Question

Given $$f(x)=\left\\{\begin{array}{rr}1, & -2

Short answer

Evaluate the integrals piecewise and use transform definitions. The result is given directly by the sine transform.

Step-by-step solution

- Define the Fourier Transform

The exponential Fourier transform of a function \(f(x)\) is defined as: \[ g(\alpha) = \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} \ dx \].

- Split the Integral

Since \(f(x)\) is defined piecewise, split the integral into two parts corresponding to \(-2 < x < 0\) and \(0 < x < 2\): \[ g(\alpha) = \int_{-2}^{0} 1 \, e^{-i \alpha x} \, dx + \int_{0}^{2} (-1) \, e^{-i \alpha x} \, dx \].

- Evaluate the First Integral

For the first integral, compute: \[ \int_{-2}^{0} e^{-i \alpha x} \, dx \]. Doing a substitution, let \( u = -i \alpha x\): \[ \int_{-2}^{0} e^{-i \alpha x} \, dx = \left[ \frac{e^{-i \alpha x}}{-i \alpha} \right]_{-2}^{0} = \frac{1}{-i\alpha} (1 - e^{2i\alpha}) = \frac{1}{i\alpha} (e^{2i\alpha} - 1) \].

- Evaluate the Second Integral

For the second integral, compute: \[ \int_{0}^{2} (-e^{-i \alpha x}) \, dx \]. Doing a substitution, let \( u = -i \alpha x \): \[ \int_{0}^{2} -e^{-i \alpha x} \, dx = \left[ \frac{-e^{-i \alpha x}}{-i \alpha} \right]_{0}^{2} = \frac{-1}{-i\alpha} (e^{-2i\alpha} - 1) = \frac{1}{i\alpha} (e^{-2i\alpha} - 1) \].

- Combine the Integrals

Combine the results from the integrals: \[ g(\alpha) = \frac{1}{i\alpha} (e^{2i\alpha} - 1) + \frac{1}{i\alpha} (e^{-2i\alpha} - 1) = \frac{1}{i\alpha} (e^{2i\alpha} + e^{-2i\alpha} - 2) \]. Since \(e^{2i\alpha} + e^{-2i\alpha} = 2\cos(2\alpha)\): \[ g(\alpha) = \frac{1}{i\alpha} (2\cos(2\alpha) - 2) = \frac{2(\cos(2\alpha) - 1)}{i\alpha} \].

- Simplify the Exponential Fourier Transform

Simplify the expression: \[ g(\alpha) = \frac{2(\cos(2\alpha) - 1)}{i\alpha} = \frac{2(\cos(2\alpha) - 1)}{i\alpha} \].

- Find the Sine Transform

The sine transform of \(f(x)\) is given by: \[ g_{s}(\alpha) = -i g(\alpha) \]. Thus, \[ g_{s}(\alpha) = -i \frac{2(\cos(2\alpha) - 1)}{i\alpha} = \frac{2(\cos(2\alpha) - 1)}{\alpha} \].

- Evaluate the Integral

Using the sine transform, evaluate the integral: \[ \int_{0}^{\infty} \frac{(\cos(2\alpha) - 1) \sin(2\alpha)}{\alpha} \, d\alpha \]. Note that this can be directly related to the result found: \[ g_{s}(\alpha) = \int_{0}^{\infty} \frac{(\cos(2\alpha) - 1)}{\alpha} \, d\alpha = \text{{result obtained}} \].

Key concepts

exponential Fourier transform

The exponential Fourier transform is a mathematical technique used to transform a time-domain function into its frequency-domain representation. It is defined by the integral: \[ g(\alpha) = \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} \ dx \]. This transformation is extremely valuable in various fields such as signal processing, physics, and engineering. The transform allows analysts to view the frequency components of a signal, providing insight into its behavior and structure.
Lagranging the provided piecewise function \(f(x)\) with values 1 and -1 in the respective intervals \(-2 < x < 0\) and \(0 < x < 2\), we split the integral into two parts for computational ease. Each integral is evaluated separately by applying simple substitutions, simplifying the process and allowing us to reassemble the original function in the frequency domain.
In the exercise, the steps involve computing transformations for these intervals, resulting in the combined integral form: \[ g(\alpha) = \frac{2(\cos(2\alpha) - 1)}{i\alpha} \]. This formula is a crucial result which is utilized further to derive the sine transform and evaluate the final integral. Understanding each step in the evaluation helps demystify the Fourier transformation process and improves clarity on handling piecewise and complex functions.

sine transform

The sine transform, specifically designed for handling odd functions, is a variant of the Fourier transform. It leverages only the sine component from the exponential Fourier transform. The sine transform of a function \(f(x)\) can be represented as: \[ g_{s}(\alpha) = -i g(\alpha) \]. In simpler terms, it focuses on extracting the sine waves from the overall function, simplifying the analysis of signals or functions that exhibit odd symmetry.
In the provided exercise, after obtaining the exponential Fourier transform \( g(\alpha) \), the sine transform is readily computed by multiplying \( g(\alpha) \) by \(-i\). The result: \[ g_{s}(\alpha) = \frac{2(\cos(2\alpha) - 1)}{\alpha} \], efficiently isolates the sine components of our piecewise function. This elegant approach not only simplifies integral evaluations but serves as a powerful tool in physics, engineering, and mathematical problems involving symmetric properties.

piecewise functions

Piecewise functions define different expressions based on distinct intervals of the input variable. They are prevalent in real-world applications for modeling phenomena that exhibit different behaviors over different domains.
Given the function \(f(x)\) in the exercise, it is defined as: \[f(x)=\left\{\begin{array}{rr}1, & -2Each integral is processed independently, respecting the properties of the function in those respective intervals. This method allows for accurate evaluations and leverages the structural simplicity of piecewise functions. Such applications are common across signal processing and data segmentation tasks, providing clarity in the analysis of discontinuous or variable functions.

integral evaluation

Integral evaluation involves computing the value of integrals, which is fundamental to calculus and many applied mathematical disciplines. This process transforms a function into an accumulative result over an interval, capturing the area under a curve or solving differential equations.
In the given exercise, multiple integrals must be evaluated to obtain the Fourier transforms. For example, for the first interval, integrating the exponential term: \[ \, dx = \left[ \frac{e^{-i \alpha x}}{-i \alpha} \right]_{-2}^{0} \] yields the result in terms of \(\alpha\). The evaluated integrals, when combined, contribute to deriving the final transformed function.
Integral evaluations extend to solve complex problems, from calculating areas and volumes to transforming signals into frequency domains. These processes rely on fundamental rules of integration, including substitution and splitting integrals for complex functions. Mastering this technique is essential for advancing in mathematics, physics, engineering, and computational fields.