Problem 60 - Calculus Single Variable

Question

Pach of the improper integrals is improper for two reasons. Determine whether the integral converges or diverges. If it converges, then evaluate it. $$ \int_{2}^{\infty} \frac{2}{x^{2} \sqrt{x^{2}-4}} d x $$

Step-by-Step Solution

Answer

The integral diverges due to the logarithm of infinity.

Step 1: Recognize and Analyze Improprieties

The integral \( \int_{2}^{\infty} \frac{2}{x^{2} \sqrt{x^{2}-4}} \, dx \) is improper for two reasons. First, the interval is unbounded as it extends to infinity. Second, the integral has a lower limit where the function potentially becomes undefined because the part under the square root, \( x^2 - 4 \), equals zero at \( x=2 \). We will need to separately address these issues to determine convergence.

Step 2: Split and Rewrite the Integral

To address the improper integral, we express it as a limit for handling the infinity and as an approach to the point where the function is undefined:\[\int_{2}^{\infty} \frac{2}{x^{2} \sqrt{x^{2} - 4}} \, dx = \lim_{a \to \infty} \int_{2}^{a} \frac{2}{x^{2} \sqrt{x^{2} - 4}} \, dx\]

Step 3: Substitution to Simplify the Integral

Use the substitution \( x = 2 \sec(\theta) \), where \( dx = 2\sec(\theta)\tan(\theta) \, d\theta \) and \( \sqrt{x^2 - 4} = 2\tan(\theta) \). The limits of integration transform as: at \( x = 2 \), \( \theta = 0 \); and as \( x \to \infty \), \( \theta \to \frac{\pi}{2} \). This simplifies the integral:\[\int_{0}^{\frac{\pi}{2}} \frac{2 \cdot 2\sec(\theta)\tan(\theta)}{(2\sec(\theta))^2 \cdot 2\tan(\theta)} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{2\cos^2(\theta)} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \sec(\theta) \, d\theta\]

Step 4: Evaluate the Transformed Integral

The integral \( \int \sec(\theta) \, d\theta \) is known to be \( \ln |\sec(\theta) + \tan(\theta)| + C \). So,\[\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \sec(\theta) \, d\theta = \frac{1}{2} \left[ \ln | \sec(\theta) + \tan(\theta) | \right]_{0}^{\frac{\pi}{2}}\]\[= \frac{1}{2} \left( \ln(\infty) - \ln(1) \right)\]The natural logarithm of infinity diverges, so the integral does not converge.

Key Concepts

Convergence

Understanding convergence is crucial when dealing with improper integrals, as it indicates whether the integral has a finite value as it approaches its limits. In the context of improper integrals, convergence depends on:

• The behavior of the function as it approaches infinite bounds.
• The behavior at points within the integration range where the function may become undefined.

In our example, the integral is improper because it has an infinite upper limit and the function is undefined at the lower limit due to a zero under the square root. The first step in evaluating such integrals is to examine these limiting behaviors. If both the limits lead to finite values, the integral converges; otherwise, it diverges, as we found with this integral.

Substitution Method

The substitution method is a key technique to simplify complex integrals, especially in cases involving trigonometric expressions. It helps to transform the original variable into another one, easing the integration process.
In the given problem, we use the substitution method with:

• Letting \( x = 2 \sec(\theta) \), transforming both the integrand and the limits of integration.
• The substitution simplifies the square root \( \sqrt{x^2 - 4} \) to \( 2\tan(\theta) \), making the integral more manageable.

This trigonometric substitution is particularly fitting for integrals involving the square root of a quadratic expression.

Transformation of Limits

When performing a substitution, it's essential to also transform the limits of the integral accordingly. This maintains the integrity of the integration.
For our integration:

• At the lower limit \( x = 2 \), the substitution \( x = 2 \sec(\theta) \) gives \( \theta = 0 \).
• As \( x \to \infty \), \( \theta \to \frac{\pi}{2} \), a pivotal part of the integral's evaluation.

These transformed limits redefine the range for the new variable \( \theta \), allowing the integration process to occur over a different but equivalent range, ensuring that the integral's original convergence behavior can be analyzed with respect to \( \theta \).

Integration by Parts

While integration by parts wasn't directly used in this example, understanding it enriches your mathematical toolkit, especially when solving integrals involving products of functions.
The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \), where you choose\( u \) and \( dv \)

This technique is invaluable, particularly when one function in the product becomes simpler upon differentiation and the other maintains its complexity under integration, paving the way for a more approachable integral.