Section E.1 Laplace Transforms
🔍 Limit comments.
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*}
🔍 Integration by parts details.
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 = \left[-\frac{t}{s}e^{-st}\right]_0^b+ \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*}
🔍 Limit comments.
Assuming \(s \gt 0\text{,}\) the limits are computed by:
-
\(\displaystyle \ds L_1 = \lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} \,\us{LH}{=}\, \lim_{b \to \infty}\frac{1}{se^{sb}} = \frac{1}{s}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0\)
-
\(\displaystyle \ds L_2 = \lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0\)
where \(LH\) denotes L’Hôpital’s Rule.
💡 Derivation of L\(_1\).
By definition, we have
\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{.}
\end{equation*}
Assuming \(s\) and \(b\) are constant, we integrate \(I\) with respect to \(t\text{:}\)
\begin{equation*}
I = \int_0^b e^{-st}dt = -\frac{1}{s}e^{-st}\Big|_{t=0}^{t=b} = -\frac{1}{s}\left[ e^{-sb} - 1 \right]\text{.}
\end{equation*}
Next, we compute the limit of \(I\) as \(b \to \infty\) (with \(s\) constant):
\begin{align*}
\lim_{b \to \infty} I
\amp = \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]
= -\frac{1}{s} \Big[ \ub{\lim_{b \to \infty} e^{-sb}}_{L} - 1 \Big]
\end{align*}
Taking \(b\to\infty\text{,}\) \(L\) must go to \(0\) if \(s\) is positive, leaving us with the transform of \(1\text{:}\)
\begin{equation*}
\lap{1} = -\frac{1}{s} [0 - 1] = \frac{1}{s} \quad \text{for } s \gt 0\text{.}
\end{equation*}
🔍 \(u\)-substitution details.
👉 Using \(u\)-substitution with
\begin{align*}
u \amp = \ob{(7-s)}^{\text{constant}}t\\
du \amp = (7-s)dt \quad \Rightarrow \quad dt = \frac{du}{7-s}
\end{align*}
the integral is computed as
\begin{align*}
\int_0^b e^{(7-s)t}\ dt = \int_{u=0}^{u=(7-s)b} e^{u}\frac{du}{7-s}
\amp = \frac{1}{7-s}\int_{u=0}^{u=(7-s)b} e^{u}\ du\\
\amp = \frac{1}{7-s} \left[ e^{u} \right]\Bigg|_{u=0}^{u=(7-s)b}\\
\amp = \frac{1}{7-s} \left[ e^{(7-s)b} - 1 \right]\text{.}
\end{align*}
💡 Derivation of L\(_2\).
By definition,
\begin{equation*}
\lap{ e^{at} }
= \int_0^{\infty} e^{-st} \cdot e^{at}\ dt
= \lim_{b \to \infty}\int_0^b e^{(a-s)t}\ dt\text{.}
\end{equation*}
For the integral, let \(u=a-s\) and apply \(u\)-substitution:
\begin{align*}
\int_0^b e^{(a-s)t}\, dt
\knowl{./knowl/xref/lt-example-2-details.html}{\text{\(\os{\large ❔}{=}\)}}
\frac{1}{a-s} e^{(a-s)t} \Bigg|_0^b
\amp = \frac{1}{a-s} \left( e^{(a-s)b} - e^{0} \right)\\
\amp = \frac{1}{a-s} \left( e^{(a-s)b} - 1 \right)
\end{align*}
Now, we compute the limit of this expression as \(b \to \infty\text{:}\)
\begin{align*}
\lim_{b \to \infty} \frac{1}{a-s} \left( e^{(a-s)b} - 1 \right)
\amp = \frac{1}{a-s} \Bigg[ \ub{\lim_{b \to \infty} e^{(a-s)b}}_{L} - \lim_{b \to \infty} 1 \Bigg]\text{.}
\end{align*}
As before, \(L\) goes to \(0\) if \(a-s\) is negative. That is, \(\ a-s \lt 0\ \) or \(\ a \lt s\text{.}\)
So, as long as \(s \gt a\text{,}\) then \(L \to 0\) and the transform becomes
\begin{equation*}
\lap{e^{at}} = \frac{1}{s - a}, \quad s \gt a\text{.}
\end{equation*}
🔍 Applying integration by parts to \(I\) shows:.
\begin{align*}
u = t^2, \quad\amp dv = e^{-st}dt, \\
du = 2t\ dt, \quad\amp v = -\frac{1}{s}e^{-st}
\end{align*}
Integration by parts gives
\begin{align*}
\int_0^b e^{-st} \cdot t^2\ dt
\amp = t^2 \cdot \left( -\frac{1}{s}e^{-st} \right)\Bigg|_0^b - \int_0^b \left( -\frac{1}{s}e^{-st} \right) 2t\ dt\\
\amp = -\frac{b^2}{s}e^{-sb} + \frac{2}{s} \int_0^b e^{-st} \cdot t\ dt
\end{align*}
🔍 Limit comments.
