Question

Concerned with differentiation of the Laplace transform. Let $$ F(s)=\int_{0}^{\infty} e^{-s i} f(t) d t $$ It is possible to show that as long as \(f\) satisfics the conditions of Theorem \(6.1 .2,\) it is legitimate to differentiate under the integral sign with respect to the parameter \(s\) when \(s>a .\) (a) Show that \(F^{\prime}(s)=\mathcal{L}\\{-l f(t)\\}\) (b) Show that \(F^{(n)}(s)=\mathcal{L}\left\\{(-t)^{n} f(t)\right\\} ;\) hence differentiating the Laplace transform corresponds to multiplying the original function by \(-t .\)

Short answer

Question: Prove the following properties of the Laplace Transform related to differentiation: (a) Show that the first derivative of the Laplace transform with respect to s is equal to the Laplace transform of -tf(t), (b) Demonstrate that the nth derivative of the Laplace transform with respect to s is equal to the Laplace transform of (-t)^n * f(t).

Step-by-step solution

Differentiate F(s) with respect to s

To demonstrate that the first derivative of F(s) is equal to the Laplace Transform of -tf(t), we will first find the derivative of F(s) with respect to s. Recall that F(s) is given as: $$ F(s)=\int_{0}^{\infty} e^{-st} f(t) dt $$ We will differentiate F(s) under the integral sign with respect to s: $$ F'(s) = \frac{d}{ds} \int_{0}^{\infty} e^{-st} f(t) dt $$

Apply the chain rule for differentiation

Apply the chain rule to differentiate the exponential term e^(-st) with respect to s. Remember that f(t) does not depend on s, so it can be treated as a constant during differentiation: $$ F'(s) = \int_{0}^{\infty} \frac{d}{ds}(e^{-st}) f(t) dt $$ $$ F'(s) = \int_{0}^{\infty} (-te^{-st}) f(t) dt $$ At this point, we have found the first derivative of F(s) with respect to s.

Compare with the definition of the Laplace Transform

To prove that the first derivative of F(s) is equal to the Laplace Transform of -tf(t), we need to compare this result with the definition of the Laplace Transform. Recalling the definition of the Laplace Transform: $$ \mathcal{L}\\{g(t)\\} = \int_{0}^{\infty} e^{-st} g(t) dt $$ From Step 2, we have found that: $$ F'(s) = \int_{0}^{\infty} (-te^{-st}) f(t) dt $$ This matches the definition of the Laplace Transform for a function g(t) = -tf(t). Therefore, we can conclude that: $$ F'(s) = \mathcal{L}\\{-tf(t)\\} $$ #Part (b)#

Differentiate F(s) n times with respect to s

To show that the nth derivative of F(s) is equal to the Laplace Transform of (-t)^n * f(t), we will first find the nth derivative of F(s) with respect to s. Recall our result from Part (a) for the first derivative of F(s): $$ F'(s) = \mathcal{L}\\{-tf(t)\\} = \int_{0}^{\infty} (-te^{-st}) f(t) dt $$ Now, we will differentiate this expression n times with respect to s: $$ F^{(n)}(s) = \frac{d^n}{ds^n} \int_{0}^{\infty} (-te^{-st}) f(t) dt $$

Apply the chain rule for differentiation repeatedly

We will apply the chain rule for differentiation repeatedly, differentiating the exponential term e^(-st) and the constant term (-t)^n. After differentiating n times, we will get: $$ F^{(n)}(s) = \int_{0}^{\infty} (-t)^n e^{-st} f(t) dt $$

Compare with the definition of the Laplace Transform

To prove that the nth derivative of F(s) is the Laplace Transform of (-t)^n * f(t), we need to compare this result with the definition of the Laplace Transform. We have found that: $$ F^{(n)}(s) = \int_{0}^{\infty} (-t)^n e^{-st} f(t) dt $$ This matches the definition of the Laplace Transform for a function g(t) = (-t)^n * f(t). Therefore, we can conclude that: $$ F^{(n)}(s)=\mathcal{L}\left\\{(-t)^{n} f(t)\right\\} $$ This completes the proof for both parts (a) and (b).

Key concepts

Laplace Transform

The Laplace Transform is a powerful tool used to analyze linear time-invariant systems and to solve differential equations. It transforms a function of time, often a signal or a system's response, into a function of complex frequency, making complex differential equations easier to work with. In essence, it shifts the perspective from the time domain to the frequency domain.

The basic definition of the Laplace Transform is given by the integral

\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt\]
where \( f(t) \) is the original time-domain function, \( e^{-st} \) is the exponential kernel, \( s \) is a complex frequency variable, and \( F(s) \) is the transform of the function in the frequency domain. The expression \( e^{-st} \) where \( s \) is complex, includes both exponential decay and oscillation for real and imaginary parts of \( s \), encompassing a wide range of signals.

When solving differential equations, this transform simplifies the process by converting derivatives into algebraic terms, which are easier to manipulate. It also has the capability to handle initial conditions directly, making it especially useful for solving initial-value problems. In engineering and physics, it is extensively used for its ability to handle both transient and steady-state analysis.

Differentiation Under the Integral Sign

Differentiation under the integral sign is a technique that allows taking the derivative of an integral with respect to a parameter. This method is used when an integral contains a parameter that is not an integration variable. It comes in handy when evaluating definite integrals that are otherwise difficult to integrate.

The rule states that for an integral of the form
\[ F(s) = \int_{a}^{b} g(x, s) dx\]
where \(s\) is a parameter, the derivative of \(F\) with respect to \(s\) can be expressed as
\[ F'(s) = \int_{a}^{b} \frac{\partial}{\partial s}g(x, s) dx\]
provided certain conditions regarding continuity are satisfied. This powerful property is applied when computing the Laplace transform of a derivative, which involves a parameter-dependent integrand. In our case, \( s \) is our parameter, and \( f(t) \) is treated as a constant with respect to \( s \), leading to simplification when differentiating.

Chain Rule for Differentiation

The chain rule is a fundamental principle in calculus that allows the calculation of the derivative of a composite function. If a variable \( u \) depends on \( t \) and \( s \) depends on \( u \), then the chain rule provides a way to relate the derivative of \( s \) with respect to \( t \) through \( u \). This concept is crucial when dealing with functions involving several layers of variables that depend on each other.

The chain rule states that
\[ \frac{d}{dt}s = \frac{ds}{du} \times \frac{du}{dt}\]
In the context of our exercise, we used the chain rule to differentiate the exponential term \( e^{-st} \), treating \( f(t) \) as a constant. We found \( F'(s) \) by applying the chain rule, which converted our problem into a more manageable form that resembles the Laplace transform of \( -tf(t) \).

Exponential Functions

Exponential functions are the backbone of the Laplace transform. These mathematical constructs are essential for modeling growth and decay processes in various fields like physics, engineering, and economics. An exponential function generally has the form \( A e^{kt} \), where \( A \) is a constant, \( t \) is the variable, and \( k \) is a constant rate. The function \( e \) is Euler's number, an irrational base of the natural logarithm.

In our exercise with the Laplace transform, the exponential function \( e^{-st} \) is the kernel that weights the function \( f(t) \) over time. The negative power indicates a decaying exponential, essential for the convergence of the integral on the interval from zero to infinity. The Laplace transform leverages the exponential function’s properties to encode information about a signal or system into a form that is simpler to analyze and manipulate algebraically.