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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.9.88b
Chapter 8, Problem 8.9.88b

88. Incorrect Calculation
b. Evaluate ∫(from -1 to 1) dx/x or show that the integral does not exist.

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1
Identify the integral to evaluate: \(\int_{-1}^{1} \frac{dx}{x}\). Notice that the integrand \(\frac{1}{x}\) is undefined at \(x=0\), which lies within the interval of integration.
Since the function is not defined at \(x=0\), split the integral into two improper integrals at the point of discontinuity: \(\int_{-1}^{1} \frac{dx}{x} = \int_{-1}^{0} \frac{dx}{x} + \int_{0}^{1} \frac{dx}{x}\).
Rewrite each integral as a limit to handle the improper nature: \(\int_{-1}^{0} \frac{dx}{x} = \lim_{t \to 0^-} \int_{-1}^{t} \frac{dx}{x}\) and \(\int_{0}^{1} \frac{dx}{x} = \lim_{s \to 0^+} \int_{s}^{1} \frac{dx}{x}\).
Evaluate the antiderivative of \(\frac{1}{x}\), which is \(\ln|x|\), and apply it to each integral before taking the limits.
Examine the behavior of the limits as \(t \to 0^-\) and \(s \to 0^+\) to determine if the integrals converge or diverge. If either limit diverges, conclude that the original integral does not exist.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

An improper integral occurs when the integrand is undefined or unbounded within the interval of integration. In this problem, the function 1/x is undefined at x = 0, which lies between -1 and 1, making the integral improper and requiring special evaluation methods.
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Cauchy Principal Value

The Cauchy principal value is a method to assign a finite value to certain improper integrals that are otherwise divergent. For ∫ from -1 to 1 of 1/x dx, the principal value considers symmetric limits approaching zero from both sides, potentially yielding a finite result despite the singularity.
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Non-Existence of the Integral

If the limits of the integral do not converge to a finite number due to the singularity, the integral is said not to exist in the usual sense. Since 1/x has an infinite discontinuity at zero, the integral from -1 to 1 diverges unless interpreted as a principal value.
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Related Practice
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