DE-BOOK Defining the Laplace Transform

Section Defining the Laplace Transform

We’ve seen how integration by parts transfers derivatives from one function to another inside an integral. Now we take the next step: defining a transform that systematically applies this principle.

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In this section, we’ll see how this transform is constructed and how it converts functions of \(t\) into functions of \(s\text{.}\) You’ll also see how to compute Laplace transforms from the definition.

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Subsection Why Use the Laplace Transform?

To see where we’re headed, imagine you need to solve the differential equation

\begin{equation*} y'(t) - 9y(t) = 10 e^{2t}. \end{equation*}

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Suppose you multiply both sides by \(e^{-st}\) and integrate with respect to \(t\text{:}\)

\begin{equation*} \int \left[y'(t) - 9y(t)\right] e^{-st} \, dt = \int 10 e^{2t} e^{-st} \, dt. \end{equation*}

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The left-hand side contains a derivative inside the integral. From the previous section, you know we can transfer this derivative to the exponential function. Doing so removes the derivative from \(y(t)\text{,}\) essentially allowing you to use algebra to isolate it.

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This is the idea behind the Laplace transform and the motivation for its definition.

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Subsection Definition of the Laplace Transform

📙 Definition 184. Laplace Transform.

Given a function \(f(t)\) defined for \(t \geq 0\text{,}\) its Laplace transform, denoted \(\lap{f(t)}\text{,}\) is defined by

\begin{equation*} \lap{f(t)} = \int_0^{\infty} e^{-st} f(t) \, dt, \end{equation*}

provided the integral converges. The result is a new function, typically written as a capitalized version of the variable you are transforming. In this case:

\begin{equation*} F(s) = \lap{f(t)}. \end{equation*}

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Checkpoint 185. 📖❓ Spot the Error in the Definition.

Click on the incorrect part of the Laplace transform definition.

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\(\ds \laplacesym\) \(\big\{\) \(f(t)\) \(\big\}\) \(=\) \(\ds \int_{0}^{\infty}\) \(\ds e^{st}\) \(\ds f(t)\) \(\ds dt\)

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Working with Improper Integrals.

Notice that the Laplace transform involves an improper integral: the upper limit is \(\infty\text{.}\) In calculus, we handle this by replacing \(\infty\) with a variable \(b\text{,}\) computing the definite integral, and then taking the limit as \(b \to \infty\text{:}\)

\begin{equation*} \lap{f(t)} = \int_0^{\infty} e^{-st} f(t)\, dt = \ub{\lim_{b \to \infty} \ob{\int_0^b e^{-st} f(t)\, dt}^{\large\text{(step 1)}}}_{\large\text{(step 2)}}\text{.} \end{equation*}

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When calculating Laplace transforms, you first compute the integral with respect to \(t\text{,}\) treating \(s\) as a constant. Then you compute the limit of the result as \(b\) approaches \(\infty\text{.}\)

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The Laplace transform exists if this limit produces a finite value. Often, exponential decay from \(e^{-st}\) guarantees convergence when \(s\) is large enough.

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Checkpoint 186. 📖❓ Convergence Requirement for Laplace Transforms.

To compute an improper integral, you write it as a limit of a definite integral. For example,

\begin{equation*} \int_0^{\infty} e^{-st} t^3 \, dt = \lim_{b \to \infty} \ub{\int_0^{b} e^{-st} t^3 \, dt}_{\ds I}\text{.} \end{equation*}

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Select variables/constants would you expect to see after computing the integral, \(I\text{,}\) but before taking the limit.

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\(\ s\)
The variable \(s\) is a constant in \(e^{-st} t^3\) so it will also appear in the antiderivative.
\(\ b\)
The variable \(b\) gets plugged into \(t\) after you find the antiderivative.
\(\ t\)
Since the variable of integration is \(t\text{,}\) you plug the limits of integration, \(0\) and \(b\text{,}\) into \(t\) once you have the antiderivative.
\(\ c\) (integration constant)
We are computing a definite integral, not an indefinite one.

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Reading Questions 👀 Quick Review: Exponential Properties

To get through the transform rule derivations, you’ll need to be familiar with some key exponential properties that will appear throughtout this chapter.

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Exponential Antiderivative.

For any constant \(K\text{,}\) we have:

\begin{equation*} \int e^{Kt}\ dt = \frac1{K} e^{Kt} + C\text{.} \end{equation*}

While this can be found by substituting \(u=Kt\text{,}\) it is such a common antiderivative that you should just know it without this extra step.

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Exponential End Behavior.

