Question

Find the Laplace transform of the given function. In Problem 27 assume that term-by-term integration of the infinite series is permissible. $$ f(t)=\left\\{\begin{array}{ll}{1,} & {0 \leq t<1} \\ {0,} & {t \geq 1}\end{array}\right. $$

Short answer

$$ f(t) = \begin{cases} 1, & 0 \leq t < 1 \\ 0, & 1 \leq t \end{cases} $$ Answer: The Laplace transform of the given function is: $$ \mathcal{L}\{f(t)\} = \frac{1}{s} - \frac{1}{s}e^{-s} $$

Step-by-step solution

Identify the regions of the piecewise function

Before applying the Laplace transform, let's identify the different parts of the function. We have two regions, one where the function is equal to 1 from 0 to 1, and the other where the function is equal to 0 from 1 to infinity.

Apply the Laplace transform definition to each region

Now that we have identified the regions of the piecewise function, we can apply the definition of the Laplace transform by taking the integral of each piece. First, we consider the part where the function is 1 from 0 to 1. $$ \mathcal{L}\{f(t)\} = \int_0^1 e^{-st} dt $$ Second, we consider the part where the function is 0 from 1 to infinity. $$ \mathcal{L}\{f(t)\} = \int_1^{\infty} e^{-st}(0) dt $$

Solve the integrals

Now we need to solve the integrals for both parts of the function. For the first integral, we have: $$ \int_0^1 e^{-st} dt = \left[-\frac{1}{s}e^{-st}\right]_0^1 = -\frac{1}{s}e^{-s} + \frac{1}{s} $$ For the second integral, since we are multiplying by 0, the integral will be 0. $$ \int_1^{\infty} e^{-st}(0) dt = 0 $$

Combine the results

Now we can combine the results to find the Laplace transform of our function: $$ \mathcal{L}\{f(t)\} = \left(-\frac{1}{s}e^{-s} + \frac{1}{s}\right) + 0 = \frac{1}{s} - \frac{1}{s}e^{-s} $$ So, the Laplace transform of the given function is: $$ \mathcal{L}\{f(t)\} = \frac{1}{s} - \frac{1}{s}e^{-s} $$

Key concepts

Understanding Piecewise Functions

In mathematical analysis, a piecewise function is one where a function is defined by multiple sub-functions, each applying to a certain interval of the main function's domain. This allows for the easy handling of functions that behave differently in various sections of their domain.
In the given problem, the piecewise function was defined by:

• For the interval \(0 \leq t < 1\), the function is constant at 1.
• For \(t \geq 1\), the function becomes 0.

This kind of function is perfect for modeling situations where a phenomenon changes behavior at specific points, like a switch turning on or off. Understanding the regions and behavior of these sub-functions is crucial before applying any mathematical transformations, like the Laplace transform.

The Basics of Integration

Integration is a fundamental concept in calculus that helps us find the area under a curve defined by a function, over a certain interval. It is the reverse process of differentiation, and it is indispensable for analyzing functions and solving differential equations.
To solve for the Laplace transform, the integral approach was crucial. We divide our piecewise function into intervals and apply integration to each interval separately:

• For \(0 \leq t < 1\), we calculated \(\int_0^1 e^{-st} dt\), which resulted in a value using the integration formula for an exponential function.
• For \(t \geq 1\), since the function is zero, the integral is simplified directly to 0.

Overall, integration allows us to sum up continuous values over an interval, producing a complete and comprehensive result that aids in analyzing the function's behavior.

Exploring Infinite Series

Infinite series is a sum of an infinite sequence of terms. They are extensively used in mathematics to describe functions through their series expansions. When encountering a problem, infinite series often let us redefine functions to express them in more manageable forms.
In context, the assumption was made that term-by-term integration of the series defining the function is permissible. This assumption is critical:

• It supports rearranging and simplifying the computations by evaluating narrower segments of a broader range.
• It fosters more straightforward manipulation of both finite and infinite form results, which is indispensable when analyzing complex functions.

By understanding infinite series, we gain insight into not only the structure and behavior of specific functions but also the methods for approximating these functions.

Analyzing Functions

Function analysis is a crucial part of higher mathematics that involves the investigation of properties and behaviors of functions. This includes understanding continuity, limits, and differentiability.
In our task, analyzing the function involved:

• Determining the layout of the piecewise components and when they are effective.
• Applying and integrating mathematical operations to each segment.
• Using series assumptions and close examination to simplify unnecessary calculations or steps.

Effectively analyzing functions ensures that the solution to the problem is both correct and concise, leveraging classic techniques like decomposition of piecewise functions and integration to make complex problems more tractable.