Question

Let \(f(t)\) be piecewise continuous and exponentially bounded on the interval \(0 \leq t<\infty\), and let \(F(s)\) denote the Laplace transform of \(f(t)\). It is shown in advanced calculus \(^{3}\) that it is possible to differentiate under the integral sign with respect to the parameter \(s\). That is, $$ \frac{d}{d s} F(s)=\frac{d}{d s} \int_{0}^{\infty} e^{-s t} f(t) d t=\int_{0}^{\infty} \frac{d}{d s}\left[e^{-s t} f(t)\right] d t . $$ (a) Use this result to show that \(\mathcal{L}\\{t f(t)\\}=-F^{\prime}(s)\). (b) Use the result of part (a) to establish formula (20) in Table 5.1.

Short answer

Q: Use the result of part (a) to establish that the Laplace transform of the function t^n (n is a natural number) is given by L{t^n} = n!/s^(n+1). A: To establish the Laplace transform of the function t^n, we need to find the Laplace transform of the power function f(t) = t^n, where n is a natural number. We can use the result of part (a), which states that the Laplace transform of t*f(t) is -F'(s). We find the Laplace transform of t^n using the definition: F(s) = ∫(0 to ∞) e^(-st) t^n dt Now, we use the result of part (a): L{t^(n+1)} = -F'(s) We then integrate by parts, and after applying this procedure repeatedly up to n times, we arrive at the required formula: L{t^n} = n! / s^(n+1)

Step-by-step solution

Prove that \(\mathcal{L}\\{t f(t)\\}=-F^{\prime}(s)\) using the given result

To prove this, first recall the definition of Laplace transform of \(f(t)\) which is given by: $$ F(s) = \int_{0}^{\infty} e^{-st} f(t) dt $$ Now, we can find the derivative of \(F(s)\) with respect to \(s\) using the provided result of differentiation under the integral sign: $$ \frac{d}{ds} F(s) = \frac{d}{ds} \int_{0}^{\infty} e^{-st} f(t) dt = \int_{0}^{\infty} \frac{d}{ds}\left[e^{-st} f(t)\right] dt $$ Observe that \(f(t)\) does not depend on \(s\), so we can differentiate the integrand with respect to \(s\): $$ \int_{0}^{\infty} \frac{d}{ds}\left[e^{-st} f(t)\right] dt = \int_{0}^{\infty} -te^{-st} f(t) dt $$ Comparing this result with the definition of the Laplace transform, we can see that the integrand represents the Laplace transform of \(t f(t)\). Therefore, we have: $$ \frac{d}{ds} F(s) = -\mathcal{L}\\{t f(t)\\} $$ Therefore, we have shown that \(\mathcal{L}\\{t f(t)\\}=-F^{\prime}(s)\).

