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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.8.52
Chapter 8, Problem 8.8.52

In Exercises 35–68, 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.
∫ from 4 to ∞ of (dx / (√x - 1))

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Identify the integral to be tested for convergence: \(\displaystyle \int_{4}^{\infty} \frac{dx}{\sqrt{x} - 1}\).
Analyze the behavior of the integrand as \(x \to \infty\). Since \(\sqrt{x}\) grows without bound, \(\sqrt{x} - 1\) behaves like \(\sqrt{x}\) for large \(x\), so the integrand behaves like \(\frac{1}{\sqrt{x}}\).
Choose a comparison function to apply the Direct Comparison Test or Limit Comparison Test. A natural choice is \(\frac{1}{\sqrt{x}}\), because it simplifies the integrand's behavior at infinity.
Recall that the integral \(\int_{4}^{\infty} \frac{1}{\sqrt{x}} \, dx\) is a p-integral with \(p = \frac{1}{2}\), which diverges since \(p \leq 1\).
Use the Direct Comparison Test or Limit Comparison Test by comparing \(\frac{1}{\sqrt{x} - 1}\) with \(\frac{1}{\sqrt{x}}\) to determine if the original integral converges or diverges.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, we take limits to handle the infinite bounds or singularities, determining whether the integral converges (has a finite value) or diverges.
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Direct Comparison Test

The Direct Comparison Test compares the given integral's integrand to a simpler function whose convergence behavior is known. If the integrand is smaller than a convergent function or larger than a divergent function on the interval, we can conclude about the original integral's convergence.
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Limit Comparison Test

The Limit Comparison Test uses the limit of the ratio of two functions as the variable approaches infinity. If this limit is a positive finite number, both integrals either converge or diverge together, allowing us to infer the behavior of a complicated integral by comparing it to a simpler one.
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