Problem 20 - A First Course in Complex Analysis with Applications

Question

Find the Fourier transform (19) of the given function. $$ f(x)=\left\\{\begin{array}{ll} \sin x, & |x| \leq \pi \\ 0, & |x|>\pi \end{array}\right. $$

Step-by-Step Solution

Answer

The Fourier transform of the function is \( F(k) = \frac{\sin(\pi(1-k))}{1-k} - \frac{\sin(\pi(1+k))}{1+k} \).

Step 1: Understand the Given Function

The given function is a piecewise function defined as \( f(x) = \sin x \) for \( |x| \leq \pi \) and \( f(x) = 0 \) for \( |x| > \pi \). It is a sinusoidal function that is bounded within the interval from \(-\pi\) to \(\pi\) and zero elsewhere.

Step 2: Recall the Fourier Transform Definition

The Fourier transform \( F(k) \) of a function \( f(x) \) is defined as \[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} \, dx \] Since \( f(x) \) is non-zero only within \( |x| \leq \pi \), the limits of integration reduce to \(-\pi\) to \(\pi\).

Step 3: Set Up the Integral for Fourier Transform

Using the definition of Fourier transform and the limits of the non-zero domain of \( f(x) \): \[ F(k) = \int_{-\pi}^{\pi} \sin x \cdot e^{-ikx} \, dx \] This reduces the Fourier transform to an evaluation over the interval \([-\pi, \pi]\).

Step 4: Evaluate the Integral

To evaluate the integral, \[ F(k) = \int_{-\pi}^{\pi} \sin x \cdot e^{-ikx} \, dx \] Use the identity \( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \). Substitute into the integral:\[ F(k) = \int_{-\pi}^{\pi} \frac{e^{ix} - e^{-ix}}{2i} \cdot e^{-ikx} \, dx \] This breaks into two integrals:\[ F(k) = \frac{1}{2i} \left( \int_{-\pi}^{\pi} e^{ix(1-k)} \, dx - \int_{-\pi}^{\pi} e^{-ix(1+k)} \, dx \right) \] The integrals are standard and evaluate using the limits of \(-\pi\) and \(\pi\).

Step 5: Solve Integrals and Simplify

Each integral has the form of \[ \int e^{iax} \, dx = \frac{1}{ia}(e^{iax}) \] Evaluating on the limits, both integrals simplify and can be computed using the exponential properties on complex numbers.Applying the limits yields:- For \( e^{ix(1-k)} \), it simplifies as: \[ \frac{e^{ix(1-k)}}{i(1-k)} \bigg|_{-\pi}^{\pi} = \frac{2i\sin(\pi(1-k))}{1-k} \]- For \( e^{-ix(1+k)} \), similarly yields: \[ \frac{2i\sin(\pi(1+k))}{1+k} \]Combine these terms with the prefactor \(\frac{1}{2i}\), cancel the imaginary units, and simplify further.

Step 6: Obtain Final Expression for Fourier Transform

Combine and simplify the results from the integrals:\[ F(k) = \frac{\sin(\pi(1-k))}{1-k} - \frac{\sin(\pi(1+k))}{1+k} \]Simplify using trigonometric identities or properties to express succinctly, if applicable.

Key Concepts

Piecewise Functions

Piecewise functions are mathematical constructs that can define a function over different intervals using different expressions. This allows for great flexibility when describing complex or non-uniform behaviors of functions. In the context of Fourier transforms, piecewise functions are very useful to model signals or functions that are only active over certain intervals.
In this exercise, our function, \( f(x) \), is piecewise because it is defined by different values based on the input \( x \): it is \( \sin x \) within the interval \( |x| \leq \pi \), and 0 outside of that range. This means that our focus, when calculating the Fourier transform, will only be on the non-zero portion of the function.
This type of function is common in signal processing, where a signal might exist only within a certain time frame. Utilizing piecewise functions allows one to effectively "switch" a function on and off over specified intervals.

Integration Techniques

Integration techniques are methods or strategies employed to solve integrals, especially those that can be complex due to their specific interval or piecewise nature. In Fourier analysis, integrals often involve complex exponential functions, making certain techniques crucial.
One important technique used in this problem is changing a trigonometric function, such as \( \sin x \), into its exponential form using Euler’s formula:

• \( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \)

This transformation is vital for simplifying the integral of \( \sin x e^{-ikx} \), because it breaks the problem into separate exponential integrals. Another integration technique involved is evaluating definite integrals with exponential functions:

• The integral \( \int e^{iax} \, dx \) within specified limits can be simplified using the result \( \frac{1}{ia}(e^{iax}) \).

Knowing these techniques allows us to compute the Fourier transform efficiently by breaking down complex integrands into simpler forms. These methods ultimately lead to terminology simplification and help avoid unnecessary computational complexity.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where the functions are defined. They are crucial in simplifying expressions, solving integrals, and performing transformations.
In the exercise, after substituting \( \sin x \) with its exponential form, we used a couple of these identities to simplify expressions:

• The conversion of trigonometric functions into exponential form using Euler's formula.
• The simplification of sine functions at specific points using properties like \( \sin(-x) = -\sin(x) \) and symmetry properties related to \( \, \pi \).

Applying these identities allows the complex exponential terms to combine and reduce to simpler terms involving sine. The final step involves carefully combining these results into the more recognizable form of a Fourier transform output. Recognizing and utilizing these identities helps in achieving concise and straightforward expressions, which is essential in understanding and interpreting results effectively.