Problem 55 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral as an infinite series. \( \int \frac {\cos x - 1}{x} dx \)

Step-by-Step Solution

Answer

Evaluate as \( \int \frac{\cos x - 1}{x} \, dx = -\frac{x^2}{4} + \frac{x^4}{96} - \cdots + C \).

Step 1: Understand the problem

We need to evaluate the indefinite integral \( \int \frac{\cos x - 1}{x} \, dx \) using its infinite series expansion.

Step 2: Expand \( \cos x \) using Taylor series

The Taylor series expansion for \( \cos x \) is \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \). Substituting this back into our integral, the expression inside becomes \( \frac{(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) - 1}{x} = \frac{- \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots}{x} \).

Step 3: Simplify the Expression

Simplify the series inside the integrand to be \( -\frac{x}{2!} + \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \). This expression will be used to find the integral term by term.

Step 4: Integrate Term-by-Term

Integrate each term in the series separately: \( \int -\frac{x}{2!} \, dx = -\frac{x^2}{2 \cdot 2} + C_1 \), \( \int \frac{x^3}{4!} \, dx = \frac{x^4}{4 \cdot 4!} + C_2 \), \( \int -\frac{x^5}{6!} \, dx = -\frac{x^6}{6 \cdot 6!} + C_3 \), and so on.

Step 5: Combine the Integrated Terms into Series

Combine the integrated results into an infinite series: \[ \int \frac{\cos x - 1}{x} \, dx = -\frac{x^2}{4} + \frac{x^4}{96} - \frac{x^6}{4320} + \cdots + C \]where \( C \) is the constant of integration.

Key Concepts

Infinite Series

An infinite series is essentially a sum of an infinite number of terms. When dealing with functions, infinite series can represent those functions in a different form, often making them easier to integrate or differentiate. In mathematics, an infinite series is represented as:\[ a_1 + a_2 + a_3 + \cdots + a_n + \cdots \]Here, each term, represented as \( a_n \), contributes to the overall value of the series.

• Infinite series are used when the function itself is difficult to work with in its original form, such as the function in our integral.


• By using the infinite series representation of a function, we can manipulate each term individually.


• Many functions, like sine, cosine, and exponential functions, are commonly represented as infinite series.

Taylor Series Expansion

The Taylor series expansion is a powerful tool that allows us to express functions as infinite sums of their derivatives at a single point. The Taylor series expansion for a function \( f(x) \) around the point \( a \) is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]For a series centered at \( a = 0 \), it is known as a Maclaurin Series, which is what we used for cosine in this problem.

• The Taylor series expansion allows complex functions to be represented as simpler polynomial series.


• For \( \cos x \), this expansion is: \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \).


• It simplifies the integration process because it reduces the function to a series of polynomial terms, each much easier to integrate than the original function.

Term-by-Term Integration

Term-by-term integration comes into play when you have a series, such as a Taylor series, and you want to integrate it. This concept is particularly useful for integrating power series because each term can be handled separately. Given a series:\[ \sum_{n=0}^{\infty} a_n x^n \]Term-by-term integration allows us to integrate each term:\[ \int \left(\sum_{n=0}^{\infty} a_n x^n \right) dx = \sum_{n=0}^{\infty} \int a_n x^n \, dx \]

• Each term in the series \( \int a_n x^n \, dx = \frac{a_n}{n+1} x^{n+1} + C_n \) is integrated individually.


• After integrating each term, the results are combined back into a series.


• This approach can simplify complex integrals by handling them as one simple polynomial at a time, providing greater control and insights into integration processes.

Constant of Integration

The constant of integration is an essential part of indefinite integrals. When we integrate a function, we're only reversing the process of differentiation. Since the derivative of a constant is zero, determining the original constant from an integral isn't possible. Hence, we add an arbitrary constant \( C \):\[ \int f(x) \, dx = F(x) + C \]

• The constant of integration ensures that the derived function accounts for all possible vertical shifts of its antiderivative.


• In series integration, after integrating term-by-term, a single constant \( C \) is added to denote this family of functions.


• In practical terms, it represents the most general solution to the indefinite integral problem, reflecting any constant value that could have existed in the original function.