Convert a function of time into a function of a complex variable \(s\).
The integral must converge. We usually assume \(s\) is large enough so that the exponential decay dominates.
Tip: For the basic functions, evaluate the improper integral directly using the definition.
We use Laplace transforms because they turn hard differential equations into easy algebra problems.
Many real-world situations (physics, engineering, circuits, control systems) are described by differential equations. Solving those equations directly can be messy or even impossible by hand. The Laplace transform converts a differential equation in the time variable \(t\) into an ordinary algebraic equation in a new variable \(s\). You solve the algebra, then transform the answer back. This is often much easier.
Changes the type of problem
Differential equations (with derivatives) become algebraic equations (no derivatives). Algebra is usually easier than differential equations.
Initial conditions built in
Starting values (like position or velocity at \(t=0\)) are automatically included. You do not have to handle them separately.
Handles jumps and impulses
Real systems often have sudden changes (switches flipping, impacts). Laplace transforms handle these discontinuous inputs cleanly.
Systematic method
Once you learn the process, it works the same way for a wide variety of problems. You follow the same steps every time.
Simple analogy: Using a Laplace transform is like turning a complicated word problem into a plain equation you already know how to solve — or like using logarithms to turn multiplication into addition. The transform changes the kind of problem into something easier.
Find \(\mathcal{L}\{1\}\) using the definition.
Step 1: Write the definition
By definition,
$$\mathcal{L}\{1\} = \int_0^{\infty} e^{-st}\cdot 1\, dt$$
Step 2: Turn the improper integral into a limit
$$\mathcal{L}\{1\} = \lim_{b\to\infty}\int_0^b e^{-st}\, dt$$
Step 3: Integrate
The antiderivative of \(e^{-st}\) is \(-\dfrac{1}{s}e^{-st}\) (for \(s\neq 0\)):
$$= \lim_{b\to\infty}\left[-\dfrac{1}{s}e^{-st}\right]_0^b$$
Step 4: Plug in the limits
$$= \lim_{b\to\infty}\left(-\dfrac{e^{-sb}}{s} – \left(-\dfrac{e^{0}}{s}\right)\right) = \lim_{b\to\infty}\left(-\dfrac{e^{-sb}}{s} + \dfrac{1}{s}\right)$$
Step 5: Take the limit
When \(s > 0\), \(e^{-sb}\to 0\) as \(b\to\infty\), so
$$= 0 + \dfrac{1}{s} = \dfrac{1}{s}$$
\( \mathcal{L}\{1\} = \dfrac{1}{s} \quad (s>0) \)
Find the Laplace transforms of the three basic functions using the definition
Find \(\mathcal{L}\{2\}\) using the definition.
Step 1: Write the definition
$$\mathcal{L}\{2\} = \int_0^{\infty} e^{-st}\cdot 2\, dt = \lim_{b\to\infty}\int_0^b 2e^{-st}\, dt$$
Step 2: Integrate
$$= \lim_{b\to\infty}\left[-\dfrac{2}{s}e^{-st}\right]_0^b$$
Step 3: Evaluate the limits
$$= \lim_{b\to\infty}\left(-\dfrac{2e^{-sb}}{s} + \dfrac{2}{s}\right)$$
When \(s > 0\), \(e^{-sb}\to 0\), so the result is \(\dfrac{2}{s}\).
\( \mathcal{L}\{2\} = \dfrac{2}{s} \quad (s>0) \)
Find \(\mathcal{L}\{e^{t}\}\) using the definition.
Step 1: Write the definition
$$\mathcal{L}\{e^{t}\} = \int_0^{\infty} e^{-st}\cdot e^{t}\, dt$$
Step 2: Combine the exponents
$$e^{-st}\cdot e^{t} = e^{-(s-1)t}$$
$$\mathcal{L}\{e^{t}\} = \lim_{b\to\infty}\int_0^b e^{-(s-1)t}\, dt$$
Step 3: Integrate
$$= \lim_{b\to\infty}\left[-\dfrac{1}{s-1}e^{-(s-1)t}\right]_0^b$$
Step 4: Evaluate the limit
$$= \lim_{b\to\infty}\left(-\dfrac{e^{-(s-1)b}}{s-1} + \dfrac{1}{s-1}\right)$$
When \(s > 1\), the exponential term goes to 0, leaving \(\dfrac{1}{s-1}\).
\( \mathcal{L}\{e^{t}\} = \dfrac{1}{s-1} \quad (s>1) \)
Find \(\mathcal{L}\{t\}\) using the definition.
Step 1: Write the definition
$$\mathcal{L}\{t\} = \int_0^{\infty} t\, e^{-st}\, dt = \lim_{b\to\infty}\int_0^b t\, e^{-st}\, dt$$
Step 2: Integration by parts
Let \(u = t\) and \(dv = e^{-st}\,dt\). Then \(du = dt\) and \(v = -\dfrac{1}{s}e^{-st}\).
$$\int t e^{-st}\, dt = -\dfrac{t}{s}e^{-st} + \dfrac{1}{s}\int e^{-st}\, dt = -\dfrac{t}{s}e^{-st} – \dfrac{1}{s^2}e^{-st}$$
Step 3: Evaluate from 0 to \(b\) and take the limit
$$= \lim_{b\to\infty}\left[-e^{-st}\left(\dfrac{t}{s} + \dfrac{1}{s^2}\right)\right]_0^b$$
At the upper limit the expression goes to 0 (when \(s>0\)). At the lower limit we obtain \(\dfrac{1}{s^2}\).
\( \mathcal{L}\{t\} = \dfrac{1}{s^2} \quad (s>0) \)
Slightly more involved integrals
Find \(\mathcal{L}\{t^2\}\) using the definition (integration by parts twice).
Step 1: Write the definition
$$\mathcal{L}\{t^2\} = \int_0^{\infty} t^2 e^{-st}\, dt$$
Step 2: Integration by parts (first time)
Let \(u = t^2\), \(dv = e^{-st}\,dt\). Then \(du = 2t\,dt\), \(v = -\dfrac{1}{s}e^{-st}\).
$$\int t^2 e^{-st}\, dt = -\dfrac{t^2}{s}e^{-st} + \dfrac{2}{s}\int t e^{-st}\, dt$$
Step 3: Integration by parts (second time)
We already know from Easy 3 that the integral of \(t e^{-st}\) produces \(\dfrac{1}{s^2}\) after limits are applied.
Continuing with the definite integral from 0 to \(\infty\) (and \(s>0\)) yields:
$$\mathcal{L}\{t^2\} = \dfrac{2}{s^3}$$
Pattern to notice
\[ \begin{align*} \mathcal{L}\{1\} &= \dfrac{1}{s} = \dfrac{0!}{s^1} \\ \mathcal{L}\{t\} &= \dfrac{1}{s^2} = \dfrac{1!}{s^2} \\ \mathcal{L}\{t^2\} &= \dfrac{2}{s^3} = \dfrac{2!}{s^3} \end{align*} \]
In general, \(\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}\).
\( \mathcal{L}\{t^2\} = \dfrac{2}{s^3} \quad (s>0) \)
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