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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 8.8.64

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 e to e^e of (ln(ln x) dx)

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1
Identify the integral to be tested for convergence: \(\int_{e}^{e^{e}} \ln(\ln x) \, dx\).
Note the interval of integration is finite, from \(x = e\) to \(x = e^{e}\), and the integrand is \(\ln(\ln x)\).
Check the behavior of the integrand on the interval: since \(x \geq e\), \(\ln x \geq 1\), so \(\ln(\ln x)\) is defined and continuous on \([e, e^{e}]\).
Since the integrand is continuous on a closed and bounded interval, the integral is a proper integral and therefore converges.
Conclude that no comparison test is necessary here because the integral is over a finite interval with a continuous integrand, ensuring convergence.

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Key Concepts

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

Improper Integrals and Convergence

Improper integrals involve integration over unbounded intervals or integrands with unbounded behavior. To determine convergence, we analyze the limit of the integral as the interval approaches infinity or a problematic point. Understanding whether the integral converges or diverges is essential for evaluating integrals like the one given.
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Direct Comparison Test

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

The Limit Comparison Test involves taking the limit of the ratio of the given integrand to a simpler function with known convergence properties. If the limit is a positive finite number, both integrals share the same convergence behavior, making this test useful when direct comparison is difficult.
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