Understanding Improper Integrals with Infinite Limits

Introduction

Key Concepts

Definition and Overview

In calculus, an improper integral refers to an integral where either the interval of integration is infinite or the integrand approaches infinity within the interval. Specifically, when dealing with infinite limits, the integral takes the form:

$$ \int_{a}^{\infty} f(x) \, dx \quad \text{or} \quad \int_{-\infty}^{b} f(x) \, dx \quad \text{or} \quad \int_{-\infty}^{\infty} f(x) \, dx $$

These integrals are evaluated as limits. For example:

$$ \int_{a}^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_{a}^{t} f(x) \, dx $$

Conditions for Convergence

Not all improper integrals converge; some diverge. Convergence occurs when the limit defining the integral exists and is finite. The primary conditions affecting convergence include:

• Behavior of the Integrand: How \( f(x) \) behaves as \( x \) approaches infinity or negative infinity.
• Speed of Decay: Whether \( f(x) \) approaches zero rapidly enough to make the area under the curve finite.

Evaluating Improper Integrals with Infinite Limits

To evaluate an improper integral with an infinite limit, follow these steps:

1. Set Up the Limit: Replace the infinite limit with a variable approaching infinity.
2. Integrate Normally: Perform the integration as you would for a definite integral.
3. Apply the Limit: Evaluate the integral by taking the limit as the variable approaches infinity.

For example, to evaluate \( \int_{1}^{\infty} \frac{1}{x^2} \, dx \):

$$ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_{1}^{t} = \lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 1 $$

The integral converges to 1.

Types of Improper Integrals with Infinite Limits

Improper integrals with infinite limits can be categorized based on the nature of their limits:

• Single Infinite Limit: Integrals where one of the limits is infinite, such as \( \int_{a}^{\infty} f(x) \, dx \).
• Both Limits Infinite: Integrals where both the lower and upper limits are infinite, such as \( \int_{-\infty}^{\infty} f(x) \, dx \).

Comparison with Improper Integrals Due to Discontinuities

While improper integrals with infinite limits deal with unbounded intervals, another category involves integrals where the integrand becomes unbounded within the interval of integration. For example, \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) is improper because \( f(x) \) approaches infinity as \( x \) approaches 0.

Applications of Improper Integrals with Infinite Limits

These integrals are essential in various fields, including:

• Probability Theory: Calculating probabilities for continuous distributions over an infinite range.
• Physics: Determining quantities like electric fields and potentials extending to infinity.
• Engineering: Analyzing systems with infinite time horizons or spatial domains.

Techniques to Determine Convergence

Several tests help determine whether an improper integral with infinite limits converges:

• Comparison Test: Compare \( f(x) \) with a function \( g(x) \) whose integral is known to converge or diverge.
• Limit Comparison Test: Examine the limit \( \lim_{x \to \infty} \frac{f(x)}{g(x)} \). If it exists and is positive, \( \int f(x) \, dx \) and \( \int g(x) \, dx \) share the same convergence behavior.
• Integral Test: Often used for series, but applicable to integrals to assess convergence based on the behavior of \( f(x) \).

Examples of Evaluating Improper Integrals

Example 1: Evaluate \( \int_{2}^{\infty} \frac{3}{x^4} \, dx \).

Solution:

$$ \int_{2}^{\infty} \frac{3}{x^4} \, dx = \lim_{t \to \infty} \int_{2}^{t} \frac{3}{x^4} \, dx = \lim_{t \to \infty} \left[ -\frac{1}{x^3} \right]_{2}^{t} = \lim_{t \to \infty} \left( -\frac{1}{t^3} + \frac{1}{8} \right) = \frac{1}{8} $$

Example 2: Determine if \( \int_{1}^{\infty} \frac{\ln(x)}{x^2} \, dx \) converges.

Solution:

Using the limit comparison test with \( g(x) = \frac{1}{x} \), since \( \int_{1}^{\infty} \frac{1}{x} \, dx \) diverges:

$$ \lim_{x \to \infty} \frac{\frac{\ln(x)}{x^2}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{\ln(x)}{x} = 0 $$

Since the limit is 0 and \( \int_{1}^{\infty} \frac{1}{x} \, dx \) diverges, the comparison test is inconclusive. However, using the integral test:

$$ \int_{1}^{\infty} \frac{\ln(x)}{x^2} \, dx = \lim_{t \to \infty} \left[ -\frac{\ln(x)}{x} - \frac{1}{x} \right]_{1}^{t} = \lim_{t \to \infty} \left( -\frac{\ln(t)}{t} - \frac{1}{t} + 1 \right) = 1 $$

The integral converges to 1.

Comparison Table

Aspect	Proper Integrals	Improper Integrals with Infinite Limits
Limits of Integration	Both limits are finite.	One or both limits are infinite.
Integrand Behavior	Function is finite over the interval.	Function may approach infinity at one or both ends.
Evaluation Method	Standard definite integral techniques.	Limit processes are used to evaluate the integral.
Convergence	Always results in a finite value.	May converge or diverge based on the integrand.

Summary and Key Takeaways