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Section 11.3 Common Transforms

Now that we’ve defined the Laplace transform, the next step is to build a basic toolkit of common transforms. These are formulas for frequently encountered functions, such as constants, exponentials, powers of \(t\text{,}\) and trigonometric functions. Learning these transforms will allow you to compute Laplace transforms quickly, without having to evaluate integrals from scratch each time.
We’ll derive these formulas using straightforward examples and then generalize the results. You’ll see how the Laplace transform simplifies for each type of function and how certain patterns repeat across cases.

Subsection Laplace Transform of a Constant

Checkpoint 201. 🤔💭 Exponential Functions Pre-Reading Questions.

(a) 📖❓ Exponential Integration Rules.

(b) 📖❓ How is \(s\) Treated During Integration?

In the Laplace transform integral, the variable \(s\) is treated as a during the integration process.
  • constant
  • Correct! When evaluating the integral, \(s\) is treated as a constant, since the integration is with respect to \(t\text{.}\)
  • variable
  • No, even though \(s\) is a variable overall, it is treated as a constant during the integration.
  • coefficient
  • No, while \(s\) acts like a coefficient in \(e^{-st}\text{,}\) it’s conceptually treated as a constant in the integration.
  • limit
  • No, \(s\) is not a limit of integration; it appears inside the integrand.

(c) 📖❓ Choosing \(u\) and \(dv\) in \(\int t^2 e^{-st} dt\).

Which functions should you choose as \(u\) and \(dv\) when applying integration by parts to \(\int t^2 e^{-st} dt\text{?}\)
  • \(u = t^2,\quad dv = e^{-st} dt\)
  • Correct! This choice simplifies with each integration by parts.
  • \(u = e^{-st},\quad dv = t^2 dt\)
  • No, this makes integration more difficult. We want to differentiate \(t^2\) and integrate the exponential.
  • \(u = st,\quad dv = t dt\)
  • These choices are not aligned with the integrand \(t^2 e^{-st}\text{.}\)
We begin with the simplest case: the constant function \(f(t) = 1\text{.}\) By the definition of the Laplace transform,
\begin{equation*} \lap{1} = \int_0^\infty e^{-st}\, dt. \end{equation*}
This integral converges when \(s \gt 0\) and leads to our first Laplace transform rule:
Although this transform is specific to the constant function \(1\text{,}\) we will see later that it will serve as the transform formula for all constant functions.

Subsection Laplace Transform of Exponentials

To get the transform formula for exponentials of the form \(f(t) = e^{at}\text{,}\) where \(a\) is a constant, we will first find it for \(e^{7t}\text{:}\)

🌌 Example 202. Laplace Transform of \(e^{7t}\).

\begin{equation*} \lap{e^{7t}} = \frac{1}{s - 7}, \quad s \gt 7 \end{equation*}
Solution.
Using the definition of the Laplace transform, we get
\begin{equation*} \lap{ e^{7t} } = \int_0^{\infty} e^{-st} \cdot e^{7t}\ dt = \lim_{b \to \infty}\int_0^b e^{(7-s)t}\ dt\text{.} \end{equation*}
Focusing just on the integral, we let \(u=(7-s)t\) and apply \(u\)-substitution:
\begin{align*} \int_0^b e^{(7-s)t}\, dt = \frac{1}{7-s} e^{(7-s)t} \Bigg|_0^b \amp = \frac{1}{7-s} \left( e^{(7-s)b} - e^{0} \right)\\ \amp = \frac{1}{7-s} \left( e^{(7-s)b} - 1 \right) \end{align*}
🔍 \(u\)-substitution details.
Now, we compute the limit of this expression as \(b \to \infty\text{:}\)
\begin{align*} \lim_{b \to \infty} \frac{1}{7-s} \left( e^{(7-s)b} - 1 \right) \amp = \frac{1}{7-s} \Bigg[ \ub{\lim_{b \to \infty} e^{(7-s)b}}_{L} - \lim_{b \to \infty} 1 \Bigg]\text{.} \end{align*}
As before, \(L\) goes to \(0\) if the power of the exponential function goes to \(-\infty\text{.}\) This only happens if \(7-s\) is negative. That is, \(\ 7-s \lt 0\ \) or \(\ 7 \lt s\text{.}\)
So, as long as \(s \gt 7\text{,}\) then \(L \to 0\) and the transform becomes
\begin{equation*} \lap{e^{7t}} = \frac{1}{s - 7}, \quad s \gt 7\text{.} \end{equation*}
Replacing \(7\) with an \(a\) in this solution leads to the general formula:
For any constant \(a\text{,}\) the Laplace transform of \(e^{at}\) is given by:

