Question

In the following integrals express the sines and cosines in exponential form and then integrate to show that: $$\int_{-\pi}^{\pi} \cos 2 x \cos 3 x d x=0$$

Short answer

The integral evaluates to zero because the integrals of the exponential terms over \(-\text{π} \) to \( \text{π} \) all equal zero.

Step-by-step solution

- Express Cosines in Exponential Form

Recall that \(\text{cos} x \) can be expressed using Euler's formula as \[ \text{cos} x = \frac{e^{ix} + e^{-ix}}{2} \]. Using this, express \(\text{cos} 2x\) and \(\text{cos} 3x\) in exponential form: \[ \text{cos} 2x = \frac{e^{i2x} + e^{-i2x}}{2}, \] \[ \text{cos} 3x = \frac{e^{i3x} + e^{-i3x}}{2} \]

- Multiply the Expressions

Multiply the two exponential forms derived in Step 1: \[ \text{cos} 2x \text{cos} 3x = \frac{1}{4} (e^{i2x} + e^{-i2x})(e^{i3x} + e^{-i3x}) \]Expand the product: \[ = \frac{1}{4} \big(e^{i(2x+3x)} + e^{i(2x-3x)} + e^{-i(2x-3x)} + e^{-i(2x+3x)}\big) \]Simplify the exponents: \[ = \frac{1}{4} \big(e^{i5x} + e^{-ix} + e^{ix} + e^{-i5x}\big) \]

- Integrate Each Term Separately

The integral can now be taken term-by-term from \(-\text{π} \) to \( \text{π} \): \[ \frac{1}{4} \bigg( \int_{-\text{π}}^{\text{π}} e^{i5x} \ dx + \int_{-\text{π}}^{\text{π}} e^{-i5x} \ dx + \int_{-\text{π}}^{\text{π}} e^{ix} \ dx + \int_{-\text{π}}^{\text{π}} e^{-ix} \ dx \bigg) \] Evaluate each integral:Since the integrals of \(e^{ikx} \) for any non-zero \( k \) from \(-\text{π} \) to \( \text{π} \) are zero due to periodicity properties of exponential functions with imaginary exponents, we find \[ \int_{-\text{π}}^{\text{π}} e^{i5x} \ dx = 0, \int_{-\text{π}}^{\text{π}} e^{-i5x} \ dx = 0, \int_{-\text{π}}^{\text{π}} e^{ix} \ dx = 0, \int_{-\text{π}}^{\text{π}} e^{-ix} \ dx = 0 \]

- Sum the Evaluated Integrals

Add up the evaluated integrals from Step 3: \[ \frac{1}{4} \big(0 + 0 + 0 + 0\big) = 0 \]

Key concepts

Euler's Formula

Euler's formula is a fundamental equation in complex analysis that states: \[ e^{ix} = \text{cos} x + i \text{sin} x \]. It connects exponential functions with trigonometric functions. This relation is crucial in Fourier analysis, where it simplifies the representation of signals.

Understanding Euler's formula helps in converting cosine and sine functions into their exponential form. This is particularly useful for integration and solving differential equations involving trigonometric functions. By transforming these functions, we can leverage the properties of exponentials to simplify complex problems. Euler's formula is foundational for expressing periodic functions using complex exponentials, paving the way for easier integration and analysis.

Exponential Form

The exponential form of trigonometric functions leverages Euler's formula. For example, \[ \text{cos} x = \frac{e^{ix} + e^{-ix}}{2} \]. Applying this to the given exercise, we transform \[ \text{cos} 2x \] and \[ \text{cos} 3x \] into exponential forms:

• \ (e^{i2x} + e^{-i2x})/2


• \ (e^{i3x} + e^{-i3x})/2

This transformation simplifies the multiplication of these functions and helps in the integration process. The exponential form is powerful because it turns trigonometric functions into sums and differences of exponentials, making it easier to handle complex calculations involving integrals or differential equations.

Integration Techniques

Integrating functions that have been expressed in exponential form often simplifies the process. In our case, after expressing \[ \text{cos} 2x \] and \[ \text{cos} 3x \] as exponentials and multiplying them, we get:

\[ \frac{1}{4} (e^{i(2x+3x)} + e^{i(2x-3x)} + e^{-i(2x-3x)} + e^{-i(2x+3x)}) \].

To integrate this, we split it term-by-term and apply the integral limits from \[ -\pi \] to \[ \pi \]:

\[ \int_{-\pi}^{\pi} e^{ikx} dx = 0 \] for any non-zero \[ k \]. This stems from the periodicity of exponential functions with imaginary exponents. Recognizing this property helps us evaluate each term in the integral separately, leading to a simpler overall integration process.

Periodic Functions

Periodic functions repeat their values in regular intervals. Sine and cosine functions are classic examples. In Fourier analysis, periodic functions are often represented as sums of sinusoids (sines and cosines) or complex exponentials.

Examining the integral problem, we leverage the periodic nature of exponential functions:

\[ \int_{-\pi}^{\pi} e^{ikx} dx = 0 \] for non-zero \[ k \]. This is because the integral of one period of a periodic function over its period is zero, assuming the function completes an integer number of periods over the interval. Understanding the periodicity of functions simplifies the integration of complex exponential forms.

Complex Exponentials

Complex exponentials are expressions of the form \[ e^{ix} \], which, by Euler’s formula, equals \[ \text{cos} x + i \text{sin} x \]. These are instrumental in Fourier analysis for representing periodic functions.

When we convert cosine and sine functions into complex exponentials, we leverage their algebraic properties to simplify mathematical operations. For instance, multiplying and integrating complex exponentials tends to be more straightforward than dealing with trigonometric forms.

In our exercise, converting \[ \text{cos} 2x \] and \[ \text{cos} 3x \] into complex exponentials before multiplication and integration showcases the power of this approach. The zero integral result reflects the orthogonality of different frequency components over a period.