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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 8.8.54
Chapter 8, Problem 8.8.54

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 0 to ∞ of (dθ / (1 + e^θ))

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Identify the integral to be tested for convergence: \(\displaystyle \int_0^{\infty} \frac{d\theta}{1 + e^{\theta}}\).
Examine the behavior of the integrand \(f(\theta) = \frac{1}{1 + e^{\theta}}\) as \(\theta \to \infty\) and as \(\theta \to 0\) to understand the nature of the integral over the interval \([0, \infty)\).
For large \(\theta\), note that \(e^{\theta}\) grows very quickly, so \(f(\theta)\) behaves like \(\frac{1}{e^{\theta}} = e^{-\theta}\). This suggests comparing the integrand to \(g(\theta) = e^{-\theta}\), a function with a known convergent integral over \([0, \infty)\).
Apply the Direct Comparison Test by checking if \(0 \leq f(\theta) \leq g(\theta)\) for sufficiently large \(\theta\). Since \(\frac{1}{1 + e^{\theta}} < \frac{1}{e^{\theta}}\) for \(\theta > 0\), the integral of \(f(\theta)\) is bounded above by the integral of \(g(\theta)\), which converges.
Conclude that since \(\int_0^{\infty} e^{-\theta} d\theta\) converges and \(f(\theta)\) is less than or equal to \(g(\theta)\) for large \(\theta\), by the Direct Comparison Test, the original integral \(\int_0^{\infty} \frac{d\theta}{1 + e^{\theta}}\) also converges.

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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 bounds approach infinity or points of discontinuity. Understanding this helps determine if the integral converges to a finite value or diverges.
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Improper Integrals: Infinite Intervals

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 less than or equal to a convergent function, the integral converges; if it is greater than or equal to a divergent function, it diverges. This test requires finding appropriate bounding functions.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test uses the limit of the ratio of two functions 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 integrand behaves similarly to a known function at infinity.
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Related Practice
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