Problem 21 - Biocalculus Calculus for the Life Sciences

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{e}^{\infty} \frac{1}{x(\ln x)^{3}} d x\)

Step-by-Step Solution

Answer

The integral is convergent, and its value is \( \frac{1}{2} \).

Step 1: Analyze the integrand

The integrand is \( \frac{1}{x(\ln x)^3} \). We need to determine if it behaves properly at the lower and upper limits, \(e\) and \(\infty\) respectively.

Step 2: Identify improper behavior

Since the integral goes to \(\infty\), it is improper. Specifically, we need to check if \(\int_{e}^{ ext{large } b} \frac{1}{x(\ln x)^3} \, dx \) converges as \(b \rightarrow \infty\).

Step 3: Use substitution to simplify

Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). Therefore, when \( x = e \), \( u = 1 \); when \( x = \infty \), \( u = \infty \). The integral becomes \( \int_1^\infty \frac{1}{u^3} \, du \).

Step 4: Evaluate the integral \(\int_1^\infty \frac{1}{u^3} \, du\)

The integral of \( \frac{1}{u^3} \) is \( \int \frac{1}{u^3} \, du = \frac{-1}{2u^2} + C \). Evaluate from 1 to \( \infty \):\[ \lim_{A \to \infty} \left( \frac{-1}{2A^2} - \frac{-1}{2 \times 1^2} \right) = \lim_{A \to \infty} \left( \frac{-1}{2A^2} + \frac{1}{2} \right) = 0 + \frac{1}{2} \].

Step 5: Determine convergence

Since the evaluated limit approaches \( \frac{1}{2} \), the integral converges.

Key Concepts

Convergence of Integrals

When solving an improper integral like \( \int_{e}^{\infty} \frac{1}{x(\ln x)^{3}} \, dx\), it is critical to determine whether the integral is convergent or divergent. **Convergence** means that the integral settles on a finite number as the bounds approach infinite limits, while **divergence** means it grows without bound.
This involves analyzing the behavior of the integrand \( \frac{1}{x(\ln x)^3} \) as \( x \) approaches infinity, since this is where the improper nature of the integral lies.
To check for convergence:

• Substitute the indefinite bounds with finite variables (say \( A \)) and take the integral from the specified lower bound to \( A \). Then observe the limit as \( A \to \infty \).
• If this limit produces a finite value, the integral converges. Otherwise, it diverges.

The convergence of this specific integral was confirmed, yielding a finite result of \( \frac{1}{2} \), which means it settles to this exact number at infinity.

Substitution Method in Integrals

To tackle the integral \( \int_{e}^{\infty} \frac{1}{x(\ln x)^3} \, dx \), it helps to simplify using the substitution method. This technique transforms a difficult problem into a more manageable form.
Here's how substitution works for this integral:

• Select a substitution variable, \( u \), that simplifies the integral. In our case, let \( u = \ln x \).
• Calculate the derivative: \( du = \frac{1}{x} \, dx \).
• Now change the bounds of integration. For this example, when \( x = e \), \( u = \ln e = 1 \), and when \( x = \infty \), \( u = \infty \).
• The integral now becomes \( \int_1^\infty \frac{1}{u^3} \, du \), a simpler form that is easier to evaluate.

Substitution turns the original integral into a form where standard integral techniques apply more easily, often transforming it into a polynomial or other straightforward integral type.

Integral Calculus

Integral calculus is a fundamental branch of mathematics that deals with the accumulation of quantities and the areas under curves. When dealing with integral calculus, you work with two primary types of integrals: definite and indefinite.
For improper integrals:

• Improper integrals have infinite limits or discontinuous integrands within the range of integration.
• These integrals require limits to resolve, turning them into a limit problem before finding a solution.

In solving integrals, a standard method includes:

• Choosing an appropriate technique (like substitution, integration by parts, or partial fractions) to simplify the problem.
• Evaluating the resulting simpler integral.
• Interpreting the mathematical results, often with limits, to address any infinities or singularities.

Through this process, integral calculus helps in solving complex problems relating to physics, engineering, and many other fields, by transforming complex, real-world data into solvable mathematical equations.