📜 Laplace Transform of \(e^{at}\).

\begin{gather} \text{L}_2:\quad\lap{e^{at}} = \frac{1}{s - a}, \quad s \gt a.\tag{11.3} \end{gather}
💡 Derivation of L\(_2\).

Checkpoint 203. 📖❓ Give the Transform.

\begin{equation*} \lap{e^{-3t}} = \fillinmath{XX} \end{equation*}
  • \(\ds\frac{1}{s+3}\)
  • Correct! The Laplace transform of \(e^{-3t}\) is \(\ds\frac{1}{s+3}\text{.}\)
  • \(\ds\frac{1}{s-3}\)
  • No, this would be the transform of \(e^{3t}\text{.}\) Double-check the sign of the exponent.
  • \(\ds\frac{1}{s-3t}\)
  • No, \(t\) should not appear in the final expression, only \(s\text{.}\)
  • \(\ds\frac{3}{s+3}\)
  • No, the numerator should be 1, not 3. This is not a scaled exponential.

Subsection Laplace Transform of Powers of \(t\)

Now let’s consider power functions, \(f(t) = t^n\) where \(n\) is a non-negative integer.
\begin{equation*} \lap{t^n} = \frac{n!}{s^{n+1}}, \quad s > 0. \end{equation*}
We already found the transforms for \(n = 0\) and \(n = 1\) since
  • \(\displaystyle n=0:\qquad\lap{t^0} = \lap{1} = \dfrac{1}{s}\)
  • \(\displaystyle n=1:\qquad\lap{t^1} = \lap{t} = \dfrac{1}{s^2}\)
We could find \(\lap{t^2}\) next, but it is more instructive to show the following relationship:

🌌 Example 204. Laplace Transform of \(t^2\).

\begin{equation*} \lap{t^2} = \frac{2}{s}\cdot \lap{t}, \quad s \gt 0 \end{equation*}
Solution.
Applying the definition and writing the integral as a limit, we have
\begin{equation*} \lap{ t^2 } = \lim_{b \to \infty} \ub{\int_0^b e^{-st} \cdot t^2\ dt}_{I}\text{.} \end{equation*}
Applying integration by parts to \(I\) shows:
\begin{align*} \lap{ t^2 } \amp = \lim_{b \to \infty} \left[-\frac{b^2}{s}e^{-sb} + \frac{2}{s}\int_0^b e^{-st}\cdot t\ dt\right]\\ \amp = -\frac{1}{s} \lim_{b \to \infty} \left[-\frac{b^2}{e^{sb}}\right] + \frac{2}{s}\lim_{b \to \infty}\int_0^b e^{-st}\cdot t\ dt\\ \amp = -\frac{1}{s}\ub{\lim_{b \to \infty}\frac{b^2}{e^{sb}}}_{L} + \frac{2}{s}\ub{\int_0^\infty e^{-st}\cdot t\ dt}_{\large\lap{t}}. \end{align*}
🔍 Applying integration by parts to \(I\) shows:.
Noticing that the improper integral is just \(\lap{t}\text{,}\) we have
\begin{equation*} \lap{ t^2 } = -\frac{1}{s}\cdot L + \frac{2}{s}\cdot \lap{t}\text{,} \end{equation*}
but \(L=0\) if \(s \gt 0\text{,}\) so we have the relationship
\begin{equation*} \lap{t^2} = \frac{2}{s}\cdot \lap{t}, \quad s \gt 0 \end{equation*}
🔍 Limit comments.
You can use this work as a template for a general recursive relationship:

