91Ó°ÊÓ

The comparison test is a crucial tool in determining the convergence or divergence of improper integrals. It works by comparing our problematic integral with a simpler "comparison" function that we already understand well. In this case, the original integral to be evaluated is \( \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \, dx \). A suitable comparison function is \( \frac{1}{x^2} \) because it simplifies the evaluation and bears a resemblance to the original integrand.
The criteria for using the comparison test effectively involve several steps:

• Identify the comparison function \( g(x) \) that simplifies your problem and has known convergence properties.
• Ensure \( f(x) \leq g(x) \) for the interval of interest, ensuring the inequality holds for all the applicable values of \( x \).
• Check that the integral of \( g(x) \) converges. If so, \( f(x) \) will converge as well, based on the properties of integrals.

Given these steps, since \( \frac{\sin^2 x}{x^2} \leq \frac{1}{x^2} \) and \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) is a convergent integral, the original integral also converges.

The concept of p-integrals is fundamental when comparing functions to determine convergence. A p-integral refers to integrals of the form \( \int_{a}^{b} \frac{1}{x^p} \, dx \), where \( p \) is a constant.
P-integrals have straightforward rules concerning their convergence:

• When \( p > 1 \), the integral \( \int_{a}^{\infty} \frac{1}{x^p} \, dx \) converges.
• When \( p \leq 1 \), the integral diverges.

In our case, the comparison function \( \frac{1}{x^2} \) corresponds to a p-integral with \( p = 2 \). As \( p > 1 \), it confirms that the integral \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges. Utilizing p-integrals simplifies the verification process in the comparison test.

Understanding bounded functions is critical in applying the comparison test. A function \( f(x) \) is considered bounded within a range if there exists a constant \( M \) such that \( |f(x)| \leq M \) for all \( x \) in that range. In the problem, \( \sin^2 x \) is inherently bounded as \( 0 \leq \sin^2 x \leq 1 \) for all \( x \). This property is actually quite helpful for comparing functions.
Here’s why boundedness matters:

• It ensures the function doesn't exceed certain limits, which can directly affect the integrability of a given expression.
• This boundedness allowed us to establish that \( \frac{\sin^2 x}{x^2} \leq \frac{1}{x^2} \), crucial for applying the comparison test.

Thus, bounded functions maintain control over the "size" of the integrand and provide assurance that we're dealing with a "manageable" problem. By confirming \( \sin^2 x \)'s bounded nature, the integral \( \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \, dx \) is more easily assessed within the framework of the comparison test.