The end behavior can be seen from the most basic exponential function:

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\begin{alignat*}{2} t \to -\infty \amp\ \ \Rightarrow\ \ \amp e^{t} \to 0\\ t \to \infty \amp\ \ \Rightarrow\ \ \amp e^{t} \to \infty \end{alignat*}

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In general, the end behavior depends on whether the power goes to \(+\infty\) or \(-\infty\text{:}\)

\(\displaystyle \quad \text{power} \to +\infty\quad\Rightarrow\quad e^{\text{power}} \to +\infty\)

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\(\displaystyle \quad \text{power} \to -\infty\quad\Rightarrow\quad e^{\text{power}} \to 0\)

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In this section, we will take limits as \(b \to \infty\text{.}\) Your goal will be to figure out if \(b \to \infty\) causes the power to go to \(+\infty\) or \(-\infty\text{.}\) Only then will you know where \(\ds e^{\text{power}}\) goes.

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1. 📖❓ Checkpoint: Exponential Properties.

(a) Exponential Antiderivative.

Evaluate: \(\ds\int_0^b e^{-3t}\ dt\)

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\(\quad\ds \frac{1 - e^{-3b}}{3}\)
\(\quad\ds \frac{1}{3}e^{-3b}\)
\(\quad\ds -\frac{1}{3}e^{-3b}\)
\(\quad\ds -\frac{1}{3}e^{3b}\)

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(b) Exponential End Behavior.

Determine if each limit is -ထ, \(0\text{,}\) or ထ.

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Click on one of the three options next to each limit to select your answer.

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Assume \(s\) is a positive constant.

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\begin{equation*} \lim_{b \to \infty} e^{\ds b} \end{equation*}

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\begin{equation*} = \fillinmath{XX} \end{equation*}

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\(\large\ -\infty\ \)
\(\large\ 0\ \)
\(\large\ +\infty\ \)

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\begin{equation*} \lim_{b \to \infty} e^{\ds -3b} \end{equation*}

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\begin{equation*} = \fillinmath{XX} \end{equation*}

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\(\large\ -\infty\ \)
\(\large\ 0\ \)
\(\large\ +\infty\ \)

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\begin{equation*} \lim_{b \to \infty} e^{\ds sb} \end{equation*}

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\begin{equation*} = \fillinmath{XX} \end{equation*}

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\(\large\ -\infty\ \)
\(\large\ 0\ \)
\(\large\ +\infty\ \)

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\begin{equation*} \lim_{b \to \infty}e^{\ds -sb} \end{equation*}

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\begin{equation*} = \fillinmath{XX} \end{equation*}

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\(\large\ -\infty\ \)
\(\large\ 0\ \)
\(\large\ +\infty\ \)

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(c) Figure out the Limits.

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Subsection First Examples of Laplace Transforms

Let’s see how the definition works by computing Laplace transforms for simple functions. We’ll start with a constant function and a linear function.

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🌌 Example 187. First Look at Laplace Transforms.

Find the Laplace transform of

\(f(t) = 5\) (constant function)
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Solution. \(\rightarrow \lap{5}\)
By the definition above, we have the improper integral

\begin{equation} \lap{5} = \int_0^{\infty} e^{-st} \cdot 5\ dt = \lim_{b \to \infty} \ub{\int_0^b 5e^{-st}dt}_{I}\text{.}\tag{44} \end{equation}

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Assuming \(s\) and \(b\) are constant, we integrate \(I\) with respect to \(t\text{:}\)

\begin{equation*} I = \int_0^b 5e^{-st}dt = -\frac{5}{s}e^{-st}\Big|_{t=0}^{t=b} = -\frac{5}{s}\left[ e^{-sb} - 1 \right]\text{.} \end{equation*}

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🔍 Next, we compute the limit of \(I\) as \(b \to \infty\) (with \(s\) constant):.

🔍 Since the limit only controls \(b\text{,}\) we can think of \(s\) as a number (like \(2\)). This means that \(\ds-\frac{5}{s}\) is also a number that can be pulled out of the limit using the limit property from calculus:

\begin{equation*} \ds\lim_{x \to \infty} c \cdot f(x) = c \cdot \lim_{x \to \infty} f(x), \quad c\ \text{is constant} \end{equation*}

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\begin{align*} \lim_{b \to \infty} I = \lim_{b \to \infty} -\frac{5}{s}\left[ e^{-sb} - 1 \right] \amp = -\frac{5}{s} \lim_{b \to \infty} \Big[ e^{-sb} - 1 \Big]\\ \amp = -\frac{5}{s} \Big[ \ub{\lim_{b \to \infty} e^{-sb}}_{L} - 1 \Big] \end{align*}
We find the limit using the end behavior of the exponential. There are two possiblities:
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\(s = 0\) is excluded since it would lead to
\begin{equation*} \int_0^{\infty}5e^{-0t}\ dt = \int_0^{\infty}5\ dt = \infty \end{equation*}

\(s\) is positive: \(\ds\quad-sb \to -\infty \ \ \Rightarrow\ \ \lim_{b \to \infty}\ e^{-sb} = 0\)

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\(s\) is negative: \(\ds\quad-sb \to +\infty \ \ \Rightarrow\ \ \lim_{b \to \infty}\ e^{-sb} = \infty\)

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Since we only get convergence when \(s\) is positive, we must require \(s \gt 0\) in the Laplace transform of \(5\text{:}\)

\begin{equation*} \lap{5} = -\frac{1}{s} [0 - 1] = \frac{1}{s} \quad \text{for } s \gt 0\text{.} \end{equation*}

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\(f(t)=t\)
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Solution. \(\rightarrow \lap{t}\)
Applying the definition and writing the integral as a limit, we have

\begin{equation*} \lap{ t } = \lim_{b \to \infty} {\ub{\int_0^b e^{-st} \cdot t\ dt}_{I}}\text{.} \end{equation*}

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🔍 Computing \(I\) with integration by parts, gives us.

