Problem 52 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{2}^{\infty} \frac{d x}{\sqrt{x^{2}-1}} $$

Step-by-Step Solution

Answer

The integral diverges.

Step 1: Analyze the Type of Integral

The given integral is \( \int_{2}^{\infty} \frac{d x}{\sqrt{x^{2}-1}} \). This is an improper integral because the upper limit of integration is infinity. The integrand is \( \frac{1}{\sqrt{x^{2}-1}} \). We must determine whether it converges or diverges.

Step 2: Simplify the Integrand

Notice that as \( x \to \infty \), the expression \( \sqrt{x^2 - 1} \approx x \). Therefore, the integrand \( \frac{1}{\sqrt{x^2 - 1}} \approx \frac{1}{x} \). This provides a rough approximation useful for comparison.

Step 3: Choose a Comparison Function

We can use the Direct Comparison Test and compare the given integrand to \( \frac{1}{x} \). Over the interval \([2, \infty)\), we have \( \frac{1}{\sqrt{x^2 - 1}} < \frac{1}{x} \) because \( \sqrt{x^2 - 1} > x \).

Step 4: Check Convergence of Comparison Integral

We know from basic calculus that the integral \( \int_{2}^{\infty} \frac{1}{x} \ dx \) is a divergent p-integral because \( p = 1 \leq 1 \). Therefore, since the integrand \( \frac{1}{\sqrt{x^2 - 1}} < \frac{1}{x} \) and the comparison integral diverges, the Direct Comparison Test is inconclusive.

Step 5: Apply Limit Comparison Test

To use the Limit Comparison Test, compute the limit: \[ L = \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^2 - 1}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 - 1}}. \]Rewriting gives:\[ L = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 - 1}} = \lim_{x \to \infty} \frac{x}{x\sqrt{1 - \frac{1}{x^2}}}. \]Simplifying, we find:\[ L = \lim_{x \to \infty} \frac{1}{\sqrt{1 - \frac{1}{x^2}}} = 1. \]

Step 6: Conclude Using Limit Comparison Test

Since \( L = 1 \) and \( \int_{2}^{\infty} \frac{1}{x} \ dx \) diverges, the Limit Comparison Test concludes that \( \int_{2}^{\infty} \frac{d x}{\sqrt{x^{2}-1}} \) also diverges.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a straightforward method to determine the convergence or divergence of improper integrals. This test involves comparing the integrand of your integral to another function whose behavior is known. If you have an integral of the form \( \int_a^b f(x) \, dx \) and you can find a function \( g(x) \) such that:

• For all \( x \) in \([a, b]\), \( 0 \leq f(x) \leq g(x) \).
• The integral \( \int_a^b g(x) \, dx \) is known to converge.

Then, the original integral \( \int_a^b f(x) \, dx \) also converges. Conversely, if \( g(x) \) diverges and \( f(x) \geq g(x) \), then \( f(x) \) also diverges.In our specific example of \( \int_2^{\infty} \frac{dx}{\sqrt{x^2 - 1}} \), we found that \( \frac{1}{\sqrt{x^2 - 1}} < \frac{1}{x} \). Since \( \int_2^{\infty} \frac{1}{x} \, dx \) is a well-known divergent integral, the Direct Comparison Test is inconclusive for convergence in this case. However, this attempt helps to establish a baseline for comparison with another test.

Limit Comparison Test

When the Direct Comparison Test returns inconclusive results, the Limit Comparison Test is a perfect alternative. It's particularly useful when the overruling behavior of the integrand matches another known function over an infinite interval. Here's how it works: for two functions \( f(x) \) and \( g(x) \), if both are positive and continuous over \([a, \infty)\), we compute the limit:\[L = \lim_{x \to \infty} \frac{f(x)}{g(x)}\]

• If \( 0 < L < \infty \), both functions either converge or diverge together.

Applying this to the integral \( \int_2^{\infty} \frac{dx}{\sqrt{x^2 - 1}} \), we compare it to \( \frac{1}{x} \). The limit was computed as:\[L = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 - 1}} = 1\]Since \( L = 1 \) is finite and non-zero, and since \( \int_2^{\infty} \frac{1}{x} \, dx \) is divergent, the Limit Comparison Test tells us that \( \int_2^{\infty} \frac{dx}{\sqrt{x^2 - 1}} \) must also diverge. It's a powerful method that provides clear answers where direct comparison fails.

Integration Techniques

When dealing with improper integrals, various integration techniques can be employed to simplify the evaluation process or to find useful comparison functions. One crucial technique involves simplifying complex expressions by recognizing dominant terms as \( x \) approaches infinity. For an integrand like \( \frac{1}{\sqrt{x^2 - 1}} \), simplifying the square root expression to approximately \( x \) gives us a comparable function: \( \frac{1}{x} \). This approximation works best when such transformations remain valid over the interval of integration. Other techniques might involve substitution, partial fraction decomposition, or even expansion into series for more complex expressions. Recognizing patterns or standard forms, like trigonometric identities or exponential integrals, are also part of this toolset. These techniques are not just about crunching numbers but understanding the behavior of functions as \( x \) goes to infinity. Together with comparison tests, they form the backbone of evaluating improper integrals effectively.