91Ó°ÊÓ

The comparison test is a useful method in determining the convergence of integrals and series. It works by comparing a given integral with another integral whose convergence properties are known. If you can establish that the function you are studying is always less than or equal to a function with a known divergent integral, you can conclude that the original function's integral also diverges.

In this problem, we utilized the comparison test by examining the integral of the function \[ f(x) = \left| \frac{\sin x}{x} \right|. \]
Since we know that \[ \left| \sin x \right| \leq 1 \]for all \(x\), we can establish that \[ \left| \frac{\sin x}{x} \right| \leq \frac{1}{x}. \]As the integral \[ \int_1^\infty \frac{1}{x} \, dx \]diverges, the comparison test tells us that \[ \int_1^\infty \left| \frac{\sin x}{x} \right| \, dx \]also diverges. Hence, \(f(x)\) is not Lebesgue integrable over \((1, \infty)\).

The sine function, denoted as \( \sin x \), is a periodic function with a fundamental period of \( 2\pi \). It oscillates between -1 and 1 for all real numbers \(x\). In this exercise, we leverage the property that \( \left| \sin x \right| \leq 1 \) to simplify our analysis.

This property of the sine function is crucial because it allows us to bound \( \frac{\sin x}{x} \) by \( \frac{1}{x} \). Here are some key points:

• The sine function's oscillatory nature means it provides cancellations, leading to some integrals involving sine to converge when certain conditions are met.
• However, the decaying term \( \frac{1}{x} \) becomes dominant in this case, leading to a divergent behavior over the interval \((1, \infty)\).

The understanding of this property helps in comprehending more complex trigonometric integrals and their behavior over infinite intervals.

A divergent integral is one that does not meet the criteria for convergence; that is, its value does not approach a finite number as the limit extends to infinity. Divergent integrals are important to identify since they imply that the function involved behaves in a way that does not allow for integration over the entire interval studied.

In this exercise, we identified the integral\[ \int_1^\infty \frac{1}{x} \, dx \]as divergent. This is a well-known result, where the antiderivative of \( \frac{1}{x} \) is \( \ln x \). Evaluating the improper integral:\[\lim_{b \to \infty} \int_1^b \frac{1}{x} \, dx = \lim_{b \to \infty} (\ln b - \ln 1) = \infty.\]Since this integral results in an unbounded value, it's categorized as divergent. Knowing such standard results is essential in calculating the behavior of other integrals, like \( \int_1^\infty \left| \frac{\sin x}{x} \right| \, dx \), which we proved to be similarly divergent using the comparison test.