Problem 61 - Thomas Calculus

Question

Logarithmic \(p\) -series $$ \begin{array}{c}{\text { a. Show that the improper integral }} \\\ {\int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant })} \\ {\text { converges if and only if } p>1} \\ {\text { b. What implications does the fact in part (a) have for the con- }} \\\ {\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ?} \\ {\text { Give reasons for vour answer. }}\end{array} $$

Step-by-Step Solution

Answer

The integral and series converge if and only if \( p > 1 \).

Step 1: Interpret the Improper Integral

We need to evaluate whether the improper integral \( \int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} \, dx \) converges or diverges. This involves checking if the integral evaluates to a finite value as the upper limit approaches infinity.

Step 2: Evaluate the Convergence for Large x

For large \( x \), consider the substitution \( v = \ln x \), this gives \( dv = \frac{1}{x} \, dx \). The limits of integration when \( x = 2 \rightarrow v = \ln 2 \), and as \( x \to \infty \rightarrow v \to \infty \). This transforms the integral to \( \int_{\ln 2}^{\infty} \frac{dv}{v^{p}} \).

Step 3: Apply the p-test for Integrals

The integral \( \int_{\ln 2}^{\infty} \frac{1}{v^{p}} \, dv \) is a standard \( p \)-integral. It converges if \( p > 1 \) and diverges if \( p \leq 1 \). Therefore, \( \int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} \, dx \) converges if and only if \( p > 1 \).

Step 4: Relate the Integral to the Series

The problem asks about the series \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} \). Generally, the convergence of an integral resembles the convergence of its corresponding series if both follow similar terms. Therefore, the series will converge if and only if the integral converges.

Step 5: Conclude Implication for the Series

Using the comparison of the series \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} \) with the integral \( \int_{2}^{\infty} \frac{1}{x(\ln x)^{p}} \, dx \), the series converges if and only if \( p > 1 \).

Key Concepts

Improper integral

An improper integral is a unique kind of integral where the integrand has an infinite interval or the integrand becomes infinite within the interval of integration. For example, in our case, the integral \(\int_{2}^{\infty} \frac{1}{x(\ln x)^{p}}\,dx\) is improper because it extends to infinity. This means that as we calculate, we must consider the possibility that the integral does not result in a finite number.
Handling improper integrals requires a special approach. One common technique is to transform or substitute parts of the integral—especially when direct calculation isn't possible. Adjusting limits and seeing if the result stabilizes reveals whether it converges or diverges.

Convergence

Convergence is a key concept in evaluating improper integrals. It refers to whether the integral sums up to a specific, finite number as the bounds approach infinity or some undefined point. In simple terms, a converging integral "settles down" to a particular value.

• If an integral converges, it means we can calculate it, and the result will be meaningful.
• If it diverges, the integral doesn't result in a finite sum and essentially "blows up."

For the integral \(\int_{\ln 2}^{\infty} \frac{1}{v^{p}} \, dv\) to converge, \( p \) must be greater than 1. This is crucial, as it reflects how the function behaves at infinity. If \( p \leq 1 \), the function is too strong to "flatten out" at infinity, leading to divergence.

p-test for integrals

The \( p \)-test is a handy tool to determine the convergence of certain types of integrals. It's particularly useful with integrals that resemble \( \int_{a}^{\infty} \frac{1}{x^p} \, dx \). The rule is straightforward:

• If \( p > 1 \), the integral converges.
• If \( p \leq 1 \), the integral diverges.

When applying the \( p \)-test to our transformed integral \(\int_{\ln 2}^{\infty} \frac{1}{v^{p}} \, dv\), it's clear that the integral converges only if \( p > 1 \). This is fundamental in deciding the convergence of similar series, such as \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} \). The series mirrors the behavior of the integral, so it will also converge when \( p > 1 \).