🌌 Example 205. Laplace Transform of \(t^n\).

\begin{equation*} \lap{t^n} = \frac{n}{s}\cdot \lap{t^{n-1}}, \quad s \gt 0 \end{equation*}
Solution.
Replacing \(2\) with \(n\) in the previous solution, we have
\begin{equation*} \lap{ t^n } = \lim_{b \to \infty} \ub{\int_0^b e^{-st} \cdot t^n\ dt}_{I}\text{.} \end{equation*}
Apply integration by parts to \(I\text{:}\)
\begin{align*} \lap{ t^{n} } \amp = \lim_{b \to \infty} \left[-\frac{b^n}{e^{sb}} + \frac{n}{s}\int_0^b e^{-st}\cdot t^{n-1}\ dt\right]\\ \amp = -\frac{1}{s} \lim_{b \to \infty} \left[-\frac{b^n}{e^{sb}}\right] + \frac{n}{s}\lim_{b \to \infty}\int_0^b e^{-st}\cdot t^{n-1}\ dt\\ \amp = -\frac{1}{s}\ub{\lim_{b \to \infty}\frac{b^n}{e^{sb}}}_{L} + \frac{n}{s}\ub{\int_0^\infty e^{-st}\cdot t^{n-1}\ dt}_{\large\lap{t^{n-1}}}. \end{align*}
🔍 Integration by parts details.
Noticing that the improper integral is just \(\lap{t^{n-1}}\text{,}\) we have
\begin{equation*} \lap{ t^{n} } = -\frac{1}{s}\cdot L + \frac{n}{s}\cdot \lap{t^{n-1}}\text{.} \end{equation*}
Since \(L=0\) if \(s \gt 0\text{,}\) we have
\begin{equation*} \lap{t^{n}} = \frac{n}{s}\cdot \lap{t^{n-1}}, \quad s \gt 0 \end{equation*}
🔍 Limit comments.
To get the general formula, we apply this relationship repeatedly for increasing \(n\text{:}\)
\begin{equation*} \begin{array}{crcl} n=1: \amp \DLO\lap{t} \amp = \amp \dfrac{1}{s}\cdot\lap{1} = \dfrac{1}{s}\cdot\dfrac{1}{s} = {\DLO\dfrac{1}{s^2}}\\ n=2: \amp \DLBa\lap{t^2} \amp = \amp \dfrac{2}{s}\cdot{\DLO\lap{t}} = \dfrac{2}{s}\cdot{\DLO\dfrac{1}{s^2}} = {\DLBa\dfrac{2\cdot 1}{s^3}}\\ n=3: \amp \DLGb\lap{t^3} \amp = \amp \dfrac{3}{s}\cdot{\DLBa\lap{t^2}} = \dfrac{3}{s}\cdot{\DLBa\dfrac{2\cdot 1}{s^3}} = {\DLGb\dfrac{3\cdot 2\cdot 1}{s^4}}\\ n=4: \amp \DLRa\lap{t^4} \amp = \amp \dfrac{4}{s}\cdot{\DLGb\lap{t^3}} = \dfrac{4}{s}\cdot{\DLGb\dfrac{3\cdot 2\cdot 1}{s^4}} = {\DLRa\dfrac{4\cdot 3\cdot 2\cdot 1}{s^5}}\\ \vdots \amp \vdots\quad\amp \amp \qquad\qquad\vdots\\ \end{array} \end{equation*}
This pattern suggests that, in general, we have
\begin{equation*} \lap{t^n} =\dfrac{n}{s}\cdot\lap{t^{n-1}} = \dfrac{n}{s}\cdot\dfrac{(n-1)\dotsm 4\cdot 3\cdot 2\cdot 1}{s^{n}}, \end{equation*}
which can be summarized as follows:

📜 Laplace Transform of \(t^n\).

\begin{gather} \text{L}_3:\quad\lap{t^n} = \frac{n!}{s^{n+1}}, \quad s \gt 0, \quad n = 0, 1, 2, \ldots.\tag{11.4} \end{gather}
💡 Derivation of L\(_3\).

