Question

Calculate the integral $$ \int_{-\pi}^{\pi}\left|\sum_{n=1}^{\infty} \frac{1}{2^{n}} e^{i n x}\right|^{2} d x . $$

Short answer

The integral evaluates to \( \frac{\pi}{\sqrt{3}} \).

Step-by-step solution

Recognize the Series

Identify the expression inside the integral as the infinite series \( \sum_{n=1}^{\infty} \frac{1}{2^{n}} e^{i n x} \). This is a geometric series in the form \( \sum_{n=1}^{\infty} a^n \), with \( a = \frac{1}{2} e^{ix} \).

Sum the Geometric Series

The formula for the sum of an infinite geometric series \( \sum_{n=0}^{\infty} a^n = \frac{1}{1-a} \) can be applied, starting from \( n=1 \). The sum in this case is \( S(x) = \frac{1}{1 - \frac{1}{2}e^{ix}} - 1 \) to adjust for starting at \( n=1 \).

Simplify the Series Expression

The series sum \( S(x) = \frac{\frac{1}{2}e^{ix}}{1 - \frac{1}{2}e^{ix}} \) simplifies to \( S(x) = \frac{e^{ix}}{2 - e^{ix}} \) by multiplying numerator and denominator by 2.

Calculate the Absolute Square

We need \( \left| S(x) \right|^2 \), which is \( \left| \frac{e^{ix}}{2 - e^{ix}} \right|^2 = \frac{1}{(2 - cos(x))^2 + sin^2(x)} \). This simplifies further due to the identity \( sin^2(x) + cos^2(x) = 1 \): \( \left| S(x) \right|^2 = \frac{1}{5 - 4cos(x)} \).

Set Up the Integral

The integral to evaluate is \( \int_{-\pi}^{\pi} \frac{1}{5 - 4cos(x)} dx \). This requires a substitution for cosine, often using trigonometric identities.

Trigonometric Substitution

Utilize the identity \( cos(x) = \frac{1-t^2}{1+t^2} \), where \( t=tan(x/2) \), transforming the integral from \( \int \frac{1}{5 - 4cos(x)} dx \) to an integral in terms of \( t \).

Solve the Transformed Integral

Solve the transformed integral \( \int \frac{2}{1 - t^2} \cdot \frac{1}{5 - \frac{4(1-t^2)}{1+t^2}} \cdot \frac{2}{1+t^2} dt \), using partial fraction decomposition if needed. This often requires recognizing simplified forms or tables.

Calculate the Solution

After integrating using \( t \), evaluate the definite integral within the limits from \( -\infty \) to \( \infty \), as \( x \) traverses from \(-\pi\) to \(\pi\). Convert back to the original variable \( x \) and solve for the original limits.

Key concepts

Geometric Series: A Key Player in Fourier Series

Geometric series are sequences of numbers where each term is a constant multiple of the previous one. Understanding geometric series is crucial for mastering Fourier series, particularly when dealing with expressions like those found in complex exponential forms. This skill is essential when breaking down periodic functions into sums of simpler sine and cosine functions.
In the exercise, the expression inside the integral initially appears daunting. However, recognizing it as a geometric series makes it more approachable. A geometric series in its infinite form can be described by \( \sum_{n=1}^{\infty} a^n = \frac{1}{1-a} \) when it converges. The key here is identifying the base term, \( a \), so you can sum the series efficiently.
In this context, knowing the geometric series formulation allows you to simplify the complex expression. This simplification is the first step that helps transform and approach more intricate aspects of the given integral.

The Power of Trigonometric Identities

Trigonometric identities are mathematical equalities involving trigonometric functions. They allow us to simplify and solve integrals and equations by converting complex expressions into more workable forms. These identities are vital tools in mathematics, especially when integrating expressions like the one in our exercise.
Our solution makes use of identities such as \( sin^2(x) + cos^2(x) = 1 \), which simplifies the square of the geometric series sum. We also use the substitution trick \( cos(x) = \frac{1-t^2}{1+t^2} \), where \( t = tan(x/2) \). This transforms an intricate integral into an easily manageable one, making trigonometric identities indispensable in evaluating and solving integrals.
Mastery of trigonometric identities not only helps simplify complex mathematical problems but also deepens your understanding of the relationships within trigonometric functions.

Integral Calculus: Evaluating the Integral

Integral calculus focuses on finding sums and areas under curves. It's a fundamental component of mathematics used extensively in Fourier analysis, where it helps in computing the integral of series expressions. In our exercise, integral calculus is employed to evaluate the square of a geometric series transformed through trigonometric identities.
Setting up the integral in our problem involves understanding how to handle the transformed expression \( \frac{1}{5 - 4\cos(x)} \). Using calculus techniques, such as substitution and manipulation of the integral limits, is crucial. Recognizing opportunities to simplify these expressions with known identities or partial fractions can reduce complexity.
Integral calculus transforms seemingly impossible tasks into methodical problems. By breaking it down step-by-step, it facilitates solving integrals in a structured, approachable manner.

Complex Exponential Function: Bridging Algebra and Trigonometry

Complex exponential functions use exponential forms to elegantly bridge algebra and trigonometry. They are essential in simplifying periodic functions and are particularly useful in the context of Fourier series.
In this problem, the complex exponential function \( e^{inx} \) is prominent. It helps in transforming products of sines and cosines into more manageable exponential forms. The substitution \( a = \frac{1}{2} e^{ix} \) takes advantage of complex exponentials, simplifying the geometric series expansion.
Understanding complex numbers and their properties is key to applying these transformations effectively. They offer a streamlined, powerful method of handling oscillatory functions that are often found in engineering and physics. By leveraging these concepts, you gain a toolkit that transforms challenging functions into straightforward calculations.