Use the result of part (a) to establish the formula (20) in Table 5.1

To establish the formula, we will consider a simple function \(f(t) = t^n\) where \(n \in \mathbb{N}\) which represents the power function. We need to prove that \(\mathcal{L}\\{t^{n}\\}= \frac{n!}{s^{n+1}}\) First, let's find out the Laplace transform of \(f(t) = t^n\) using the definition: $$ F(s) = \int_{0}^{\infty} e^{-st} t^n dt $$ Now, we will recall the result of part (a) which states that \(\mathcal{L}\\{t f(t)\\}=-F^{\prime}(s)\). In our case, this would become: $$ \mathcal{L}\\{t \cdot t^n\\}=-F^{\prime}(s) $$ This simplifies to: $$ \mathcal{L}\\{t^{n+1}\\}=-F^{\prime}(s) $$ Now, we will integrate by parts, taking \(u' = e^{-st}\) and \(v = t^n\). Then, \(u = -\frac{1}{s}e^{-st}\) and \(v' = nt^{n-1}\). Using integration by parts formula: $$ -\frac{d}{ds}F(s) = \left[-\frac{1}{s}e^{-st} t^n\right]_{0}^{\infty} - \int_{0}^{\infty}\left(-\frac{1}{s^2}e^{-st}\right)nt^{n-1} dt $$ At \(t = 0\), the term \(e^{-st} t^n = 0\) and at \(t = \infty\), the exponential term dominates, making the whole term \(0\). Thus, the left boundary term vanishes. We have, $$ -\frac{d}{ds}F(s) = \frac{n}{s^2}\int_{0}^{\infty}e^{-st} t^{n-1} dt $$ This is the Laplace Transform of \(t^{n-1}\), so we can write it as: $$ -\frac{d}{ds}F(s) = \frac{n}{s^2}\mathcal{L}\\{t^{n-1}\\} $$ Now, we will apply this procedure repeatedly up to \(n\) times to get the required formula. On the \(n\)-th derivative, we will have: $$ (-1)^n F^{(n)}(s) = n! \mathcal{L}\\{1\\} $$ Since \(\mathcal{L}\\{1\\}= \frac{1}{s}\), we can write: $$ (-1)^n F^{(n)}(s) = \frac{n!}{s} $$ Knowing that this is the \(n\)-th derivative, the integral must be of the form \(\frac{n!}{s^{n+1}}\) (due to the integration of the exponentials). Therefore, we have shown that: $$ \mathcal{L}\\{t^n\\} = \frac{n!}{s^{n+1}} $$ And we have established the formula (20) in Table 5.1.

Key concepts

Differentiation under the Integral Sign

The concept of differentiation under the integral sign can seem daunting at first, but it's essentially a powerful technique that allows us to differentiate an integral with respect to a parameter. This becomes particularly handy in situations where the traditional rules of differentiation are difficult to apply.

In the context of Laplace Transforms, this technique involves differentiating the integral of a function that is multiplied by an exponential term. The general form here is the Laplace transform given by:

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

Here, \(e^{-st}\) acts as a scaling factor depending on the parameter \(s\). By differentiating under the integral sign with respect to \(s\), we are essentially finding how the entire transform changes as \(s\) changes. This form of differentiation is crucial in proving identities involving Laplace Transforms, such as \(\mathcal{L}\{t f(t)\} = -F'(s)\).

To master this technique, it's vital to keep in mind:

• The function \(f(t)\) does not depend on \(s\).
• The exponential term \(e^{-st}\) is the component that changes with \(s\).
• Apply the derivative to the exponential part first, then integrate the modified function.

Piecewise Continuous Functions

Piecewise continuous functions are those which may not be continuous over their entire domain, but are continuous within specific intervals. This type of function is very useful in real-world scenarios where processes are segmented.

We often encounter piecewise continuous functions in engineering and physics problems, where an action might change abruptly at certain points. For example, turning a switch on or off leads to different behaviors in a circuit. In the context of Laplace Transforms, defining a piecewise continuous function \(f(t)\) on the interval \([0, \infty)\) allows us to compute the transform reliably.

Key characteristics include:

• Continuity within given intervals, but allows for a finite number of jumps or discontinuities.
• Functions are well-defined around each discontinuity point.

When working with Laplace Transforms, piecewise functions allow for more flexibility and accuracy. Moreover, as long as the function is continuous within its segments and doesn't grow too fast (like having infinite spikes), it remains a good candidate for transformation.

Exponential Bounding

Exponential bounding refers to the condition that a function must satisfy to ensure the Laplace Transform exists. It ensures that our function does not grow faster than an exponential function as \(t\) approaches infinity.

For a function \(f(t)\) to be exponentially bounded, it must be possible to find constants \(M\) and \(c\) such that:

• \(|f(t)| \leq Me^{ct}\) for all sufficiently large \(t\).

This condition implies that as \(t\) goes to infinity, \(f(t)\) won't explode in value uncontrollably. Instead, its growth will be contained to a manageable scale. This restriction is crucially implemented in the definition of the Laplace Transform, which is

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

Here, the term \(e^{-st}\) effectively "dampens" the growth of the function \(f(t)\), allowing the integral to converge for suitable values of \(s\). The condition of exponential bounding ensures the Laplace Transform will yield a finite, meaningful result, which forms the backbone of using these transforms for solving differential equations and modeling time-based systems.