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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 8.8.60
Chapter 8, Problem 8.8.60

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 π to ∞ of ((1 + sin x) / x² dx)

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Identify the integral to be tested for convergence: \(\displaystyle \int_{\pi}^{\infty} \frac{1 + \sin x}{x^{2}} \, dx\).
Note that the integrand is \(\frac{1 + \sin x}{x^{2}}\). Since \(\sin x\) oscillates between \(-1\) and \(1\), the numerator \(1 + \sin x\) oscillates between \(0\) and \(2\).
To apply the Direct Comparison Test, find a simpler function that bounds the integrand from above and below. Since \(0 \leq 1 + \sin x \leq 2\), we have \(0 \leq \frac{1 + \sin x}{x^{2}} \leq \frac{2}{x^{2}}\) for \(x \geq \pi\).
Recall that the integral \(\int_{\pi}^{\infty} \frac{1}{x^{2}} \, dx\) converges because it is a p-integral with \(p=2 > 1\). Therefore, by the Direct Comparison Test, since \(\frac{1 + \sin x}{x^{2}} \leq \frac{2}{x^{2}}\) and the integral of \(\frac{2}{x^{2}}\) converges, the original integral converges.
Conclude that the integral \(\int_{\pi}^{\infty} \frac{1 + \sin x}{x^{2}} \, dx\) converges by the Direct Comparison Test.

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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 an infinite interval or where the integrand is unbounded. To determine convergence, we evaluate the limit of the integral as the upper bound approaches infinity. If this limit exists and is finite, the integral converges; otherwise, it diverges.
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Direct Comparison Test

The Direct Comparison Test compares the given integral to a second integral with a known convergence behavior. If the integrand is less than or equal to a function with a convergent integral, the original integral also converges. Conversely, if it is greater than or equal to a function with a divergent integral, it diverges.
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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, allowing us to conclude about the original integral's convergence.
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Related Practice
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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.

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