Question

Integrating a Discontinuous Integrand Evaluate \(\int_{-1}^{1} \frac{1}{x^{3}} d x\). State whether the improper integral converges or diverges.

Short answer

The integral diverges as both processed integrals diverge.

Step-by-step solution

Identify the Points of Discontinuity

The integrand \( \frac{1}{x^3} \) is discontinuous at \( x = 0 \). This occurs because when \( x = 0 \), the denominator becomes zero, making the function undefined. This indicates we have an improper integral and must consider it as a limit.

Split the Integral at the Discontinuity

Due to the discontinuity at \( x = 0 \), split the original integral into two separate integrals:\[\int_{-1}^{0} \frac{1}{x^3} \, dx + \int_{0}^{1} \frac{1}{x^3} \, dx\].

Evaluate Each Improper Integral Using Limits

Evaluate each integral separately using limits to handle the discontinuity at \( x = 0 \).For \( \int_{-1}^{0} \frac{1}{x^3} \, dx \), use:\[\lim_{a \to 0^{-}} \int_{-1}^{a} \frac{1}{x^3} \, dx\].For \( \int_{0}^{1} \frac{1}{x^3} \, dx \), use:\[\lim_{b \to 0^{+}} \int_{b}^{1} \frac{1}{x^3} \, dx\].

Calculate the First Integral's Limit

Calculate \( \lim_{a \to 0^{-}} \int_{-1}^{a} \frac{1}{x^3} \, dx \): The antiderivative of \( \frac{1}{x^3} \) is \( -\frac{1}{2x^2} \). Thus,\[\lim_{a \to 0^{-}} \left[ -\frac{1}{2x^2} \right]_{-1}^{a} = \lim_{a \to 0^{-}} \left( -\frac{1}{2a^2} - \left(-\frac{1}{2(-1)^2}\right) \right) = \lim_{a \to 0^{-}} \left( -\frac{1}{2a^2} + \frac{1}{2} \right)\].As \( a \to 0^{-} \), \( -\frac{1}{2a^2} \to -\infty \). This diverges.

Calculate the Second Integral's Limit

Calculate \( \lim_{b \to 0^{+}} \int_{b}^{1} \frac{1}{x^3} \, dx \): Similarly, using the antiderivative \( -\frac{1}{2x^2} \),\[\lim_{b \to 0^{+}} \left[ -\frac{1}{2x^2} \right]_{b}^{1} = \lim_{b \to 0^{+}} \left( -\frac{1}{2(1)^2} - \left(-\frac{1}{2b^2}\right) \right) = \lim_{b \to 0^{+}} \left( -\frac{1}{2} + \frac{1}{2b^2} \right)\].As \( b \to 0^{+} \), \( \frac{1}{2b^2} \to \infty \). This diverges.

Determine Overall Convergence or Divergence

Since both limits in Steps 4 and 5 go to infinity and, therefore, each part of the split integrals diverges, the original integral \( \int_{-1}^{1} \frac{1}{x^3} \, dx \) also diverges. An integral diverges if it does not settle at a finite number, which is the case here.

Key concepts

Discontinuous Functions

In calculus, a discontinuous function is a function that is not continuous at one or more points in its domain. Continuity of a function means that small changes in the input result in small changes in the output, and there are no abrupt jumps or breaks. Discontinuous functions, however, have at least one point where the function is undefined or the limits do not match. For example, the function \( f(x) = \frac{1}{x^3} \) is discontinuous at \( x = 0 \). This is because when \( x \) equals zero, the denominator becomes zero, making the overall expression undefined.

When working with integrals involving discontinuous functions, it's crucial to identify these points of discontinuity. It often means that the integral is improper. In such cases, improper integrals must be approached using limits to evaluate them correctly.

Discontinuous points require notable attention, especially when determining the convergence or divergence of an integral. They can be seen as physical obstacles in the graph where the function suddenly breaks. Understanding the nature of these points is essential in the analysis and computation of improper integrals.

Limits and Convergence

The concept of limits is fundamental in handling improper integrals, especially when dealing with discontinuous functions. A limit in calculus describes the value that a function or sequence "approaches" as the input or index approaches some point. In the context of improper integrals, limits help us redefine the integral in a way that bypasses the point of discontinuity.

To determine whether an improper integral converges or diverges, evaluate it using limits. This means we need to split the integral at the point of discontinuity and use limits to handle the sections separately. When integrating \( \int_{-1}^{1} \frac{1}{x^3} \, dx \), we handle \( \int_{-1}^{0} \) and \( \int_{0}^{1} \) using limits approaching zero from the left and right, respectively.

Convergence occurs when the limit exists and results in a finite number. This means that as the bounds of the integral approach the point of discontinuity, the area under the curve adds up to a specific value. On the other hand, divergence occurs when the limit does not exist or tends to infinity. In such cases, the integral does not sum up to a finite area, indicating that it extends indefinitely.

Divergence

Divergence is a term used in calculus to describe a situation where an expression, such as an improper integral, does not settle at a finite value. When evaluating the improper integral \( \int_{-1}^{1} \frac{1}{x^3} \, dx \), we found that both parts of the split integral diverge.

- In the first integral from \(-1\) to \(0\), as we move towards zero, the result tends toward negative infinity - In the second integral from \(0\) to \(1\), as we move away from zero, the expression tends toward positive infinity
Because both limits approach infinity, neither part converges. Therefore, the original integral is said to diverge. Divergence indicates that the values of the function grow beyond any bounds, which prevents the integral from attaining a specific value. Understanding divergence is crucial because it helps determine the behavior of functions over intervals, especially when they contain discontinuities. By recognizing divergence, one can conclude that the integral does not have a meaningful finite area under the curve, which can have significant implications in real-world applications involving such mathematical models.