{e^{at}\}=\frac{1}{s-a},s-a>0\iff s>a.}$$

PICTURE. The input $e^{at}$ tries to lift the graph; the weight $e^{-st}$ tries to crush it. Only the net exponent $-(s-a)$ decides who wins. If $s>a$ the crushing wins and the area is finite; the closer $s$ creeps to $a$, the taller the leftover pile and the bigger $1/(s-a)$ grows — blowing up as $s\to a^{+}$.

• Net exponent $-(s-a)t$: sign decided by $s-a$. This single quantity carries every convergence condition on the exponential.
• This "shift the knob" idea is exactly the Linearity and First Shifting Theorem in embryo.

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Step 5 — Climb the ladder: $\mathcal{L}\{t^n\}$ by recursion

WHAT. Now feed a power $f(t)=t^n$ (with $n=0,1,2,\dots$). Name the answer $I_n={\int }_0^{\infty }{e}^{-st}{t}^{n}dt$. We attack it with Integration by Parts: $\int udv=uv-\int vdu$, choosing

$$u=t^n(\Rightarrow du=n{t}^{n-1}dt),dv=e^{-st}dt(\Rightarrow v=\frac{e^{-st}}{-s}).$$

WHY this choice? Differentiating $t^n$ lowers the power to $t^{n-1}$. That means the problem shrinks each time — a ladder we can climb down to something we already know ($I_0=\mathcal{L}\{1\}=1/s$).

$$I_n=\underset{\text{boundary term}}{\underbrace{{\left[t^n\frac{e^{-st}}{-s}\right]}_0^{\infty }}}+\frac{n}{s}\underset{=I_{n-1}}{\underbrace{{\int }_0^{\infty }{e}^{-st}{t}^{n-1}dt}}.$$

The boundary term is zero:

• At $t\to \infty$: $e^{-st}$ (for $s>0$) shrinks faster than any polynomial $t^n$ grows, so the product $\to 0$.
• At $t=0$: $t^n=0$ for $n\ge 1$.

So we get the ladder rule and unroll it:

$$I_n=\frac{n}{s}{I}_{n-1}=\frac{n}{s}\cdot \frac{n-1}{s}\cdots \frac{1}{s}\cdot \underset{I_0}{\underbrace{\frac{1}{s}}}=\frac{n!}{s^{n+1}}.$$ $$\boxed{\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}},s>0.}$$

PICTURE. Each step of the ladder multiplies by $\frac{n}{s}$; stacking all the rungs multiplies $n\cdot (n-1)\cdots 1=n!$ into the numerator and adds one power of $s$ to the bottom each time.

• $n!$ (read "n factorial") $=n\times (n-1)\times \cdots \times 1$. It is not decoration — it is the product of every rung you stepped on. Forgetting it is a classic error flagged in the parent note.

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Step 6 — The wiggle: $\sin at$ and $\cos at$ via Euler

WHAT. For the oscillating inputs we borrow Euler's Formula $e^{i\theta}$: $e^{iat}=\cos at+i\sin at$, where $i^2=-1$. Instead of integrating a wiggle directly, we reuse Step 4 with $a$ replaced by $ia$:

$$\mathcal{L}\{e^{iat}\}=\frac{1}{s-ia}.$$

WHY. A raw wiggle needs integration by parts twice. But a complex exponential integrates as easily as a real one — it is still just $e^{(\text{stuff})t}$. So we do the easy complex integral, then split it into a real part (cos) and an imaginary part (sin). To read those parts off, make the denominator real by multiplying top and bottom by the conjugate $s+ia$:

$$\frac{1}{s-ia}\cdot \frac{s+ia}{s+ia}=\frac{s+ia}{\underset{=s^2+a^2}{\underbrace{(s-ia)(s+ia)}}}=\frac{s}{s^2+a^2}+i\frac{a}{s^2+a^2}.$$

• $(s-ia)(s+ia)=s^2-(ia{)}^2=s^2-i^2{a}^2=s^2+a^2$ — the $i^2=-1$ flips the minus into a plus. That plus sign is why trig transforms carry $s^2+a^2$.

Because $\mathcal{L}$ is linear, $\mathcal{L}\{e^{iat}\}=\mathcal{L}\{\cos at\}+i\mathcal{L}\{\sin at\}$. Match parts:

$$\boxed{\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2},\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2},s>0.}$$

PICTURE. The complex number $\frac{1}{s-ia}$ is an arrow in the plane. Its horizontal shadow is the cos-transform; its vertical shadow is the sin-transform. One integral, two answers read off as shadows.

