Problem 49 - Calculus Single Variable

Question

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{-1}^{1} \ln \left(1-x^{2}\right) d x\)

Step-by-Step Solution

Answer

The integral diverges due to its logarithmic behavior at the endpoints.

Step 1: Recognize the Type of Integral

The given integral is \(\int_{-1}^{1} \ln(1-x^2) \, dx\), which is an improper integral because the integrand \ln(1-x^2)\ becomes undefined at both endpoints \(x = -1\) and \(x = 1\). This means we need to check the convergence of the integral.

Step 2: Analyze the Singularity

Check the behavior of the function \ln(1-x^2)\ near the points where it becomes undefined. As \(x\) approaches \(-1\) and \(1\), \(\ln(1-x^2)\) tends to \(-\infty\), indicating singularities at these points.

Step 3: Break the Integral into Limits

Since the singularities occur at both \(x = -1\) and \(x = 1\), we split the integral into two separate limits to handle these singularities: \[\lim_{a \to -1^+} \int_{a}^{0} \ln(1-x^2) \, dx + \lim_{b \to 1^-} \int_{0}^{b} \ln(1-x^2) \, dx\]

Step 4: Evaluate the Integral Using Symmetry

Since the function \(\ln(1-x^2)\) is symmetric about \(x = 0\), evaluate one integral and double it:\[\lim_{b \to 1^-} 2\int_{0}^{b} \ln(1-x^2) \, dx\]Use substitution \( x = \sin(\theta) \, dx = \cos(\theta) \, d\theta \), then transform the limits of integration from \(0\) to \(\arcsin(b)\).

Step 5: Transform and Integrate

Substitute \( x = \sin(\theta) \):\[\int \ln(1-\sin^2(\theta)) \cos(\theta) \, d\theta = \int \ln(\cos^2(\theta)) \cos(\theta) \, d\theta = 2\int \ln(\cos(\theta)) \cos(\theta) \, d\theta\]Evaluate this integral using integration by parts or another suitable method.

Step 6: Analyze the Integral

The integral of \(\ln(\cos(\theta)) \cos(\theta) \, d\theta\) can be rewritten and solved. Evaluate it over the appropriate limits to finally conclude whether the improper integral converges or diverges. However, due to its symmetry and behavior of logarithmic function, this integral leads to divergence. After integration, infinite values occur at the endpoints, indicating divergence.

Key Concepts

Convergence and Divergence of Improper Integrals

Improper integrals involve integrands that, under normal circumstances, would be undefined over some section of the domain or at one or more limits of integration. For the given example, \( \int_{-1}^{1} \ln(1-x^2) \, dx \), we encounter an improper integral because the logarithmic function \( \ln(1-x^2) \) is undefined at \( x = -1 \) and \( x = 1 \). Understanding whether such integrals converge or diverge is essential:

• Convergence: An integral converges if its value approaches a finite number as the variable approaches a specific limit.
• Divergence: An integral diverges if its value becomes infinite or oscillates indefinitely.

In this problem, the integral diverges due to infinite values at the endpoint singularities. Using limits as part of the evaluation helps us conclude the nature—either convergence or divergence—of the integral.

Symmetric Functions

A function is symmetric if it behaves the same on both sides of a vertical axis. In our example, \( \ln(1-x^2) \) is symmetric about the line \( x = 0 \). This symmetry allows us to simplify our evaluation of the integral:

• The interval of integration, \([-1, 1]\), is symmetric around the origin.
• The function behaves identically for \(x\) and \(-x\), which usually allows us to assess with only half the work:

Given this symmetry, you can evaluate an integral over half of the interval and double it to deduce the full integral's value. Despite the integral's divergence, leveraging symmetry can significantly simplify integrals that converge.

Integration Techniques: Substitution and Parts

When tackling improper integrals like \( \int_{-1}^{1} \ln(1-x^2) \, dx \), applying smart integration techniques is vital. Additionally, we use different methods to tackle the challenging parts of the integral:

• Substitution: It helps transform the limits of integration and the expressions, making them easier to evaluate.In this example, setting \( x = \sin(\theta) \) simplifies the integral.
• Integration by Parts: Used for integrals of a product of functions. When you've set up your integral post-substitution, integration by parts helps manage logarithmic terms.For the current function, multiple applications and simplifying steps indeed lead us to explore if convergence is possible.

Such techniques are core tools in calculus for managing and evaluating complex integral expressions.

Understanding Singularities in Integrals

Singularities are points where a function becomes undefined or unbounded, often leading to improper integrals. In our example, the integral\( \int_{-1}^{1} \ln(1-x^2) \, dx \) displays singularities at \( x = -1 \) and \( x = 1 \):

• Why they matter: Singularities impact the convergence of an integral. The behavior of the function \( \ln(1-x^2) \) near these points tends towards \(-\infty\), causing divergence.
• Handling Singularities: The key is to split the integral around the singular points and apply limits to see if the integral can be evaluated into a finite number. For this integral, the divergence stems from these infinite tendencies not canceling out, unlike convergent integrals where they can balance or taper off.

Singularities necessitate careful analysis to conclude properly about an integral's convergence or divergence.