Visual walkthrough — Properties — linearity, first - second shift theorems, scaling

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Step 0 — When is any of this even allowed? (prerequisites on $f$ and $s$)

WHAT. Before feeding anything to the machine we must know the machine's input rules — the hypotheses that make the integral a finite number. Two conditions on $f$, and one condition on $s$.

WHY these two. The fading curtain $e^{-st}$ can only tame $f$ if $f$ does not outrun it. If $|f(t)|\le M{e}^{{\sigma }_{0}t}$ and $\mathrm{Re}(s)>{\sigma }_0$, then $|e^{-st}f(t)|\le M{e}^{-(\mathrm{Re}(s)-{\sigma }_0)t}$, which decays — so the area is finite. Piecewise continuity guarantees the pieces are integrable in the first place.

WHAT IT CHANGES DOWNSTREAM. Keep the ROC in your pocket: in Step 6 the delay tag $e^{-as}$ (with $a\ge 0$, $\sigma =\mathrm{Re}(s)>0$) satisfies $|e^{-as}|=e^{-a\sigma }\le 1$, so it never enlarges the region where things converge — the ROC of the delayed signal is the same $\sigma >{\sigma }_0$ as the original. Delaying a signal cannot make its transform converge in a smaller place; the tag is harmless to convergence.

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Step 1 — What is the machine we are feeding? (the integral, from zero)

WHAT. The Laplace transform is a machine. You feed it an admissible function of time $f(t)$ (a wiggle drawn on a time-axis), and it hands back a new function $F(s)$ of the complex variable $s$, valid for $\mathrm{Re}(s)>{\sigma }_0$:

$$\underset{\text{"transform of }f\text{"}}{\underbrace{\mathcal{L}\{f(t)\}}}=\underset{\text{output recipe}}{\underbrace{F(s)}}={\int }_0^{\infty }\underset{\text{fading weight}}{\underbrace{e^{-st}}}\underset{\text{your wiggle}}{\underbrace{f(t)}}dt.$$

WHY the pieces. Read the integrand right-to-left of the picture:

• $f(t)$ is the signal you care about (admissible, per Step 0).
• $e^{-st}$ is a fading curtain: on the real slice $\sigma =\mathrm{Re}(s)>0$ it starts at $1$ (at $t=0$) and decays toward $0$. It weights early time heavily and late time barely at all.
• ${\int }_0^{\infty }\dots dt$ means "add up the product over all future time." One number comes out for each valid $s$ — that collection of numbers is $F(s)$.

PICTURE. Below: the wiggle $f(t)$ in blue, the fading curtain $e^{-st}$ in yellow, and their product (pink, shaded) whose total area is the single value $F(s)$.

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Step 2 — What does "delay by $a$" look like? (build the unit step)

WHAT. We want the transform of a signal that starts late. To describe "starts at time $a$, nothing before," we need a switch. First define the plain Heaviside unit step $u(t)$, then shift it.

WHY we need it. Multiplying any function by $u(t-a)$ erases everything before $t=a$ and leaves the rest untouched — it is a clean on-switch flipped at $t=a$.

The delayed signal we transform is $f(t-a)u(t-a)$:

• $f(t-a)$ is the same shape as $f$, but slid right by $a$ (every feature happens $a$ seconds later).
• $u(t-a)$ guarantees it is silent (zero) until the switch flips at $t=a$.

PICTURE. Left: original $f(t)$. Middle: $f(t-a)$ — the identical shape shifted right by $a$ (the green arrow shows the slide). Right: the on-switch $u(t-a)$ that keeps it zero before $a$.

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Step 3 — Feed the delayed signal into the machine

WHAT. Write the definition from Step 1, but with our delayed signal $f(t-a)u(t-a)$ plugged in:

$$\mathcal{L}\{f(t-a)u(t-a)\}={\int }_0^{\infty }{e}^{-st}\underset{\text{slid shape}}{\underbrace{f(t-a)}}\underset{\text{on-switch}}{\underbrace{u(t-a)}}dt.$$

WHY. Nothing clever yet — we are just applying the definition. The whole derivation is: simplify this one integral until we recognise it.

PICTURE. The fading curtain $e^{-st}$ (yellow) now multiplies a signal that is flat-zero until $t=a$, then wiggles. The shaded product only exists to the right of $a$.

