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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 8.8.58

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 2 to ∞ of ((x dx) / √(x⁴ - 1))

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First, write down the integral clearly: \(\int_{2}^{\infty} \frac{x}{\sqrt{x^{4} - 1}} \, dx\).
To determine convergence, analyze the behavior of the integrand as \(x \to \infty\). Simplify the expression inside the square root for large \(x\): \(\sqrt{x^{4} - 1} \approx \sqrt{x^{4}} = x^{2}\).
Using this approximation, the integrand behaves like \(\frac{x}{x^{2}} = \frac{1}{x}\) for large \(x\). This suggests comparing the integral to \(\int_{2}^{\infty} \frac{1}{x} \, dx\).
Apply the Direct Comparison Test or Limit Comparison Test with the function \(g(x) = \frac{1}{x}\). Since \(\int_{2}^{\infty} \frac{1}{x} \, dx\) diverges (harmonic integral), use the Limit Comparison Test to confirm the behavior of the original integral.
Compute the limit \(\lim_{x \to \infty} \frac{\frac{x}{\sqrt{x^{4} - 1}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{x^{2}}{\sqrt{x^{4} - 1}}\). Simplify this limit to determine if it is a finite positive number, which will indicate that both integrals share the same convergence behavior.

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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 the limit of the integral as the upper bound approaches infinity. Determining whether the integral converges or diverges is essential for understanding the behavior of functions over unbounded domains.
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Direct Comparison Test

The Direct Comparison Test determines convergence by comparing the given integral to a simpler integral with known behavior. 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 identifying appropriate bounding functions.
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Limit Comparison Test

The Limit Comparison Test compares two functions 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. This test is useful when direct comparison is difficult but the asymptotic behavior of the integrand is known.
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