Chapter 8: Problem 65
Find the values of \(p\) for which each integral converges. a. \(\int_{1}^{2} \frac{d x}{x(\ln x)^{p}}\) b. \(\int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}}\)
Short Answer
Expert verified
Integral (a) converges for all \( p \); (b) converges for \( p > 1 \).
Step by step solution
01
Understand the Problem Statement
The problem involves evaluating the convergence of two integrals, \( \int_{1}^{2} \frac{d x}{x(\ln x)^{p}} \) and \( \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \) for different values of \( p \). We need to find for what values of \( p \), these integrals either converge or diverge.
02
Analyze Integral Part (a)
Consider \( \int_{1}^{2} \frac{d x}{x(\ln x)^{p}} \). This is a definite integral on a closed interval \([1,2]\). Generally, such integrals converge unless there is a singularity within the interval. Here, \(\ln x\) does not have a singularity in \([1,2]\) as \(\ln x\) is well-defined and continuous.Since there's no singularity and the interval is finite, the integral will converge for any real \( p \).
03
Analyze Integral Part (b)
Consider \( \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \). As \( x \to \infty \), \( \ln x \) becomes large and \( (\ln x)^{p} \) dominates. To check convergence at infinity, compare with a known convergent series using the limit comparison test or other means.Substitute \( u = \ln x \), from which \( du = \frac{1}{x} dx \). The integral becomes \( \int \frac{du}{u^{p}} \) as \( x \to \infty \), \( u \to \infty \). The integral \( \int \frac{du}{u^{p}} \) converges for \( p > 1 \). Therefore, the integral converges for \( p > 1 \).
04
Conclude on Convergence for Part (a) and (b)
Based on our analysis, integral (a) converges for all real \( p \) since there are no singularities in the range. For integral (b), it converges if \( p > 1 \), as the comparison with the convergent p-series shows convergence.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence Tests
Convergence tests are essential tools in calculus to determine if a given integral, especially an improper integral, converges or diverges. Convergence of an integral essentially means that the integral has a finite value when evaluated over its entire domain.
When dealing with convergence tests, several techniques can be employed:
When dealing with convergence tests, several techniques can be employed:
- The Comparison Test allows us to compare an integral or series with a known convergent or divergent one. If the integral in question is smaller than a known convergent integral, it converges. Similarly, if it's larger than a known divergent integral, it diverges.
- The Limit Comparison Test, like its name suggests, compares the limit of the ratio of two functions. If the limit is neatly finite and non-zero, both integrals will either converge or diverge together.
- The p-Test is particularly useful with series and improper integrals of the form \( \int \frac{1}{x^p} \, dx \). It states that the integral converges if \( p > 1 \) and diverges for \( p \leq 1 \).
Improper Integrals
Improper integrals are an extension of definite integrals and occur in two scenarios: when the limits of integration are infinite, or when the integrand becomes unbounded within the interval.
For example, in the exercise, part (b) involves evaluating an improper integral from 2 to infinity. Such an integral can diverge if the area under the curve continues indefinitely.
When dealing with infinite limits, transform the integral by making a substitution or by using known tests:
For example, in the exercise, part (b) involves evaluating an improper integral from 2 to infinity. Such an integral can diverge if the area under the curve continues indefinitely.
When dealing with infinite limits, transform the integral by making a substitution or by using known tests:
- Substitutions can simplify the problem by changing variables, as seen with \( u = \ln x \) in our example.
- The Comparison Test or Limit Comparison Test can then be applied to determine convergence.
Limit Comparison Test
The Limit Comparison Test is a powerful method for evaluating the convergence of integrals, particularly suitable when you want to compare your integral to another one whose convergence is well understood.
In essence, this test involves:
In essence, this test involves:
- Choosing a comparison function \( g(x) \) that closely resembles the behavior of \( f(x) \) at infinity or near a point of interest.
- Calculating the limit \( \lim_{x \to \infty} \frac{f(x)}{g(x)} \). If this limit is a non-zero constant (i.e., \( 0 < L < \infty \)), then both integrals \( \int f(x) \, dx \) and \( \int g(x) \, dx \) converge or diverge together.