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Step 4 — The switch cuts the integral's floor up to $t=a$

WHAT. For $t<a$, the factor $u(t-a)=0$, so the entire integrand is zero there. Those parts contribute nothing to the area. So we may raise the lower limit from $0$ to $a$:

$${\int }_0^{\infty }{e}^{-st}f(t-a)u(t-a)dt={\int }_a^{\infty }{e}^{-st}f(t-a)dt.$$

Notice $u(t-a)$ has done its job and vanished — for $t\ge a$ it just equals $1$.

WHY. Cutting off a region where the function is exactly zero can never change an integral. This is the single most important move: it turns a step-function problem into an ordinary integral, just with a shifted lower limit. (This is exactly where $a\ge 0$ is used: the new lower limit $a$ must sit inside the domain $[0,\infty )$ — see Step 7b for what breaks if $a<0$.)

PICTURE. The zero region (left of $a$) is greyed out; the live region (right of $a$) is the only area that matters. The lower limit visibly jumps from $0$ to $a$.

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Step 5 — Slide the time-axis: the substitution $\tau =t-a$

WHAT. The integrand has $f(t-a)$, but $F$ was defined as ${\int }_0^{\infty }{e}^{-s\tau }f(\tau )d\tau$ with a plain argument. To match, we rename the clock so the shift disappears. Let

$$\tau =t-a⟹t=\tau +a,dt=d\tau ,$$

and the limits move: when $t=a$, $\tau =0$; when $t\to \infty$, $\tau \to \infty$.

WHY this tool (substitution). We use $u$-substitution — the reverse of the chain rule — specifically to absorb the delay $a$ into the variable, so $f(t-a)$ becomes $f(\tau )$. That is the only way to make the integral literally the definition of $F$ again.

Every piece transforms:

$${\int }_a^{\infty }\underset{e^{-s(\tau +a)}}{\underbrace{e^{-st}}}\underset{f(\tau )}{\underbrace{f(t-a)}}\underset{d\tau }{\underbrace{dt}}={\int }_0^{\infty }{e}^{-s(\tau +a)}f(\tau )d\tau .$$

PICTURE. The two axes stacked: the old $t$-axis (start at $a$) and, directly beneath, the new $\tau$-axis with the same wiggle but now starting at $0$. The whole picture just slid left by $a$.

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Step 6 — Split the exponential and pull the delay outside (constant-factor rule)

WHAT. Use the exponent law $e^{-s(\tau +a)}=e^{-s\tau }\cdot e^{-as}$. The factor $e^{-as}$ has no $\tau$ in it, so by the constant-factor rule for integrals, $\int g(\tau )[\text{const}]d\tau =[\text{const}]\int g(\tau )d\tau$, it slides out front:

$${\int }_0^{\infty }{e}^{-s\tau }{e}^{-as}f(\tau )d\tau \stackrel{\text{const. out}}{=}{e}^{-as}\underset{=F(s)\text{(Step 1 definition!)}}{\underbrace{{\int }_0^{\infty }{e}^{-s\tau }f(\tau )d\tau }}=e^{-as}F(s).$$

WHY (explicit). The equality ${\int }_0^{\infty }cg(\tau )d\tau =c{\int }_0^{\infty }g(\tau )d\tau$ for a constant $c=e^{-as}$ is the linearity of the integral applied to a scalar multiple. Since $e^{-as}$ does not vary with the integration variable $\tau$, it is a legitimate constant $c$ and comes straight out — no approximation, an exact algebraic move. Splitting first isolated the time-part $e^{-s\tau }$ (which rebuilds $F$) from the pure delay-part $e^{-as}$ (the "I came late" stamp).

PICTURE. The integral collapses into the definition of $F(s)$ (the same blue-and-yellow area from Step 1), with a yellow tag $e^{-as}$ pinned to the front.

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Step 7 — Edge case $a=0$: no delay, no tag

WHAT. Set $a=0$. Then $u(t-0)=u(t)=1$ for all $t\ge 0$, and $f(t-0)=f(t)$. The formula must reduce to the plain transform:

$$\mathcal{L}\{f(t)\cdot 1\}=e^{-0\cdot s}F(s)=1\cdot F(s)=F(s).✓$$

WHY it matters. A good formula must survive its trivial input. Zero delay ⇒ tag $e^{-as}=e^0=1$ ⇒ nothing changes. Consistent.

PICTURE. The delay slider dragged to $a=0$: the "shifted" curve sits exactly on the original, the tag reads $1$.

