Question

Determine the Fourier transform of $$ f(x)= \begin{cases}1, & 00\), and use it to calculate $$ \int_{0}^{\infty} \frac{\cos a x-1}{x} \sin b x d x, \quad b>0 . $$

Short answer

The integral evaluates to 0.

Step-by-step solution

Define the Fourier Transform

The Fourier transform of a function \( f(x) \) is given by \( F(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx \). In this step, we'll apply this formula to the given piecewise function for \( f(x) \).

Split the Integral Based on the Piecewise Definition

The function \( f(x) \) is defined in three parts. We divide the integral accordingly: \( F(k) = \int_{-a}^{0} - e^{-ikx} \, dx + \int_{0}^{a} e^{-ikx} \, dx \), ignoring the intervals where the function is zero.

Compute the Fourier Transform for \([-a, 0] \)

Calculate \( \int_{-a}^{0} - e^{-ikx} \, dx = \left[ \frac{e^{-ikx}}{-ik} \right]_{-a}^{0} = \frac{1}{ik} (1 - e^{ika}) \).

Compute the Fourier Transform for \([0, a] \)

Calculate \( \int_{0}^{a} e^{-ikx} \, dx = \left[ \frac{e^{-ikx}}{-ik} \right]_{0}^{a} = \frac{1}{ik} (e^{-ika} - 1) \).

Combine the Results

Sum the results from Steps 3 and 4:\[ F(k) = \frac{1}{ik} (1 - e^{ika}) + \frac{1}{ik} (e^{-ika} - 1) = \frac{1}{ik} (e^{-ika} - e^{ika}) = \frac{1}{ik}(-2i \sin(ka)) \].So, \( F(k) = \frac{2 \sin(ka)}{k} \).

Relate the Fourier Transform to the Integral

The integral to evaluate is \( \int_{0}^{\infty} \frac{\cos(ax)-1}{x} \sin(bx) \, dx \). Noticing it resembles the Fourier transform form, let \( k = b \) and use properties of sine and cosine.

Evaluate the Integral

Relating it to the calculated Fourier transform:\[ \int_{0}^{\infty} \frac{\cos(ax)-1}{x} \sin(bx) \, dx = \text{Real part of } F(b)\].Thus:\[ \text{Real part of } \frac{2 \sin(ba)}{b} = \frac{2 \sin(ba)}{b} \cdot 0 = 0 \].The integral evaluates to 0.

Key concepts

Understanding Piecewise Functions

A piecewise function is a mathematical concept used to define a function that behaves differently based on the input. In simple terms, it means that depending on which "piece" of the function's domain you're considering, you might end up with different rules of calculation. This is quite useful because not all phenomena in nature or engineering can be described by a single continuous function.
For example, in the given exercise, the function \( f(x) \) is defined as:

• \( f(x) = 1 \) for \( 0 < x \leq a \)
• \( f(x) = -1 \) for \( -a \leq x \leq 0 \)
• \( f(x) = 0 \) otherwise

This function changes its value based on the interval in which \( x \) falls. Understanding how to split the function into these intervals is key when working with piecewise functions, particularly in context with integrals or transforms like the Fourier transform.

Integral Calculations with Piecewise Functions

Calculating integrals in the context of a piecewise function involves dealing with each defined segment of the function separately. Here's how to do it:
When given a piecewise function, you need to compute its integral by breaking it up according to the intervals defined by the piecewise parts. In the exercise, the integral for the Fourier transform is broken down into segments of \( [-a, 0] \) and \( [0, a] \).

The goal is to calculate the integral separately for each segment where the function is non-zero:

• Integrate \( -e^{-ikx} \) over the interval \([-a, 0]\)
• Integrate \( e^{-ikx} \) over the interval \([0, a]\)

After solving each part, you add the results together. This involves using integration techniques like evaluating indefinite integrals at boundary values, getting the total contribution from each piece. This process is essential in frequency analysis often encountered in fields like signal processing.

The Role of Complex Exponential

Complex exponentials are at the heart of Fourier transforms. In the formulation of the Fourier transform, we use the complex exponential \( e^{-ikx} \). This remarkable function ties together trigonometry and calculus in a way that simplifies the analysis of waveforms or periodic functions.

Complex exponentials work by expressing trigonometric functions as exponentials, thanks to Euler's formula: \( e^{ix} = \cos(x) + i\sin(x) \). Thus, turning complex mathematics into something more manageable and beautiful.

• The Fourier transform of a real-valued function translates the function into a frequency domain - essentially describing how much of each frequency is present in the original signal.
• In the exercise, \( e^{-ikx} \) turns the piecewise function into a tool for calculating its spectrum over various frequencies \( k \).

Understanding this connection helps in devising solutions to differential equations, making complex analysis tools like Fourier analysis applicable across the sciences.