Again, if \(s \lt 0\text{,}\) the limit goes to \(+\infty\text{.}\) So we must have \(s \gt 0\text{.}\) Applying L’Hôpital’s rule twice, we get
\begin{align*}
L = \lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b^2}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}\
\amp\us{LH}{=}\ \lim_{b \to \infty}\frac{2b}{se^{sb}}\
= \frac{2}{s}\lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}\\
\amp\us{LH}{=}\ \frac{2}{s}\lim_{b \to \infty}\frac{1}{se^{sb}}
= \frac{2}{s^2}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}
= 0
\end{align*}
🔍 Integration by parts details.
Using the substitution:
\begin{align*}
u = t^n, \quad\amp dv = e^{-st}dt, \\
du = nt^{n-1}\ dt, \quad\amp v = -\frac{1}{s}e^{-st} \text{,}
\end{align*}
integration by parts gives
\begin{align*}
\lap{ t^{n} }
\amp = \lim_{b \to \infty} \left[-\frac{t^{n}}{s}e^{-sb}\Bigg|_0^b - \int_0^b \left(-\frac{1}{s}e^{-st}\right)\cdot nt^{n-1}\ dt\right]\\
\amp = \lim_{b \to \infty} \left[\left(-\frac{b^n}{s}e^{-sb} - 0\right) + \frac{n}{s}\int_0^b e^{-st}\cdot t^{n-1}\ dt\right]\\
\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]
\end{align*}
🔍 Limit comments.
The reasoning is the same as in previous solution except we apply L’Hôpital’s rule n times. So we have
\begin{align*}
\lim_{b \to \infty}\frac{b^n}{e^{sb}}\
\amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}}
\lim_{b \to \infty}\frac{nb^{n-1}}{se^{sb}}
= \frac{n}{s}\lim_{b \to \infty}\frac{b^{n-1}}{e^{sb}}\\
\amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}}
\frac{n}{s}\lim_{b \to \infty}\frac{(n-1)b^{n-2}}{se^{sb}}
= \frac{n(n-1)}{s^2}\lim_{b \to \infty}\frac{b^{n-2}}{e^{sb}}\\
\amp \qquad\qquad \vdots \quad LH\ \ n-2 \text{ more times}\\
\amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}}
\frac{n!}{s^n}\lim_{b \to \infty}\frac{1}{se^{sb}}
= \frac{n!}{s^{n+1}}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}
= 0
\end{align*}
💡 Derivation of L\(_3\).
Most of the ideas were described above, but the formal derivation follows by induction on \(n\text{,}\) which we will not cover here.
🔍 Limit comments.
👉 If \(s > 0\text{,}\) then we know that \(e^{-sb} \to 0\) as \(b \to \infty\) and both \(\cos(3b)\) and \(\sin(3b)\) oscillate between \(-1\) and \(1\text{,}\) thus their product must approach zero. That is,
\begin{equation*}
\lim_{b \to \infty} e^{-sb} \cos(3b) = 0 \quad \text{and} \quad \lim_{b \to \infty} e^{-sb} \sin(3b) = 0
\end{equation*}
💡 Derivation of L\(_4\) & L\(_5\).
The derivation of L\(_5\) mirrors the solution to \(\lap{\cos(3t)}\text{.}\) Just swap \(3\) with \(b\) and \(9\) with \(b^2\text{.}\)
Note: \(s\) cannot be \(0\).
When \(s=0\text{,}\) the integral becomes
\begin{align*}
\amp
= \lim_{b \to \infty} \int_0^b t\ dt
= \lim_{b \to \infty} \dfrac{t^2}{2}\Big|_0^b
= \dfrac{1}{2} \lim_{b \to \infty} b^2 = \infty.
\end{align*}
Therefore, we must have \(s\ne 0\) for this integral to be finite.
Limit comments.
If \(s \lt 0\text{,}\) then as \(b \to \infty\text{,}\) we would have \(e^{-sb} \to \infty\) and so
\begin{equation*}
\lim_{b \to \infty} \os{\infty}{\os{\uparrow}{\boxed{b}}}\ \us{\infty}{\us{\downarrow}{\boxed{e^{-sb}}}} = \infty.
\end{equation*}
This shows the Laplace transform would not exist if \(s \lt 0\text{.}\) Therefore, we must require \(s \gt 0\text{.}\)
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