🔍 Choosing

\begin{align*} u = t, \quad\amp dv = e^{-st}dt, \\ du = dt, \quad\amp v = -\frac{1}{s}e^{-st} \end{align*}

integration by parts gives

\begin{align*} \int_0^b e^{-st} \cdot t\ dt \amp = -\frac{t}{s}e^{-st} - \int_0^b \left( -\frac{1}{s}e^{-st} \right) dt\\ \amp = -\frac{t}{s}e^{-st}+ \frac{1}{s}\int_0^b e^{-st} dt\\ \amp = \left(-\frac{t}{s}e^{-st}- \frac{1}{s^2}e^{-st}\right) \Bigg|_0^b\\ \amp = \left(-\frac{b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb}\right) - \left(0 - \frac{1}{s^2}\right)\\ \amp = -\frac{b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb} + \frac{1}{s^2} \end{align*}

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\begin{align*} {I} \amp\ { = -\frac{b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb} + \frac{1}{s^2}} \end{align*}

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Next, we compute the limit of \(I\) as \(b \to \infty\text{:}\)

\begin{align*} \lim_{b \to \infty} {I} \amp = \lim_{b \to \infty} \left({ -\frac{b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb} + \frac{1}{s^2}}\right)\\ \amp = -\frac{1}{s}\lim_{b \to \infty} be^{-sb} - \frac{1}{s^2}\lim_{b \to \infty} e^{-sb} + \lim_{b \to \infty} \frac{1}{s^2}\\ \amp = -\frac{1}{s}\ub{\lim_{b \to \infty} \frac{b}{e^{sb}}}_{L_1} - \frac{1}{s^2}\ub{\lim_{b \to \infty} e^{-sb}}_{L_2} + \frac{1}{s^2} \end{align*}

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🔍 As long as \(s \gt 0\text{,}\) both \(L_1\) and \(L_2\) go to zero.

🔍 Assuming \(s \gt 0\text{,}\) the limits are computed by:

\(\displaystyle \ds L_1 = \lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} \,\us{LH}{=}\, \lim_{b \to \infty}\frac{1}{se^{sb}} = \frac{1}{s}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0\)
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\(\displaystyle \ds L_2 = \lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0\)
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where \(LH\) denotes L’Hôpital’s Rule was applied to the limit.

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Therefore, the Laplace transform of \(t\) is

\begin{equation*} \lap{ t } = \frac{1}{s^2}, \quad s > 0\text{.} \end{equation*}

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These examples show the typical process: set up the improper integral, compute it as a definite integral, then take the limit as \(b \to \infty\text{.}\) You’ll see how \(s\) emerges naturally from this process.

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In upcoming sections, we’ll build a library of Laplace transforms for the most common functions and learn properties that make these computations even faster.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The Laplace transform converts a function of \(t\) into a function of \(s\) using
\begin{equation*} \lap{f(t)} = \int_0^\infty e^{-st} f(t)\, dt. \end{equation*}

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The Laplace transform is an improper integral evaluated by rewriting it as
\begin{equation*} \int_0^{\infty} e^{-st} f(t)\, dt = \lim_{b \to \infty} \int_0^b e^{-st} f(t)\, dt\text{.} \end{equation*}

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The result is a function of \(s\) and exists only when the integral converges.

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Section Defining the Laplace Transform

Subsection Why Use the Laplace Transform?

Subsection Definition of the Laplace Transform

📙 Definition 184. Laplace Transform.

Checkpoint 185. 📖❓ Spot the Error in the Definition.

Working with Improper Integrals.

Checkpoint 186. 📖❓ Convergence Requirement for Laplace Transforms.

Reading Questions 👀 Quick Review: Exponential Properties

Exponential Antiderivative.

Exponential End Behavior.

1. 📖❓ Checkpoint: Exponential Properties.

(a) Exponential Antiderivative.

(b) Exponential End Behavior.

(c) Figure out the Limits.

Subsection First Examples of Laplace Transforms

🌌 Example 187. First Look at Laplace Transforms.

🔍 Next, we compute the limit of \(I\) as \(b \to \infty\) (with \(s\) constant):.

🔍 Computing \(I\) with integration by parts, gives us.

🔍 As long as \(s \gt 0\text{,}\) both \(L_1\) and \(L_2\) go to zero.

Subsection 📤 Wrap-Up