Problem 28 - Calculus: One and Several Variables

Question

Find a power series representation for the improper integral. $$\int_{0}^{x} \frac{1-\cos t}{t^{2}} d t$$

Step-by-Step Solution

Answer

The power series representation for the given improper integral is \( \frac{x}{2!} - \frac{x^3}{3*4!} + \frac{x^5}{5*6!} - ...\).

Step 1: Power Series Expansion for Cosine

The Taylor Series expansion for the cosine function is given by \(\cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \ldots\). Hence, \(1-\cos t\) expands to \(\frac{t^2}{2!} - \frac{t^4}{4!} + \frac{t^6}{6!}\) - ...\).

Step 2: Division by \(t^2\)

Now, divide each term by \(t^2\), to get \(\frac{t^2}{2!t^2} - \frac{t^4}{4!t^2} + \frac{t^6}{6!t^2} - ...\). This simplifies to \(\frac{1}{2!} - \frac{t^2}{4!} + \frac{t^4}{6!} - ...\).

Step 3: Integration

Take the integral from 0 to x for each term to get \( \int_{0}^{x}(\frac{1}{2!} dt - \frac{t^2}{4!} dt + \frac{t^4}{6!} dt - \ldots dt)\). By integrating term-by-term, we get \(\frac{x}{2!} - \frac{x^3}{3*4!} + \frac{x^5}{5*6!} - ...\).

Key Concepts

Taylor Series Expansion

Understanding the Taylor Series is key to mastering calculus, as it enables us to approximate complex functions with polynomials. The Taylor Series expansion represents a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. Let's take, for example, the cosine function, \( \cos(t) \), which can be expressed as a Taylor Series around the point 0:

\[ \cos(t) = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \ldots \]
Each term in this series includes a derivative of the cosine function evaluated at 0, divided by the factorial of the number of the derivative. Notice the alternating signs and increasing degree of \( t \) — this pattern continues indefinitely. This expansion is not only fascinating but also quite useful when we attempt to integrate or differentiate functions that are otherwise not straightforward to handle.

Integration of Power Series

Integrating a power series involves applying the integral term-by-term to the power series representation of a function. This is a powerful technique as it turns a potentially complex integration problem into a series of much simpler integrations.

For the given problem, we've already simplified \(1 - \cos(t)\) over \(t^2\), to a power series:

\[\frac{1}{2!} - \frac{t^2}{4!} + \frac{t^4}{6!} - \ldots \]
Following this, integrating term-by-term from 0 to \(x\) gives us:

\[\int_{0}^{x}\left(\frac{1}{2!} - \frac{t^2}{4!} + \frac{t^4}{6!} - \ldots \right)dt\]
This results in a series expression for the integral, where the integration of each term simplifies to a straightforward polynomial expression in \(x\). The beauty of this approach is that it applies the fundamental theorem of calculus to power series, allowing us to handle each term individually rather than grappling with the initial complicated integral.

Cosine Function

The cosine function, \(\cos(t)\), is one of the most fundamental elements of trigonometry and calculus. It describes the horizontal coordinate of a point on a unit circle's circumference as it moves around the circle.

In the context of our problem, the cosine function is manipulated by subtracting it from 1, and then dividing by \(t^2\), which essentially explores the behavior of the series near 0. This manipulation of the cosine function gives us a new function to integrate:

\[\frac{1-\cos(t)}{t^2}\]
Understanding the behavior and properties of \(\cos(t)\) not only enhances our grasp of trigonometry but also plays a vital role when dealing with series expansions and their integrals. Its periodic nature and symmetry make it an interesting function to study and can significantly simplify solving integrals that involve trigonometric functions, especially when combined with power series.