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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 8.8.68
Chapter 8, Problem 8.8.68

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 -∞ to ∞ of ((dx) / (e^x + e^(-x)))

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1
Rewrite the integral to better understand the integrand. Notice that \(e^x + e^{-x}\) can be expressed in terms of hyperbolic cosine: \(e^x + e^{-x} = 2\cosh(x)\). So the integral becomes \(\int_{-\infty}^{\infty} \frac{dx}{2\cosh(x)}\).
Recognize that the integrand \(\frac{1}{2\cosh(x)}\) is an even function because \(\cosh(x)\) is even. This allows us to simplify the integral over \((-\infty, \infty)\) to twice the integral over \((0, \infty)\): \(2 \int_0^{\infty} \frac{dx}{2\cosh(x)} = \int_0^{\infty} \frac{dx}{\cosh(x)}\).
To test for convergence, analyze the behavior of the integrand as \(x \to \infty\). Since \(\cosh(x) \approx \frac{e^x}{2}\) for large \(x\), the integrand behaves like \(\frac{1}{\cosh(x)} \approx 2e^{-x}\), which is similar to an exponential decay function.
Use the Direct Comparison Test by comparing \(\frac{1}{\cosh(x)}\) to \(2e^{-x}\) for large \(x\). Since \(\int_0^{\infty} e^{-x} dx\) converges, and \(\frac{1}{\cosh(x)} \leq 2e^{-x}\) for sufficiently large \(x\), the integral converges on \((0, \infty)\).
Similarly, analyze the behavior as \(x \to 0\) and \(x \to -\infty\). Near zero, \(\cosh(x)\) is finite and positive, so no issues with convergence there. As \(x \to -\infty\), use the evenness of the function or the same comparison to \$2e^{x}$ (since \(\cosh(x) = \cosh(-x)\)). This confirms convergence over the entire real line.

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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 infinite intervals or integrands with infinite discontinuities. To evaluate them, we consider limits of definite integrals as the bounds approach infinity or points of discontinuity, determining if the integral converges to a finite value.
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Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral's integrand to a simpler function with known behavior. If the integrand is smaller than a convergent function or larger than a divergent one, we can conclude about the integral's convergence accordingly.
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Limit Comparison Test

The Limit Comparison Test compares the integrand to a known function 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, aiding in determining the original integral's behavior.
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