Subsection 9.2.1 Constant Function, \(1\)
Example 3.
\(\ \ \)\(\ds \lap{1}\text{.}\)Solution.
By the definition above, we have the improper integral
\begin{equation}
\lap{1} = \int_0^{\infty} e^{-st} \cdot 1\ dt = \lim_{b \to \infty} \ub{\int_0^b e^{-st}dt}_{I}\text{.}\tag{46}
\end{equation}
To evaluate this, we treat \(s\) as a constant and integrate with respect to \(t\text{.}\)
\begin{align*}
I = \int_0^b e^{-st}dt
=\amp\ -\frac{1}{s}e^{-st}\Big|_{t=0}^{t=b}\\
=\amp\ -\frac{1}{s}\left[ e^{-sb} - e^{-s\cdot 0} \right]\\
=\amp\ -\frac{1}{s}\left[ e^{-sb} - 1 \right]
\end{align*}
Now that we have \(I\text{,}\) we can plug it into the limit above.
\begin{align*}
\lim_{b \to \infty} -\frac{1}{s}\left[ e^{-sb} - 1 \right]
=\amp\ -\frac{1}{s} \lim_{b \to \infty} \Big[ e^{-sb} - 1 \Big]
\qquad \leftarrow \knowl{./knowl/xref/reason-factor-out-s.html}{\text{why?}}\\
=\amp\ -\frac{1}{s} \Big[ \ub{\lim_{b \to \infty} e^{-sb}}_{L} - \us{=\ 1}{\ub{\lim_{b \to \infty} 1}} \Big]
\end{align*}
The value of the remaining limit, \(L\text{,}\) depends on whether \(s\) is positive or negative. So we consider both cases:
\(s < 0 \text{ (negative) and } b\to \infty \implies -sb \to \infty \implies e^{-sb} \to \infty\)
\(s > 0 \text{ (positive) and } b\to \infty \implies -sb \to -\infty \implies e^{-sb} \to 0\)
Therefore, when \(s > 0\) the Laplace transform exists and evaluates to
\begin{equation*}
\lap{1} = -\frac{1}{s} [0 - 1] = \frac{1}{s} \quad \text{for } s \gt 0.
\end{equation*}
Now that we’ve worked through an example, here are a few key takeaways:
The Laplace transform always results in a function of \(s\text{.}\) We typically denote the Laplace transform using the capitalized letter of the function we are transforming. For instance,
\begin{equation*}
\lap{f(t)} = F(s) \quad \text{or} \quad \lap{q(t)} = Q(s).
\end{equation*}
The values of \(s\) for which the integral converges are crucial. While these values usually don’t affect the solution to the differential equation, we will make a habit of noting them as we build our Laplace transform toolbox.
The notation \(\lap{}\) includes curly braces to indicate the function being transformed.
With this, we’ve introduced our first Laplace transform. In the next few sections, we will explore more of the common transforms that are used to solve differential equations.
Common Laplace Transform \(L_1\).
- \({\LARGE \vphantom{\int}}L_1\)
- \(\ds \lap{ 1 } = \frac{1}{s}, \quad s > 0 \)
Reading Questions Check-Point Questions
1. What is the result of the Laplace transform of the constant function 1?
\(\ds \frac{1}{t}\)
No, \(\ds \frac{1}{t}\) is not the result of the Laplace transform of 1.
\(\ds \frac{1}{s}\)
Correct! The Laplace transform of the constant function 1 is \(\ds \frac{1}{s}\) for \(s > 0\text{.}\)
\(\ds \frac{1}{t+s}\)
No, this is not the correct expression for the Laplace transform of 1.
\(\ds \frac{1}{t-s}\)
No, this expression does not represent the Laplace transform of 1.
2. In the Laplace transform integral, the variable \(s\) is treated as a during the integration process.
constant
Correct! The variable \(s\) is treated as a constant when performing the integration in the Laplace transform.
variable
No, while \(s\) is technically a variable, it is treated as a constant during the integration with respect to \(t\text{.}\)
coefficient
No, \(s\) is treated as a constant, not as a coefficient, during the integration process.
limit
No, \(s\) does not represent a limit in this context; it is treated as a constant.
3. Based on this discussion, \(\ds\lap{f(t)}\) is a .
function of \(t\)
No, while \(f(t)\) is a function of \(t\text{,}\) \(\ds\lap{f(t)}\) is not a function of \(t\text{.}\)
function of \(s\) and \(t\)
No, \(\ds\lap{f(t)}\) is not a function of \(t\text{.}\)
function of \(s\)
Correct! The Laplace transform of a function \(f(t)\) is a function of \(s\text{.}\)
constant
No, \(\ds\lap{f(t)}\) is not a constant.
4. In a Laplace transform, where does the variable \(s\) come from?
It is an integration constant
No, \(s\) is not an integration constant.
It is a placeholder for the function \(f(t)\)
No, \(s\) is not a placeholder for the function \(f(t)\text{.}\)
It comes from \(e^{-st}\) in the definition
Correct! The variable \(s\) is introduced as a parameter in the Laplace transform definition to transform the function \(f(t)\text{.}\)
\(s\) replaces \(\infty\) in an improper integral
No, \(s\) is not a replacement for \(\infty\text{.}\)
