Problem 47 - Calculus Single Variable

Question

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{8} \frac{1}{x^{1 / 3}} \ln \left(\frac{1}{x^{1 / 3}}\right) d x\)

Step-by-Step Solution

Answer

The integral converges to \(-6\ln(2) + 3\).

Step 1: Understand the Integral

We need to determine the convergence of the improper integral \(\int_{0}^{8} \frac{1}{x^{1/3}} \ln(\frac{1}{x^{1/3}}) \, dx\). The potential issue is at \(x = 0\), as \(x \to 0^+\) could lead to a singularity.

Step 2: Substitute Variables

To analyze the behavior near \(x = 0\), perform a substitution. Let \(u = x^{1/3}\), so \(x = u^3\) and when \(x = 0\), \(u = 0\), and when \(x = 8\), \(u = 2\). The derivative \(dx = 3u^2 \, du\), which gives the integral \(\int_{0}^{2} \frac{1}{u} \ln(\frac{1}{u}) \cdot 3u^2 \, du\). Simplify to \(3 \int_{0}^{2} u\ln(\frac{1}{u}) \, du\).

Step 3: Simplify the Integral

The integral simplifies to \(3\int_{0}^{2} u(-\ln(u)) \, du = -3\int_{0}^{2} u\ln(u) \, du\). We need to evaluate \(-3\int_{0}^{2} u\ln(u) \, du\).

Step 4: Integrate by Parts

Use integration by parts, where \( \int u\ln(u) \, du\) can be set with \(v = \ln(u)\), \(dw = u \, du\). Then \(dv = \frac{1}{u} \, du\) and \(w = \frac{u^2}{2}\). The integral becomes \(-3 \left( \left. \frac{u^2}{2} \ln(u) \right|_0^2 - \int \frac{u^2}{2} \cdot \frac{1}{u} \, du \right)\).

Step 5: Evaluate the Boundary Terms

Compute the boundary terms: \(\left. \frac{u^2}{2} \ln(u) \right|_0^2\). At \(u = 2\), it is \(\frac{4}{2} \ln(2) = 2\ln(2)\). At \(u = 0\), since \(u\ln(u) \to 0\) as \(u \to 0^+\), it approaches 0. Thus, it's \(2\ln(2)\).

Step 6: Solve Remaining Integral

Compute \(\int u \, du = \frac{u^2}{2}\), so \(-3\left( 2\ln(2) - \int \frac{u}{2} \, du \right) = -3[2\ln(2) - \frac{u^2}{4}]\). Evaluate from 0 to 2, yielding \(-3[2\ln(2) - 1]\).

Step 7: Convergence and Result

The integral converges to \(-3 \times (2\ln(2) - 1) = -6\ln(2) + 3\). Thus, the integral is convergent.

Key Concepts

Convergence

When you encounter an improper integral like \(\int_{0}^{8} \frac{1}{x^{1/3}} \ln \left(\frac{1}{x^{1/3}}\right) \, dx\), one of your first tasks is to determine if it *converges* or *diverges*. An improper integral is termed "convergent" if it reaches a finite value when evaluated. Otherwise, it is called "divergent".

• Always check for singularities—points where the function could potentially "blow up." In our example, as \(x \to 0^+\), we have potential singular behavior.
• If the area under the curve of your integral over the specified limits does not tend toward infinity, then it's converging.

If you confirm convergence, as is in the given example, you can then proceed to evaluate its value. Oftentimes, a comprehensive understanding of singularity behavior allows deeper insights into the nature of such integrals.

Integration by Parts

Integration by parts is a vital technique for handling integrals involving products of functions, based on the formula:\[\int u \, dv = uv - \int v \, du\]In our exercise, we use this method to evaluate the integral \(\int u \ln(u) \, du\) where:

• Set \(v = \ln(u)\) and \(dw = u \, du\).
• Then, \(dv = \frac{1}{u} \, du\) and \(w = \frac{u^2}{2}\).

By rearranging and substituting back into the formula, this allows us to break down complex integrals into more manageable parts. When you apply integration by parts, it often simplifies equations that seem complex at first glance. This strategy is key in tackling pieces of integrals that involve logarithms and algebraic expressions.

Substitution Method

The substitution method—or *change of variables*—is a fundamental tool aimed at simplifying the expression of an integral. It's about rewriting an integral in terms of a new variable, making it easier to evaluate.

• Start by selecting a substitution that simplifies the major component of your integral. In this case, the function \(u = x^{1/3}\) was chosen.
• Find the derivative to replace the \(dx\). For our example, \(dx = 3u^2 \, du\).

This substitution effectively transforms the original limits of integration and helps deal with tricky expressions or limits—especially singularities. After substituting, you will often find a simpler or standard-form integral that is more straightforward to solve.

Boundary behavior of integrals

Understanding the boundary behavior of integrals is crucial for resolving the integration process, particularly for improper integrals. When evaluating an integral from one boundary to another, you often deal with potentially undefined behavior at the endpoints.

• Carefully evaluate the limits of integration. In this example, special attention is needed as we consider the behavior at \(x = 0\), where singularities may occur.
• Computations should include checking boundary terms separately. For instance, \(\left. \frac{u^2}{2} \ln(u) \right|_0^2\) needs both endpoints evaluated.

It's vital to ensure that terms approach meaningful, finite values at all limits. Especially, you should not assume that boundary terms automatically equal zero without checking. This thorough methodology ensures the integral's result is both accurate and interpretable.