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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.PE.67

Which of the improper integrals in Exercises 63–68 converge and which diverge?
∫ from −∞ to ∞ of (2 / (e^x + e^(−x))) dx

Guida verificata passo dopo passo
1
First, recognize that the integral is improper because the limits of integration are from \(-\infty\) to \(\infty\).
Rewrite the integrand \(\frac{2}{e^x + e^{-x}}\) in a simpler form. Notice that \(e^x + e^{-x} = 2 \cosh x\), so the integrand becomes \(\frac{2}{2 \cosh x} = \frac{1}{\cosh x}\).
Express the integral as \(\int_{-\infty}^{\infty} \frac{1}{\cosh x} \, dx\). This is a standard integral involving the hyperbolic secant function, since \(\frac{1}{\cosh x} = \operatorname{sech} x\).
To determine convergence, analyze the behavior of \(\operatorname{sech} x\) as \(x \to \pm \infty\). Since \(\cosh x\) grows exponentially, \(\operatorname{sech} x\) approaches zero rapidly, suggesting the integral may converge.
Set up the integral as the sum of two limits: \(\lim_{a \to -\infty} \int_a^0 \operatorname{sech} x \, dx + \lim_{b \to \infty} \int_0^b \operatorname{sech} x \, dx\). Evaluate these integrals using the antiderivative of \(\operatorname{sech} x\), which is \(2 \arctan(\tanh(\frac{x}{2}))\), to check for convergence.

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