Problem 48 - 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. $$ \int_{4}^{\infty} \frac{d x}{\sqrt{x}-1} $$

Step-by-Step Solution

Answer

The integral diverges by the Direct Comparison Test.

Step 1: Identify the Type of Improper Integral

The given integral is \( \int_{4}^{\infty} \frac{dx}{\sqrt{x}-1} \). This is an improper integral because the upper limit of integration is infinite.

Step 2: Choose a Suitable Comparison Function

The term \( \sqrt{x} \) at large values of \( x \) dominates the expression \( \sqrt{x} - 1 \). So, compare \( \frac{1}{\sqrt{x}-1} \) with \( \frac{1}{\sqrt{x}} \) as \( x \to \infty \). This is because \( \frac{1}{\sqrt{x}} \) is similar to \( \frac{1}{\sqrt{x}-1} \) when \( x \) is large, and it is easier to check for convergence.

Step 3: Test the Comparison Function for Convergence

The integral \( \int_{4}^{\infty} \frac{1}{\sqrt{x}} \, dx \) is equivalent to \( 2\int_{4}^{\infty} x^{-1/2} \, dx \). This is a \( p \)-integral with \( p = 1/2 \), which is known to diverge for \( p \leq 1 \).

Step 4: Use Direct Comparison Test

Since \( \frac{1}{\sqrt{x}-1} \geq \frac{1}{\sqrt{x}} \) for all \( x \geq 4 \), and since \( \int_{4}^{\infty} \frac{1}{\sqrt{x}} \, dx \) diverges, the Direct Comparison Test implies that \( \int_{4}^{\infty} \frac{dx}{\sqrt{x}-1} \) also diverges.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a powerful tool used in the analysis of improper integrals. This test is applied when you want to understand if an integral converges or diverges, by comparing it to another integral whose convergence properties are already known. In this particular problem, let's consider two positive, continuous functions: \( f(x) \) and \( g(x) \).

• If \( f(x) \leq g(x) \) for all \( x \) in the interval under consideration, and if \( \int g(x) \, dx \) converges, then \( \int f(x) \, dx \) also converges.
• Conversely, if \( f(x) \geq g(x) \) for all \( x \), and \( \int g(x) \, dx \) diverges, then \( \int f(x) \, dx \) also diverges.

For our problem, the function \( \frac{1}{\sqrt{x}-1} \) was compared to \( \frac{1}{\sqrt{x}} \). By observing that \( \frac{1}{\sqrt{x}-1} \geq \frac{1}{\sqrt{x}} \) for \( x \geq 4 \), and knowing the integral of \( \frac{1}{\sqrt{x}} \) diverges, we apply the Direct Comparison Test to conclude that our integral also diverges.

Limit Comparison Test

The Limit Comparison Test is another method to determine the convergence or divergence of integrals and series. This test is particularly useful when the Direct Comparison Test is not applicable, either because it's difficult to find a straight-forward inequality, or the functions don't perfectly align.Here's how the Limit Comparison Test works:

• Select two functions \( f(x) \) and \( g(x) \), both positive and continuous over the interval.
• Compute the limit \( L = \lim_{x \to \infty} \frac{f(x)}{g(x)} \).
• If \( 0 < L < \infty \), then both integrals \( \int f(x) \, dx \) and \( \int g(x) \, dx \) share the same behavior; they either both converge or both diverge.

In our exercise, although we did not use the Limit Comparison Test, it might be suitable to apply if you're considering other functions close to \( \frac{1}{\sqrt{x}-1} \), as it guarantees convergence information when direct inequalities are tough to prove.

Convergence and Divergence

Understanding the concepts of convergence and divergence is crucial in calculus, especially when dealing with improper integrals. Convergence of an integral means that as the variable approaches infinity, the area under the curve is finite and measurable. Conversely, divergence indicates an infinite area.In the realm of improper integrals, evaluating convergence involves several steps:

• Identify the integral as improper by looking at its bounds or points of discontinuity.
• Choose a suitable comparison function if applying a test like the Direct or Limit Comparison Tests.
• Analyze the behavior of these functions as they approach the limits of integration.

In conclusion, the integral \( \int_{4}^{\infty} \frac{1}{\sqrt{x}-1} \, dx \) diverges, meaning the area under the curve from 4 to infinity is infinite. Recognizing whether an integral converges or diverges helps you determine the type of behavior you can expect from a function across its limits.