Problem 44 - Essential Calculus Early Transcendentals

Question

Evaluate the indefinite integral as an infinite series. $$\int \frac{e^{x}-1}{x} d x$$

Step-by-Step Solution

Answer

The indefinite integral is \( x + \frac{x^2}{4} + \frac{x^3}{18} + \cdots + C \).

Step 1: Understanding the Integral

The integral given is \( \int \frac{e^x - 1}{x} \, dx \). This integral is challenging to solve in terms of elementary functions, so we'll expand the function \( \frac{e^x - 1}{x} \) into an infinite series.

Step 2: Series Expansion of \( e^x \)

Recall the Taylor series expansion for \( e^x \): \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] Thus, \( e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \).

Step 3: Divide Each Term by \( x \)

Substitute \( e^x - 1 \) in the integral, giving:\[ \frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \]We now have an infinite series representation of the integrand.

Step 4: Integrate Each Term of the Series

Integrate term by term:1. \( \int 1 \, dx = x \)2. \( \int \frac{x}{2} \, dx = \frac{x^2}{4} \)3. \( \int \frac{x^2}{6} \, dx = \frac{x^3}{18} \)Continue this pattern, using the formula: \[ \int \frac{x^n}{(n+1)!} \, dx = \frac{x^{n+1}}{(n+1)!(n+1)} \]for each term in the series.

Step 5: Write the Infinite Series Solution

As a result of integrating each term, the indefinite integral can be expressed as: \[ \int \frac{e^x - 1}{x} \, dx = x + \frac{x^2}{4} + \frac{x^3}{18} + \cdots + C \]where \( C \) is the constant of integration.

Key Concepts

Indefinite Integral

An indefinite integral, often referred to as an antiderivative, represents a family of functions whose derivative is the given function. When you see an integral sign, like \( \int f(x) \, dx \), it asks you to determine a function whose derivative results in \( f(x) \). The process of finding this function is known as integration. Unlike definite integrals, which give a numerical value, indefinite integrals produce a function plus a constant of integration, \( C \). This constant arises because differentiating a constant produces zero. Therefore, any constant added to an antiderivative leads to another valid solution.Consider the example: the indefinite integral of \( \int \frac{e^x-1}{x} \, dx \). Since this integral involves a complex expression, it isn't directly solvable using simple algebraic techniques. Instead, it can be expressed in terms of infinite series, allowing for integration on each individual term. Understanding this concept helps you tackle intricate integrals that don't lend themselves easily to standard methods.

Infinite Series

Infinite series are sums of infinite sequences. When you encounter an integral like \( \int \frac{e^x-1}{x} \, dx \), finding a solution may involve representing the integrand as an infinite series. This allows you to address each part separately. In our original exercise, the series expansion was found by first considering the Taylor series for \( e^x \):

• \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)
• Subtracting 1: \( e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)
• Dividing by \( x \): \( \frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \)

This new expression as an infinite series simplifies the original function, making it more manageable for integration, as each term in the series becomes a straightforward power function. Understanding infinite series is essential in calculus as they enable solutions to complex integral problems that would otherwise be unsolvable using standard algebraic methods.

Taylor Series Expansion

The Taylor series expansion allows you to express complex functions as infinite sums of polynomial terms. It's an incredible tool for calculus, simplifying the integration or differentiation of complex functions. The Taylor series expansion of a function \( f(x) \) around a point \( a \) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\]For \( e^x \), which is a classic function frequently expanded with a Taylor series, the series is given by:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]Our exercise involved expanding \( \frac{e^x - 1}{x} \) by subtracting 1 from the usual series, generating a new function conveniently in series form. Using the Taylor series for \( e^x \) helps in simplifying the calculation and obtaining a series representation of the integral.By expanding functions using Taylor series, calculus can tackle problems involving otherwise-intractable integrals. Knowing how to derive and use such expansions is crucial for both theoretical and applied mathematics in calculus.