Problem 44 - 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. $$ \int_{0}^{2} \frac{d x}{1-x} $$

Step-by-Step Solution

Answer

The integral diverges due to the asymptote at \( x = 1 \).

Step 1: Understand the Integral

We want to determine whether the integral \( \int_{0}^{2} \frac{dx}{1-x} \) converges. The integrand is \( \frac{1}{1-x} \), which becomes undefined at \( x = 1 \). Hence, we need to consider this when evaluating convergence.

Step 2: Identify the Type of Discontinuity

The integrand \( \frac{1}{1-x} \) has a vertical asymptote at \( x = 1 \). This means the integral range \([0, 2]\) includes a point (specifically \( x = 1 \)) where the function is not defined, indicating a potential improper integral.

Step 3: Split the Integral into Two Parts

Split the integral at the point of discontinuity: \[\int_{0}^{2} \frac{dx}{1-x} = \int_{0}^{1} \frac{dx}{1-x} + \int_{1}^{2} \frac{dx}{1-x}\]Now, evaluate each of these limits separately to determine convergence.

Step 4: Evaluate the First Integral

The first integral to consider is \( \int_{0}^{1} \frac{dx}{1-x} \). Let's use a substitution \( u = 1-x \), resulting in \( du = -dx \). Hence,\[\int_{x=0}^{x=1} \frac{dx}{1-x} = -\int_{u=1}^{u=0} \frac{du}{u} = \left[-\ln |u| \right]_1^0 = -(-\infty) + 0 = \infty\]This integral diverges as it approaches infinity.

Step 5: Evaluate the Second Integral

Now consider \( \int_{1}^{2} \frac{dx}{1-x} \). Replace \( x \) with a limit approaching 1 from the right-hand side:\[\int_{1}^{2} \frac{dx}{1-x} = \lim_{b \to 1^+} \int_{b}^{2} \frac{dx}{1-x}\]Evaluating this, \[= \lim_{b \to 1^+} [-\ln|1-x|]_{b}^{2} = -\ln|1-2| + \ln|1-b| \]This simplifies as: \[\lim_{b \to 1^+} \ln|1-b| = -\infty\]The integral also diverges here.

Step 6: Conclusion on Convergence

Since both parts of the integral, \( \int_{0}^{1} \) and \( \int_{1}^{2} \), diverge, the entire integral \( \int_{0}^{2} \frac{dx}{1-x} \) must diverge as well.

Key Concepts

Convergence Tests

When we talk about improper integrals, one of the main concerns is to determine whether they converge or diverge. Convergence would mean that as we calculate the area under the curve over a specific interval, the total area is finite. Divergence, however, indicates that the area is infinite and doesn't settle to a finite number. There are several convergence tests available to help make a determination about integrals:

• p-integral test: Often used when dealing with integrals of the form \( \int \frac{1}{x^p} dx \), this tells us that the integral converges if \( p > 1 \) and diverges if \( p \leq 1 \).
• Comparison tests: These involve comparing your integral to another function with a known convergence behavior.
• Limit comparison test: Similar to the comparison test but uses limits to compare behaviors when direct comparison isn't straightforward.

Each of these tests helps assess whether the integral's area accumulates to a set amount or extends indefinitely. In our example, the integral does not settle to a finite value, indicating divergence.

Direct Comparison Test

The Direct Comparison Test is a handy tool when you need to determine the convergence or divergence of an integral by comparing it with another function that is easier to evaluate. The idea is to use a second function that bounds your function from above or below. Here's how it works:

• If the function \( g(x)\) is such that \( 0 \leq f(x) \leq g(x) \) for all \( x \) in your interval, and if \( \int g(x) \, dx \) converges, then \( \int f(x) \, dx \) also converges.
• Conversely, if \( \int f(x) \, dx \) diverges, you could show \( \int g(x) \, dx \) diverges as well when \( f(x) \geq g(x) \).

In the exercise example provided, using comparison might reveal that near the asymptote, the function behaves similarly to well-known divergent functions, indicating that the integral of interest also diverges.

Limit Comparison Test

The Limit Comparison Test is essential when a direct comparison isn't clear-cut. It leverages the power of computing limits to compare your integral's function with another function. Here’s how to use it:

• Select a comparison function \( g(x) \), which has a known convergence behavior.
• Compute the limit: \( \lim_{x \to a^+} \frac{f(x)}{g(x)} \).
• If the limit \( c \) is positive and finite (\( 0 < c < \infty \)), then \( \int f(x) \, dx \) and \( \int g(x) \, dx \) either both converge or both diverge.

This test is particularly useful when the integral function has a similarly complex structure around a singularity, like an asymptote observed in the exercise. It helps to anchor the understanding of an unknown integral in terms of a known one.

Asymptotes

Asymptotes play a critical role in understanding the behavior of functions within an integral. They indicate where the function goes to infinity, which is important when assessing the convergence of integrals. In essence:

• Vertical asymptotes: Occur when a function approaches infinity as \( x \) approaches a particular value from the left or the right. For example, the integrand \( \frac{1}{1-x} \) has a vertical asymptote at \( x = 1 \).
• Horizontal asymptotes: These occur as \( x \) approaches infinity, indicating the function levels out to a certain value.
• A key strategy is to split the integral around the asymptote, identifying where the integrand becomes undefined, as seen in our example exercise.

Identifying asymptotes allows you to break down the integral into sections that can be evaluated more easily, and you can determine where these sections converge or diverge. In our exercise, the presence of an asymptote indicated trouble with convergence, leading to divergence on both sides of the function.