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Step 7b — Edge case $a<0$: why the theorem forbids it

WHAT. Suppose someone tries a negative shift $a<0$ (an advance, i.e. the signal starts before $t=0$). Look back at Step 4: we replaced the lower limit $0$ by $a$. If $a<0$, then $a$ lies outside the integration domain $[0,\infty )$ — the Laplace integral simply has no information about $t<0$, since it starts at $0$. The move "raise the floor to $a$" becomes "lower the floor below $0$," which the definition cannot do.

WHY it fails, concretely. For $a<0$ the switch $u(t-a)=1$ for all $t\ge 0$ (it fired in the past), so $u(t-a)$ does nothing — but $f(t-a)$ now needs values of $f$ at negative arguments $t-a$ which range into $t=0\Rightarrow$ argument $-a>0$... yet also the early part of the advanced signal (the part between the true start and $t=0$) is thrown away by the lower limit $0$. The clean substitution collapses; you cannot pull out a single tag. Hence the theorem is stated only for $a\ge 0$.

PICTURE. The delay slider dragged to $a<0$: the shape pushes left past the $t=0$ wall; the shaded early lobe crosses into $t<0$ and is chopped off by the integral's hard floor at $0$.

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Step 8 — The classic trap: the argument must already read $(t-a)$

WHAT. The theorem needs the shape written as $f(t-a)$ — a shifted argument. If you see $u(t-a)$ next to a function whose argument is plain $t$, you must first rewrite it.

Example. Transform $t^{2}u(t-2)$. The argument is $t^2$, not $(t-2{)}^2$, so we cannot grab the rule yet. Rewrite around $a=2$:

$$t^2=((t-2)+2{)}^2=(t-2{)}^2+4(t-2)+4.$$

Now every piece has argument $(t-2)$, and we apply $e^{-2s}F(s)$ term-by-term with $\mathcal{L}\{t^2\}=\frac{2}{s^3},\mathcal{L}\{t\}=\frac{1}{s^2},\mathcal{L}\{1\}=\frac{1}{s}$:

$$\mathcal{L}\{t^{2}u(t-2)\}=e^{-2s}\left(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s}\right).$$

WHY. The naive answer $e^{-2s}\cdot \frac{2}{s^3}$ silently assumes the shape was $(t-2{)}^2$ — a different signal. The rewrite forces the argument into legal form: only a term already written in the delayed variable $(t-2)$ is allowed to inherit the $e^{-2s}$ tag, so we expand first and transform each shifted piece separately.

PICTURE. Two different curves after the switch at $t=2$: the wrong reading $(t-2{)}^2$ versus the actual $t^2$ — visibly not the same, which is exactly why the rewrite is mandatory.

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The one-picture summary

Everything compressed: with $f$ admissible and $\mathrm{Re}(s)>{\sigma }_0$, a delay of $a\ge 0$ in the time world slides the shape right and switches it on with $u(t-a)$; the substitution $\tau =t-a$ un-slides it back to the plain shape; the constant-factor rule frees a single tag $e^{-as}$ into the s world. Shape $F(s)$ and its ROC preserved, tag attached.

Recall Feynman retelling (plain words)

Picture a machine that squishes a time-wiggle into a compact recipe $F(s)$, by multiplying it with a fading yellow curtain and adding up the area. The machine only accepts reasonable wiggles — ones with only finite jumps and that grow no faster than some exponential — and even then only for knob settings $s$ far enough to the right (its real part $\sigma$ big enough) that the area is finite. Now take the same wiggle but start it $a$ seconds late (with $a$ not negative), using a light-switch $u(t-a)$ that stays off until then. When you feed this late signal in, the switch means all the early "nothing" contributes nothing — so you might as well start counting area at $t=a$. Then you slide your clock back by $a$ (call it $\tau$); the wiggle looks exactly like the original, so its recipe is again $F(s)$. The only leftover from sliding the clock is a constant $e^{-as}$ — and because it's a constant it walks straight out of the integral and stands outside as a sticker meaning "I arrived $a$ seconds late." Delay in time = sticker $e^{-as}$ in $s$, and since the sticker never grows, the region of good knob settings is unchanged. If the delay is zero the sticker is just $1$. If someone tries a negative delay, the trick breaks — part of the signal would fall before $t=0$ and get chopped. And beware: the rule only works if your wiggle is genuinely written as the shifted shape $f(t-a)$ — if it says plain $t$, expand it first.

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Connections

• Laplace Transform — Definition and Existence
• Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
• Unit Step and Dirac Delta Functions
• Solving ODEs with Laplace Transforms
• Inverse Laplace Transform and Partial Fractions
• Convolution Theorem