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7.4 Derivative, Integral, Multiplication of Laplace Transform

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

let's look at \( F'(s) \). What happens in \( t \)-domain?

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

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

this can help us with, for example, \( \mathcal{L} \{ t \cosh(6t) \} \)

Note: \( t \) is an extra factor; \( \cosh(6t) \) is on the table.

look at \( F'(s) = \mathcal{L} \{ -t f(t) \} \)

\[ F'(s) = \mathcal{L} \{ t \cdot (-\cosh(6t)) \} \]
deriv. in s-domain

\( f(t) \to F(s) \) (table)

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

If \( F(s) = \ln \left( \frac{1}{s^2 - 16} \right) \), what is \( f(t) \)?

Note: This form is not on the table.

We use the property: \( F'(s) = \mathcal{L} \{ -t f(t) \} \)

Step 1: Simplify \( F(s) \) using logarithm properties

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

Step 2: Differentiate \( F(s) \)

\[ F'(s) = -\frac{1}{s+4} - \frac{1}{s-4} \]
Note: These terms are now on the table.

Step 3: Solve for \( f(t) \)

Since \( F'(s) = \mathcal{L} \{ -t f(t) \} \), then:

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

Now integration in s-domain, specifically \( \int_{s}^{\infty} F(\sigma) d\sigma \) where \( \sigma \) is a dummy variable.

Start with the definition of the Laplace transform:

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

Then integrate both sides with respect to \( \sigma \):

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

Swap integration order

\[ \begin{aligned} \int_{s}^{\infty} F(\sigma) d\sigma &= \int_{0}^{\infty} f(t) \left( \int_{s}^{\infty} e^{-\sigma t} d\sigma \right) dt \\ &= \int_{0}^{\infty} f(t) \left[ -\frac{1}{t} e^{-\sigma t} \right]_{\sigma=s}^{\sigma=\infty} dt \\ &= \int_{0}^{\infty} f(t) \left( 0 + \frac{1}{t} e^{-st} \right) dt \\ &= \int_{0}^{\infty} \left( \frac{f(t)}{t} \right) e^{-st} dt \end{aligned} \]
\[ \int_{s}^{\infty} F(\sigma) d\sigma = \mathcal{L} \left\{ \frac{f(t)}{t} \right\} \]
\( \lim_{t \to 0^+} \frac{f(t)}{t} \) must exist
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Laplace Transform of Functions Divided by \(t\)

Useful for Laplace transform of something over \(t\).

Example Calculation

For example, consider \(\mathcal{L} \left\{ \frac{1 - \cos(t)}{t} \right\} \). Note that the numerator \(1 - \cos(t)\) has a Laplace transform available on the table.

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

Check if \(\lim_{t \to 0^+} \frac{1 - \cos(t)}{t}\) exists:

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

From table lookup:

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

Applying the property:

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

Let's revisit the Laplace transform of a derivative:

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

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

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

Final value theorem

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Homework 5: Final Value Theorem Application

On HW 5: Evaluate the following integral using the final value theorem:

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

Define the function \( g(t) \) as:

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

Find the limit:

\[ \lim_{t \to \infty} g(t) \]