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7.4 Derivative, integral, and multiplication of Laplace Transforms

we know:

\[ \mathcal{L} \{ f'(t) \} = s F(s) - f(0) \]\[ \mathcal{L} \left\{ \int_{0}^{t} f(\tau) d\tau \right\} = \frac{F(s)}{s} \]

today: deriv. and integral in s-domain

\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]\[ F'(s) = \frac{d}{ds} \int_{0}^{\infty} f(t) e^{-st} dt \]\[ = \int_{0}^{\infty} f(t) \left( \frac{d}{ds} e^{-st} \right) dt = \int_{0}^{\infty} f(t) (-t \cdot e^{-st}) dt \]\[ = \int_{0}^{\infty} [-t \cdot f(t)] e^{-st} dt \]
\[ F'(s) = \mathcal{L} \{ -t \cdot f(t) \} \]
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repeat: \( F''(s) = \mathcal{L} \{ t^2 f(t) \} \) each differentiation gets a factor of -t

\[ F^{(n)}(s) = \mathcal{L} \{ (-t)^n f(t) \} \]

useful when finding Laplace transform of \( t \) times something on the table

for example, \( \mathcal{L} \{ t \cosh(6t) \} \) is not on the table but \( \mathcal{L} \{ \cosh(6t) \} \) is

\[ \mathcal{L} \{ -t f(t) \} = F'(s) \]\[ \mathcal{L} \{ t \cosh(6t) \} = \mathcal{L} \{ -t (-\cosh(6t)) \} = F'(s) \]

where \( f(t) = -\cosh(6t) \)

find \( F(s) = -\frac{s}{s^2 - 36} \)

\[ \frac{d}{ds} \left( -\frac{s}{s^2 - 36} \right) = \dots = \frac{s^2 + 36}{(s^2 - 36)^2} = \mathcal{L} \{ t \cosh(6t) \} \]
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Inverse Laplace Transform via Differentiation

We can also use it the other way:

\[ F'(s) = \mathcal{L} \{ -t f(t) \} \]
\[ \frac{\mathcal{L}^{-1} \{ F'(s) \}}{-t} = f(t) \]

For example, find inverse of \( \ln \left( \frac{1}{s^2 - 16} \right) \). (Note: not on table)

Step-by-Step Solution

Rewrite:

\[ \begin{aligned} \ln \left( \frac{1}{s^2 - 16} \right) &= \ln(1) - \ln(s^2 - 16) \\ &= 0 - \ln[(s+4)(s-4)] \\ F &= -\ln(s+4) - \ln(s-4) \end{aligned} \]

Differentiate with respect to \( s \):

\[ F' = -\frac{1}{s+4} - \frac{1}{s-4} \]

Apply the inverse Laplace transform to the derivative:

\[ \mathcal{L}^{-1} \left\{ -\frac{1}{s+4} - \frac{1}{s-4} \right\} = -e^{-4t} - e^{4t} \]

Then find \( f(t) \):

\[ f(t) = \frac{\mathcal{L}^{-1} \{ F' \}}{-t} = \frac{e^{-4t} + e^{4t}}{t} \]
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Integration in the s-domain

Now integral in \( s \)-domain, specifically \( \int_{s}^{\infty} F(\sigma) d\sigma \).

Start with:

\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]
\[ \int_{s}^{\infty} F(\sigma) d\sigma = \int_{s}^{\infty} \int_{0}^{\infty} f(t) e^{-\sigma t} dt d\sigma \]

Swap integration order:

\[ \begin{aligned} &\int_{0}^{\infty} \int_{s}^{\infty} f(t) e^{-\sigma t} d\sigma dt \\ &= \int_{0}^{\infty} f(t) \cdot \left. \frac{1}{-t} e^{-\sigma t} \right|_{\sigma=s}^{\sigma=\infty} dt \\ &= \int_{0}^{\infty} -\frac{1}{t} f(t) (0 - e^{-st}) dt = \int_{0}^{\infty} \left[ \frac{f(t)}{t} \right] e^{-st} dt \end{aligned} \]

\[ \int_{s}^{\infty} F(\sigma) d\sigma = \mathcal{L} \left\{ \frac{f(t)}{t} \right\} \]

Condition: \( \lim_{t \to 0^+} \frac{f(t)}{t} \) must exist.

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Laplace Transform of Division by t

Useful to transform something over \(t\) when \(\mathcal{L}\{\text{something}\}\) is known.

\[\mathcal{L} \left\{ \frac{1 - \cos(t)}{t} \right\} = \int_{s}^{\infty} F(\sigma) d\sigma\]

Note: \(f(t) \to F(s)\). The condition \(\lim_{t \to 0^+} \frac{1 - \cos(t)}{t}\) must exist.

\[= \lim_{t \to 0^+} \frac{\sin(t)}{1} = 0 \text{ (exists)}\]

First, find the Laplace transform of the numerator:

\[\mathcal{L} \{1 - \cos(t)\} = \frac{1}{s} - \frac{s}{s^2 + 1} = F(s)\]

Now, integrate from \(s\) to \(\infty\):

\[\begin{aligned} \int_{s}^{\infty} \left( \frac{1}{\sigma} - \frac{\sigma}{\sigma^2 + 1} \right) d\sigma &= \left. \ln \sigma - \frac{1}{2} \ln(\sigma^2 + 1) \right|_{\sigma=s}^{\sigma=\infty} \\ &= \left. \ln \sigma - \ln \sqrt{\sigma^2 + 1} \right|_{\sigma=s}^{\sigma=\infty} \\ &= \left. \ln \left( \frac{\sigma}{\sqrt{\sigma^2 + 1}} \right) \right|_{\sigma=s}^{\sigma=\infty} \\ &= \ln(1) - \ln \left( \frac{s}{\sqrt{s^2 + 1}} \right) \end{aligned}\]
\[\mathcal{L} \left\{ \frac{1 - \cos(t)}{t} \right\} = -\ln \left( \frac{s}{\sqrt{s^2 + 1}} \right)\]
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Final Value Theorem Derivation

Revisit the Laplace transform of a derivative:

\[\begin{aligned} \mathcal{L} \{ f'(t) \} &= s F(s) - f(0) \\ &= \int_{0}^{\infty} f'(t) e^{-st} dt \end{aligned}\]

Taking the limit as \(s \to 0\):

\[\lim_{s \to 0} [s F(s) - f(0)] = \lim_{s \to 0} \int_{0}^{\infty} f'(t) e^{-st} dt\]
\[\lim_{s \to 0} s F(s) - f(0) = \int_{0}^{\infty} f'(t) dt = \lim_{t \to \infty} \int_{0}^{t} f'(\tau) d\tau\]

By the Fundamental Theorem of Calculus:

\[= \lim_{t \to \infty} f(t) - f(0)\]

Canceling \(f(0)\) from both sides, we obtain the Final Value Theorem:

\[\lim_{s \to 0} s F(s) = \lim_{t \to \infty} f(t)\]
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Homework Problem Analysis

#5 on HW:

\[ \int_{0}^{\infty} \frac{\sin(t)}{t} dt \]

Let \( g(t) = \int_{0}^{t} \frac{\sin(\tau)}{\tau} d\tau \)

Equivalent to asking \( \lim_{t \to \infty} g(t) \)