The Laplace transform turns a function of time into a function of a new variable $s$. Its power, which you will see in the next topics, is that it turns derivatives into multiplication by $s$, so differential equations become algebra. It is built from an improper integral, and most table entries come from integration by parts.
Key idea
$$\mathcal{L}\{f\}(s) = F(s) = \int_0^{\infty} e^{-st}f(t)\,dt,$$ defined for the $s$ where the integral converges. The transform is linear: $\mathcal{L}\{af + bg\} = aF + bG$.
| $f(t)$ | $F(s)$ | valid for |
|---|---|---|
| $1$ | $\frac{1}{s}$ | $s > 0$ |
| $e^{at}$ | $\frac{1}{s - a}$ | $s > a$ |
| $t^{n}$ | $\frac{n!}{s^{n+1}}$ | $s > 0$ |
| $\sin bt$ | $\frac{b}{s^{2} + b^{2}}$ | $s > 0$ |
| $\cos bt$ | $\frac{s}{s^{2} + b^{2}}$ | $s > 0$ |
Why it works
For $e^{at}$: $\int_0^{\infty} e^{-(s - a)t}\,dt$ $= \left[-\frac{e^{-(s - a)t}}{s - a}\right]_0^{\infty} = \frac{1}{s - a}$, provided $s > a$ so the exponential dies out. For powers, integrating by parts gives $\mathcal{L}\{t^{n}\} = \frac{n}{s}\mathcal{L}\{t^{n-1}\}$, and repeating down to $\mathcal{L}\{1\} = \frac{1}{s}$ gives $\frac{n!}{s^{n+1}}$. For sine and cosine, put $a = ib$: $\frac{1}{s - ib} = \frac{s + ib}{s^{2} + b^{2}}$, and by Euler's formula the real and imaginary parts are the transforms of $\cos bt$ and $\sin bt$.
Worked example: From the definition
Find $\mathcal{L}\{e^{3t}\}$ and say where it is valid.
- $F(s) = \int_0^{\infty} e^{-st}e^{3t}\,dt$ $= \int_0^{\infty} e^{-(s - 3)t}\,dt$.
- For $s > 3$ the integrand decays and the integral is $\frac{1}{s - 3}$; for $s \le 3$ it diverges.
Answer: $F(s) = \frac{1}{s - 3}$ for $s > 3$
Worked example: Linearity with the table
Find $\mathcal{L}\{4 - 2t + 5\sin 3t\}$.
- Transform term by term: $\frac{4}{s} - 2\cdot\frac{1}{s^{2}} + 5\cdot\frac{3}{s^{2} + 9}$.
Answer: $\frac{4}{s} - \frac{2}{s^{2}} + \frac{15}{s^{2} + 9}$
Worked example: A pulse
Find the transform of $f(t) = 2$ for $0 \le t < 3$ and $f(t) = 0$ for $t \ge 3$.
- The integral stops at 3: $\int_0^{3} 2e^{-st}\,dt = 2\left[-\frac{e^{-st}}{s}\right]_0^{3}$.
- $= \frac{2}{s}(1 - e^{-3s})$.
Answer: $F(s) = \frac{2(1 - e^{-3s})}{s}$
Common mistake
Treating $s$ and $t$ alike. The integral is over $t$, so $s$ acts as a constant inside it and survives in the answer, while $t$ disappears. Also keep the convergence condition in mind: $\frac{1}{s - 3}$ represents $\mathcal{L}\{e^{3t}\}$ only for $s > 3$.
Which functions have a transform? It is enough that $f$ is piecewise continuous and of exponential order: $|f(t)| \le Me^{at}$ for large $t$. Then the integral converges for every $s > a$, because the factor $e^{-st}$ wins. Polynomials, exponentials, sines and step functions all qualify; $e^{t^{2}}$ grows too fast and has no transform. For such functions $F(s) \to 0$ as $s \to \infty$, so an expression like $\frac{s^{2}}{s + 1}$ is not the transform of any ordinary function.