Problem 39 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x $$

Step-by-Step Solution

Answer

The integral diverges by the Direct Comparison Test.

Step 1: Determine the Type of Integral

The integral given is \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \). This is an improper integral because as \( x \to 0^+ \), the term \( x^{-2} \) becomes unbounded. Therefore, we need to check for convergence at the lower limit \( x = 0 \).

Step 2: Choose a Comparison Function

To apply the comparison test, we need to choose a function to compare. Note that for small \( x \), \( e^{-1/x} \approx 1 \), so the function \( x^{-2} \) is a good choice to compare with, since \( x^{-2} e^{-1/x} < x^{-2} \) for all \( x > 0 \).

Step 3: Check the Integral of the Comparison Function

Consider the integral of the comparison function \( \int_{0}^{\ln 2} x^{-2} \, dx \). This can be rewritten as \( \lim_{a\to 0^+} \int_{a}^{\ln 2} x^{-2} \, dx \). Upon integration, we get:\[ \lim_{a\to 0^+} \left[ -x^{-1} \right]_{a}^{\ln 2} = \lim_{a\to 0^+} \left( -1/{\ln 2} + 1/a \right) \]The integral diverges as \( a \to 0^+ \) because \( 1/a \to \infty \).

Step 4: Apply the Direct Comparison Test

Since \( x^{-2} e^{-1/x} < x^{-2} \) and the integral \( \int_{0}^{\ln 2} x^{-2} \, dx \) diverges, by the Direct Comparison Test, the integral \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \) also diverges.

Key Concepts

Direct Comparison Test

The Direct Comparison Test is a helpful tool for determining the convergence or divergence of improper integrals. An improper integral is often tricky to evaluate directly because it involves infinity or an unbounded interval. In using the Direct Comparison Test, we choose a simpler function that is easy to integrate, serving as a benchmark for comparison.
Take the integral \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \) used in our exercise. When \( x \) becomes very small, \( e^{-1/x} \) approaches 1. Therefore, \( x^{-2} e^{-1/x} \) is always less than \( x^{-2} \), which is a simpler function.
The key idea here is:

• Find a simpler function, \( g(x) = x^{-2} \), that satisfies \( f(x) < g(x) \) for all \( x \) in the range.
• If \( \int_{0}^{\ln 2} g(x) \, dx \) diverges, then \( \int_{0}^{\ln 2} f(x) \, dx \) also diverges.

This comparison tells us that since \( \int_{0}^{\ln 2} x^{-2} \, dx \) diverges due to the behavior at \( x = 0 \), the original integral also diverges.

Convergence of Integrals

Understanding the convergence of integrals is crucial, especially when dealing with improper integrals. Convergence of an integral indicates that the area under the curve is finite. Improper integrals have one or more infinite limits, making direct evaluation hard. We often resort to tests like the Direct Comparison Test to discern convergence.
When encountering an integral such as \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \), our focus is on the behavior as \( x \) approaches the boundary—in this case, \( 0 \).
Steps to check convergence include:

• Estimate how the integral behaves as it approaches its limits, often using asymptotic approximations.
• Use tests to compare the function with known integrable functions.
• If a comparison shows a larger function's integral diverges, our original integral likely does the same.

Thus, convergence tests simplify the challenge of determining whether an integral evaluates to a finite value.

Limit Comparison Test

The Limit Comparison Test is another strategic method to analyze the convergence of an improper integral. It is particularly useful when direct comparisons are not clear-cut. This test involves calculating the limit of the ratio of our function to a known benchmark function.
For example, if the Direct Comparison Test is inconclusive, one could use the Limit Comparison Test for \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \) by considering:

• Select a function \( g(x) = x^{-2} \) comparable to \( f(x) = x^{-2} e^{-1/x} \).
• Compute \( \lim_{x \to 0^+} \frac{f(x)}{g(x)} \).
• If the limit is a positive finite number, the two integrals will both converge or both diverge.

This test confirms the behavior at points where the Direct Comparison Test may falter. However, in the given exercise, the easy assessment through the Direct Comparison Test efficiently demonstrated divergence without needing the Limit Comparison Test.