We aim to express the function

$\frac{\cos(x) - 1}{x}$

as a power series and then evaluate its indefinite integral as an infinite series. Let’s go step by step.

1. Expand $\cos(x)$ into its power series:

The power series for $\cos(x)$ is:

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

Subtract $1$ from both sides:

$\cos(x) - 1 = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

Divide this by $x$:

$\frac{\cos(x) - 1}{x} = -\frac{x}{2!} + \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots$

This is the power series representation of the given function:

$\frac{\cos(x) - 1}{x} = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n)!}.$

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2. Indefinite integral of $\frac{\cos(x) - 1}{x}$:

The indefinite integral is:

$\int \frac{\cos(x) - 1}{x} \, dx = \int \left( -\frac{x}{2!} + \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \right) dx.$

Integrate term by term:

$\int \frac{\cos(x) - 1}{x} \, dx = -\frac{x^2}{2 \cdot 2!} + \frac{x^4}{4 \cdot 4!} - \frac{x^6}{6 \cdot 6!} + \cdots + C.$

Simplify the coefficients:

$\int \frac{\cos(x) - 1}{x} \, dx = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n}}{2n \cdot (2n)!} + C.$

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Final Results:

1. Power series representation of the function: $\frac{\cos(x) - 1}{x} = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n)!}.$

2. Indefinite integral as a power series: $\int \frac{\cos(x) - 1}{x} \, dx = \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n}}{2n \cdot (2n)!} + C.$

Would you like more details on any step or explanation of specific terms?

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Related Questions:

1. What is the radius of convergence for these power series?
2. How does this series behave for large values of $x$?
3. Can this integral be expressed in terms of special functions?
4. How does the series compare with numerical approximations for small $x$?
5. Can the result be generalized for other trigonometric functions like $\sin(x)$ or $\tan(x)$?

Tip:

When dealing with series, always verify convergence to ensure the validity of term-by-term integration or