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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 8.8.42
Chapter 8, Problem 8.8.42

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 1 to 2 of (dx / (x ln x))

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Identify the integral to be tested for convergence: \(\int_{1}^{2} \frac{dx}{x \ln x}\).
Recognize that the integrand has a potential issue at the lower limit \(x=1\) because \(\ln 1 = 0\), which makes the denominator zero and the integrand undefined there.
To analyze the behavior near \(x=1\), consider the limit of the integrand as \(x\) approaches 1 from the right: examine \(\lim_{x \to 1^+} \frac{1}{x \ln x}\).
Use a comparison test by comparing the integrand to a simpler function that behaves similarly near \(x=1\). For example, since \(\ln x \approx x-1\) near 1, compare \(\frac{1}{x \ln x}\) to \(\frac{1}{x (x-1)}\) or simply \(\frac{1}{x-1}\) near 1.
Determine if the integral \(\int_{1}^{2} \frac{dx}{x \ln x}\) converges by checking if the integral of the comparison function converges near \(x=1\). If the comparison integral converges, then by the Direct Comparison Test or Limit Comparison Test, the original integral converges as well.

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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 interval where the function is unbounded or the interval is infinite. To determine convergence, we evaluate limits approaching the problematic points. In this problem, the integral has a potential issue near x = 1 because ln(x) approaches zero, making the integrand unbounded.
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Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test compares the given integral to a known integral with a simpler integrand. If the simpler integral converges and the original integrand is smaller, the original integral converges. Conversely, if the simpler integral diverges and the original integrand is larger, the original integral diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test uses the limit of the ratio of two functions to determine convergence. If the limit of the ratio of the given integrand to a known integrand is a positive finite number, both integrals either converge or diverge together. This test is useful when direct comparison is inconclusive.
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Limit Comparison Test
Related Practice
Textbook Question

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