Problem 62 - 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. $$ \int_{1}^{\infty} \frac{1}{e^{x}-2^{x}} d x $$

Step-by-Step Solution

Answer

The integral converges by the Direct Comparison Test.

Step 1: Understand the Integral

We are asked to determine whether the integral \( \int_{1}^{\infty} \frac{1}{e^{x}-2^{x}} \, dx \) converges or diverges. To do this, we need to analyze the behavior of the integrand as \( x \to \infty \).

Step 2: Analyze the Behavior of the Integrand

For large values of \( x \), both \( e^x \) and \( 2^x \) grow exponentially, but \( e^x \) grows faster than \( 2^x \). Thus, as \( x \to \infty \), \( e^x - 2^x \approx e^x \). Therefore, the integrand behaves like \( \frac{1}{e^x} \) for large \( x \).

Step 3: Choose a Comparison Function

A suitable comparison function is \( \frac{1}{e^x} \). We know that \( \int_{1}^{\infty} \frac{1}{e^x} \, dx \) can be evaluated using an antiderivative, and it converges because it transforms to \( -e^{-x} \mdoteq 0-(-e^{-1}) \) in the limit as \( x \to \infty \).

Step 4: Apply the Direct Comparison Test

Since \( e^x - 2^x \geq e^x \) for all \( x \geq 1 \), it follows that \( \frac{1}{e^x - 2^x} \leq \frac{1}{e^x} \). By the Direct Comparison Test, because \( \int_{1}^{\infty} \frac{1}{e^x} \, dx \) converges, \( \int_{1}^{\infty} \frac{1}{e^x - 2^x} \, dx \) also converges.

Key Concepts

Understanding Convergence Tests

Convergence tests are essential tools in calculus for determining whether an integral or series converges or diverges. If it converges, the integral has a finite value; if it diverges, it does not approach a limit.

These tests analyze the behavior of the integrand or sequence as it tends toward infinity. Some of the popular tests include the Direct Comparison Test, the Limit Comparison Test, and others like the Ratio or Root Tests.

For improper integrals, selecting the correct test involves examining how the terms grow or shrink over the interval of integration.

Exploring the Direct Comparison Test

The Direct Comparison Test is often used when you can compare your given function to a simpler function that is easier to integrate, and whose convergence properties are known.

Here's how it works: Given two functions, say \( f(x) \) and \( g(x) \), if \( 0 \leq f(x) \leq g(x) \) for all \( x \) in the interval under consideration, then the integral computes as follows:

• If \( \int g(x) \, dx \) converges, \( \int f(x) \, dx \) also converges.
• If \( \int f(x) \, dx \) diverges, \( \int g(x) \, dx \) also diverges.

This effectively allows the behavior of the simpler integral to guide conclusions about the complex or untested integral.

In the given problem, the function \( \frac{1}{e^x} \) was used as a comparison because it's known to converge, and the inequality between the two functions supports using Direct Comparison.

Limit Comparison Test Explained

The Limit Comparison Test is another useful way to determine the convergence behavior of integrals or series. This test involves comparing the given function with another, but it uses limits instead of inequalities.

To apply the Limit Comparison Test, you follow these steps:

• Choose a function \( g(x) \) such that both \( \int f(x) \, dx \) and \( \int g(x) \, dx \) are positive and comparable.
• Compute the limit: \[ \lim_{{x \to \infty}} \frac{f(x)}{g(x)} = L \]
• If the limit \( L \) is a positive finite number, then either both integrals converge or both diverge.

The key is finding a \( g(x) \) that is manageable and having clear behavior as \( x \) approaches infinity.

This test is particularly useful when the straightforward comparisons required for the Direct Comparison Test are hard to verify.