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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.8.62

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 2 to ∞ of ((1 / ln x) dx)

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Identify the integral to test for convergence: \(\displaystyle \int_{2}^{\infty} \frac{1}{\ln x} \, dx\).
Recognize that the integral has an infinite upper limit, so we are dealing with an improper integral and need to analyze its behavior as \(x \to \infty\).
Consider comparing the integrand \(\frac{1}{\ln x}\) to a simpler function to determine convergence. Since \(\ln x\) grows without bound but slowly, think about how \(\frac{1}{\ln x}\) compares to functions like \(\frac{1}{x^p}\) or \(\frac{1}{x}\) for large \(x\).
Use the Direct Comparison Test or the Limit Comparison Test by choosing a suitable comparison function. For example, compare \(\frac{1}{\ln x}\) with \(\frac{1}{x}\) because \(\ln x\) grows slower than \(x\), so \(\frac{1}{\ln x}\) decreases slower than \(\frac{1}{x}\).
Evaluate the limit of the ratio of the integrands as \(x \to \infty\): \(\lim_{x \to \infty} \frac{\frac{1}{\ln x}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{x}{\ln x}\). Since this limit goes to infinity, the behavior of \(\frac{1}{\ln x}\) is not dominated by \(\frac{1}{x}\), so use this information to conclude about convergence.

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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 convergence, we consider limits of definite integrals as the interval approaches infinity or the point of discontinuity. Understanding how to set up and interpret these limits is essential for determining if the integral converges or diverges.
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Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral to another integral with a known behavior. If the integrand is smaller than a convergent integral or larger than a divergent integral, conclusions about convergence or divergence can be drawn. This test requires identifying suitable comparison functions that bound the original integrand.
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Limit Comparison Test

The Limit Comparison Test compares two functions by examining the limit of their ratio as the variable approaches infinity. If the limit is a positive finite number, both integrals either converge or diverge together. This test is useful when direct comparison is difficult but the asymptotic behavior of the integrand is similar to a simpler function.
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