Problem 22 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Determine whether the integral is convergent or divergent. Evaluate all convergent integrals. Be efficient. If \(\lim _{x \rightarrow \infty} \neq 0\), then \(\int_{a}^{\infty} f(x) d x\) is divergent. \(\int_{e}^{\infty} \frac{1}{x \ln x} d x\)

Step-by-Step Solution

Answer

The given integral \( \int_{e}^{\infty} \frac{1}{x \ln x} dx \) is divergent.

Step 1: Identify a simpler function

Identify a simpler function that behaves in a similar manner to our function \( \frac{1}{x \ln x} \), especially as \( x \) approaches infinity. A good choice is \( g(x) = \frac{1}{x} \), because \( \ln x \) is always less than or equal to \( x \) for \( x \ge e \). This implies that \( \frac{1}{x \ln x} \ge \frac{1}{x} \) for \( x \ge e \).

Step 2: Perform the integral test

Establish whether the integral of \( g(x) \) from \( e \) to infinity is convergent. This serves as a comparison to our function. To evaluate this, calculate \(\int_{e}^{\infty} \frac{1}{x} dx = \left[ \ln|x| \right]_{e}^{\infty} \). The result is \( \infty - 1 = \infty \), thus the integral is divergent.

Step 3: Apply the Comparison Test

As per the Comparison Test, if \( 0 \le f(x) \le g(x) \) for \( x \ge a \) and the integral of \( g(x) \) from \( a \) to infinity diverges, then the integral of \( f(x) \) also diverges. Here, we have \( f(x) = \frac{1}{x \ln x} \) and \( g(x) = \frac{1}{x} \), and since the integral of \( g(x) \) from \( e \) to infinity diverges, \( \int_{e}^{\infty} f(x) dx \) also diverges.

Key Concepts

Convergence and Divergence

Understanding whether an improper integral converges or diverges is crucial in calculus. When we talk about convergence, it means that the value of the integral stabilizes as we extend the limits towards infinity. Conversely, divergence indicates that the integral grows without bounds or doesn't settle to a number. For instance, the integral \( \int_{e}^{\infty} \frac{1}{x \ln x} dx \) requires us to determine if it converges or diverges to figure out if there's a finite area under the curve beyond a certain point.
If the limit of the function as it reaches infinity isn't zero, the integral immediately diverges, according to the limit comparison test. In this case, checking if \( \lim _{x \rightarrow \infty} \frac{1}{x \ln x} \) would help in determining divergence.
Using techniques like the Comparison Test can efficiently reveal whether integrals are convergent or not by considering similar functions.

Comparison Test

The Comparison Test is a valuable method used to determine if an improper integral converges or diverges. This test involves comparing the integral of a complex function to that of a simpler one. By doing so, if we know the simpler function diverges or converges, we can often infer the behavior of the more complex function.
For instance, in the exercise where \( f(x) = \frac{1}{x \ln x} \) was compared to \( g(x) = \frac{1}{x} \), we observed that \( g(x) \) diverges. The Comparison Test tells us if \( 0 \leq f(x) \leq g(x) \) and the integral of \( g(x) \) diverges, then the integral of \( f(x) \) must also diverge.
This method simplifies computations, making convergence checks quicker without needing to directly solve the complicated integral. Always remember, when using the Comparison Test:

• Align function inequalities correctly: \( 0 \leq f(x) \leq g(x) \)
• Ensure the simpler function is familiar and its convergence/divergence is known.

Integral Test

The Integral Test is another technique that aids in determining whether a series converges or diverges. It applies particularly to series that can be associated with functions that are positive, continuous, and decrease over an interval.
For improper integrals presented in series form, here's how the Integral Test works:

• Associate the series with a function \( f(x) \)
• Check that \( f(x) \) is positive, continuous, and decreases over your chosen interval
• If \( \int_{1}^{\infty} f(x)dx \) converges, then the corresponding series converges
• If \( \int_{1}^{\infty} f(x)dx \) diverges, then the series diverges too

This test provides a reliable way to analyze convergence and divergence, testament to the power of analysis within calculus. With such tools, we can confidently classify the behavior of complicated functions and series in mathematical analysis.