Transforms of standard functions — proofs

The one definition everything comes from

HOW to read it: $s$ is a parameter, $t$ is the integration variable. After integrating, $t$ disappears and we are left with a function of $s$ only.

1. Transform of $f(t)=1$

Derivation (from scratch). $\mathcal{L}\{1\} = \int_0^\infty e^{-st}\cdot 1\,dt = \left[\frac{e^{-st}}{-s}\right]_0^\infty.$

• Why this step? Antiderivative of $e^{-st}$ w.r.t. $t$ is $e^{-st}/(-s)$ (chain rule in reverse).

At $t\to\infty$: $e^{-st}\to 0$ provided $s>0$ (else it blows up — this is where the condition $s>0$ is born). At $t=0$: $e^{0}=1$. So $= 0 - \frac{1}{-s} = \frac{1}{s}.$

2. Transform of $f(t)=t^n$ (and $e^{at}$ along the way)

Derive $\mathcal{L}\{e^{at}\}$ first (it is the easiest and most reusable): $\mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st}e^{at}\,dt = \int_0^\infty e^{-(s-a)t}\,dt = \left[\frac{e^{-(s-a)t}}{-(s-a)}\right]_0^\infty = \frac{1}{s-a},$

• Why $s>a$? We need $s-a>0$ so the exponential decays at infinity. This is just the $\mathcal{L}\{1\}$ proof with $s\to s-a$.

Derive $\mathcal{L}\{t^n\}$ by a recursion (integration by parts). Let $I_n = \int_0^\infty e^{-st}t^n\,dt$. Use $\int u\,dv = uv - \int v\,du$ with $u=t^n,\ dv=e^{-st}dt$: $I_n = \left[t^n\frac{e^{-st}}{-s}\right]_0^\infty + \frac{n}{s}\int_0^\infty e^{-st}t^{n-1}\,dt.$

• Why this choice? Differentiating $t^n$ lowers its power, building a recursion.

The boundary term vanishes: at $\infty$, $e^{-st}$ beats any polynomial ($s>0$); at $0$, $t^n=0$ for $n\ge1$. Hence $I_n = \frac{n}{s}I_{n-1}.$ Unrolling from $I_0=\mathcal{L}\{1\}=1/s$: $I_n = \frac{n}{s}\cdot\frac{n-1}{s}\cdots\frac{1}{s}\cdot\frac1s = \frac{n!}{s^{n+1}}.$

3. Transforms of $\sin at$ and $\cos at$

Derivation (the slick complex way). Use $e^{iat} = \cos at + i\sin at$. From step 2 with $a\to ia$: $\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia} = \frac{s+ia}{(s-ia)(s+ia)} = \frac{s+ia}{s^2+a^2}.$

• Why this step? Multiply top and bottom by the conjugate $s+ia$ to make the denominator real.

Now match real and imaginary parts. Since $\mathcal{L}$ is linear, $\mathcal{L}\{e^{iat}\}=\mathcal{L}\{\cos at\}+i\,\mathcal{L}\{\sin at\}$: $\underbrace{\mathcal{L}\{\cos at\}}_{\text{real}} = \frac{s}{s^2+a^2}, \qquad \underbrace{\mathcal{L}\{\sin at\}}_{\text{imag}} = \frac{a}{s^2+a^2}.$

4. Transforms of $\sinh at$ and $\cosh at$

Derivation. Write hyperbolic functions in exponentials and use linearity + step 2: $\mathcal{L}\{\cosh at\}=\mathcal{L}\!\left\{\tfrac{e^{at}+e^{-at}}{2}\right\}=\tfrac12\!\left(\frac{1}{s-a}+\frac{1}{s+a}\right)=\frac{s}{s^2-a^2}.$ $\mathcal{L}\{\sinh at\}=\mathcal{L}\!\left\{\tfrac{e^{at}-e^{-at}}{2}\right\}=\tfrac12\!\left(\frac{1}{s-a}-\frac{1}{s+a}\right)=\frac{a}{s^2-a^2}.$

• Why $s>|a|$? We need both $e^{at}$ and $e^{-at}$ pieces to converge, i.e. $s>a$ and $s>-a$.

Forecast-then-Verify

Common mistakes (steel-manned)

Active-recall flashcards

Connections

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

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Hinglish (regional understanding)