• Real part → cos (an even function, worth $1$ at $t=0$, hence the $s$ on top). Imaginary part → sin (an odd function, hence the $a$ on top).

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Step 7 — The degenerate cases (nothing is left unshown)

WHAT & WHY. A derivation is only honest if it says what happens at the edges.

• $s=0$ for $\mathcal{L}\{1\}$: the weight is a flat line, area is infinite. $1/s$ is undefined at $s=0$ — the formula warns you by blowing up. No transform.
• $s\le a$ for $\mathcal{L}\{e^{at}\}$: net exponent $-(s-a)\ge 0$, weight no longer decays, area diverges. $1/(s-a)$ blows up as $s\to a^{+}$.
• $n=0$ in $\mathcal{L}\{t^n\}$: then $t^0=1$ and $0!=1$, giving $\frac{0!}{s^{0+1}}=\frac{1}{s}$ — it collapses exactly back to Step 3. Beautiful consistency check.
• $a=0$ in the wiggle: $\sin 0=0$ everywhere, and indeed $\frac{a}{s^2+a^2}{|}_{a=0}=0$. Also $\cos 0=1$, matching $\frac{s}{s^2+0}=\frac{1}{s}=\mathcal{L}\{1\}$. All consistent.

PICTURE. One "convergence map": on the horizontal $s$-axis, shade the region where each transform lives. $\mathcal{L}\{1\}$ lives on $s>0$; $\mathcal{L}\{e^{at}\}$ starts at $s>a$; the wiggle again on $s>0$. The left edge of each shaded strip is the abscissa of convergence.

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The one-picture summary

Everything above is a single idea used four times: multiply by the fading weight $e^{-st}$, add up the area, and note how fast the weight must fade.

$$\underset{\text{area of weight}}{\underbrace{1\to \frac{1}{s}}}\Rightarrow \underset{\text{shift the knob}}{\underbrace{e^{at}\to \frac{1}{s-a}}}\Rightarrow \underset{\text{climb the ladder}}{\underbrace{t^n\to \frac{n!}{s^{n+1}}}}\Rightarrow \underset{\text{take shadows of }\frac{1}{s-ia}}{\underbrace{\cos /\sin \to \frac{s,a}{s^2+a^2}}}$$

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

You have a machine that "weighs" a function. Every moment in time gets a weight that starts big and fades — the fading speed is set by a knob called $s$. To transform a function you multiply it by that fading weight and add up the total over all time; the total is your answer for that knob setting.

• For a flat line the answer is just the area under the fading weight: it comes out $1/s$ — turn the knob up (fade faster), get less area.
• For a growing exponential $e^{at}$, the growth fights the fade; only the leftover fade matters, so the answer is the same picture with the knob shifted: $1/(s-a)$.
• For a power $t^n$ you peel one power off at a time (integration by parts), and each peel multiplies by a number — stack them all and you collect $n!$, giving $n!/s^{n+1}$.
• For a wiggle (sine/cosine) you cleverly use a spinning complex arrow $1/(s-ia)$; its flat shadow is the cosine answer and its upright shadow is the sine answer, both over $s^2+a^2$. The single warning throughout: the knob $s$ must be big enough that the weight really does fade, or the total runs off to infinity and there is no answer.

Recall Quick self-test

Why is the top boundary term in the $t^n$ integration by parts equal to zero? ::: Because for $s>0$ the weight $e^{-st}$ shrinks faster than the polynomial $t^n$ grows, so their product tends to $0$ at infinity (and it is $0$ at $t=0$ for $n\ge 1$). Where does the $+a^2$ in the trig denominator come from? ::: From $(s-ia)(s+ia)=s^2-i^2{a}^2=s^2+a^2$; the $i^2=-1$ turns a minus into a plus. Set $n=0$ in $n!/s^{n+1}$. What do you get and why is that reassuring? ::: $0!/s^1=1/s=\mathcal{L}\{1\}$, matching Step 3 — the power formula contains the flat-line case.

Connections

• Parent: Transforms of standard functions — proofs
• Laplace Transform — Definition and Existence · Improper Integrals
• Linearity and First Shifting Theorem · Transform of Derivatives — solving ODEs
• Integration by Parts · Euler's Formula $e^{i\theta}$ · Inverse Laplace Transform — Partial Fractions