Textbook Question
77–86. Comparison Test Determine whether the following integrals converge or diverge.
81. ∫(from 1 to ∞) (sin²x) / x² dx
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12. Techniques of Integration
Improper Integrals
2:40 minutesProblem 8.9.88b
Textbook Question
88. Incorrect Calculation
b. Evaluate ∫(from -1 to 1) dx/x or show that the integral does not exist.
Verified step by step guidance1
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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Video duration:
2mKey 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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Improper Integrals: Infinite Intervals
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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