Problem 27 - Essential Calculus Early Transcendentals

Question

Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int x^{2} \ln (1+x) d x $$

Step-by-Step Solution

Answer

The indefinite integral is expressed as a power series: \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n+3}}{n(n+3)} + C\). The radius of convergence is 1.

Step 1: Expand the logarithm into a power series

The first step is to express \( \ln(1+x) \) as a power series. This is done using the expansion:\[\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}\]This expansion is valid for \(-1 < x \leq 1\).

Step 2: Multiply with \(x^2\)

Substitute the power series expansion of \(\ln(1+x)\) into the integral and multiply each term by \(x^2\):\[x^2 \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n+2}}{n}\]This transforms the problem into integrating a power series term by term.

Step 3: Integrate the power series term by term

Integrate each term of the series:\[\int x^2 \ln(1+x) \, dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \int x^{n+2} \, dx\]This results in:\[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cdot \frac{x^{n+3}}{n+3} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n+3}}{n(n+3)}\]Add the constant of integration \(C\) to the result.

Step 4: Determine the radius of convergence

Since the original expansion of \(\ln(1+x)\) is valid for \(-1 < x \leq 1\), we need to assess the convergence of the resulting power series. The radius of convergence does not change after integrating or multiplying by \(x^2\), so the radius of convergence remains \(R=1\).

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when working with power series. It determines the interval within which a power series converges and provides a bounded area for valid calculations.

The original series expansion for the natural logarithm \(\ln(1+x)\) has a radius of convergence of 1. This means the series will sum to a finite value whenever \(-1 < x \leq 1\). After integrating the power series or multiplying by \(x^2\), the radius of convergence doesn’t change.

Practical implications of this are:

• The series remains valid and usable only within this interval of \(-1 < x \leq 1\).
• Operations like integration or multiplication by polynomial factors do not affect the radius.

Always check the radius before performing operations to ensure accurate results.

Series Expansion

Series expansion is a method to express functions as an infinite sum of terms. It's especially useful for functions that are difficult to handle directly, like \( \ln(1+x) \).

In this problem, the logarithmic function is expanded using its power series:

• \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \]
• This representation is handy for evaluating integrals or derivatives since it converts them into simpler operations on polynomials.

Once we have the series, each term is an individual polynomial, making techniques like integration and differentiation more straightforward to apply.

This approach is a powerful tool across calculus and helps simplify complex expressions and their manipulations.

Indefinite Integral

The indefinite integral represents the collection of all antiderivatives of a function. It is the reverse operation of differentiation, where instead of finding the rate of change, you find the original function.

In this exercise, we find the indefinite integral of \( x^2 \ln(1+x) \), represented as an infinite series. The process works as follows:

• After expanding \( \ln(1+x) \) into a series, multiply each term by \( x^2 \).
• Integrate each term separately, which simplifies the process.
• The integration is done term by term, following the standard rule \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} \].
• Don’t forget to add a constant \( C \) at the end, as it represents any constant value that helps meet initial conditions when solving real problems.

This method allows you to tackle more complex functions by breaking them down into manageable parts.