Subsection Laplace Transforms of Sine and Cosine

Finally, we derive the Laplace transforms of \(\sin(bt)\) and \(\cos(bt)\text{.}\)
To make the derivation easier, we will find the transform of \(\cos(3t)\) first.

🌌 Example 206. Laplace Transform of \(\cos(3t)\).

\begin{equation*} \lap{\cos(3t)} = \frac{s}{s^2 + 3^2}, \quad s \gt 0 \end{equation*}
Solution.
Applying the definition and writing the integral as a limit, we have
\begin{equation*} \lap{ \cos(3t)} = \int_0^{\infty} e^{-st} \cdot \cos(3t)\ dt = \lim_{b \to \infty} \ub{\int_0^{b} e^{-st} \cos(3t)\ dt}_{I}\text{.} \end{equation*}
Treating \(s\) as constant, we apply integration by parts to \(I\) as follows:
\begin{equation*} I = -\frac{e^{-st}}{s} \cos(3t) \Bigg|_0^b - \frac{3}{s}\int_0^{b} e^{-st} \sin(3t)\ dt\text{,} \end{equation*}

📝: Integration by Parts.

but this requires a second integration by parts:
\begin{align*} I \amp = -\frac{e^{-st}}{s} \cos(3t) \Bigg|_0^b - \frac{3}{s} \left[ -\frac{e^{-st}}{s} \sin(3t)\Bigg|_0^b + \frac{3}{s} \int_0^{b} e^{-st} \cos(3t)\ dt \right]\\ \amp = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b - \frac{9}{s^2} \ub{\int_0^{b} e^{-st} \cos(3t)\ dt}_{I}\text{.} \end{align*}
Since the integral \(I\) reappeared, we actually have
\begin{equation*} I = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b - \frac{9}{s^2} I\text{.} \end{equation*}
Evaluating the limits of integration and solving for \(I\text{,}\) we find
\begin{align*} I + \frac{9}{s^2} I \amp = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b\\ I\left(\frac{s^2+9}{s^2}\right) \amp = \left(-\frac{e^{-sb}}{s} \cos(3b) + \frac{3e^{-sb}}{s^2} \sin(3b) + \frac{1}{s} \right)\\ I \amp = \frac{s^2}{s^2+9}\left(-\frac{e^{-sb}}{s} \cos(3b) + \frac{3e^{-sb}}{s^2} \sin(3b) + \frac{1}{s} \right) \end{align*}
Finally, the limit of \(I\) as \(b \to \infty\) is the sum of the limits
\begin{align*} \lim_{b \to \infty} I = \amp\\ \frac{s^2}{s^2+9} \amp \Bigg( - \frac{1}{s}\ub{\lim_{b \to \infty} e^{-sb} \cos(3b)}_{L_1} + \frac{3}{s^2}\ub{\lim_{b \to \infty} e^{-sb} \sin(3b)}_{L_2} + \frac{1}{s} \Bigg) \end{align*}
Since \(L_1 \to 0\) & \(L_2 \to 0\) if \(s \gt 0\text{,}\) we can conclude that
\begin{equation*} \lap{\cos(3t)} = \frac{s}{s^2 + 9}, \quad s \gt 0 \end{equation*}
🔍 Limit comments.
This computation, along with a similar one for sine, leads to the following general formulas:

📜 Laplace Transforms of Sine and Cosine.

\begin{gather} \text{L}_4:\quad\lap{\sin(bt)} = \frac{b}{s^2 + b^2}, \quad s \gt 0.\tag{11.5} \end{gather}
\begin{gather} \text{L}_5:\quad\lap{\cos(bt)} = \frac{s}{s^2 + b^2}, \quad s \gt 0,\tag{11.6} \end{gather}
💡 Derivation of L\(_4\) & L\(_5\).
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