Section 7-8: Improper Integrals

Welcome to the fascinating world of improper integrals! In this section, we'll delve into integrals where either the interval of integration is infinite or the function being integrated has an infinite discontinuity within the interval.

What are Improper Integrals?

An improper integral is a definite integral that has one or both limits infinite or an integrand that is discontinuous at a point within the interval of integration. Understanding improper integrals allows us to evaluate areas under curves that extend to infinity or have vertical asymptotes.

Types of Improper Integrals

We can classify improper integrals into two main types:

  • Type 1: Infinite Intervals: These integrals have one or both limits of integration extending to infinity. For example, $$\int_{a}^{\infty} f(x) \, dx$$ or $$\int_{-\infty}^{b} f(x) \, dx$$.
  • Type 2: Discontinuous Integrands: These integrals have a discontinuity (usually a vertical asymptote) within the interval of integration. For example, if $f(x)$ has a discontinuity at $c$ where $a < c < b$, then we need to consider $$\int_{a}^{b} f(x) \, dx$$.

Type 1: Improper Integrals - Infinite Intervals

To evaluate an improper integral with an infinite interval, we use limits. Here's the general approach:

  1. Replace the infinite limit with a variable (e.g., $t$).
  2. Evaluate the definite integral with the variable limit.
  3. Take the limit as the variable approaches infinity.

Formally:

  • If $\int_{a}^{t} f(x) \, dx$ exists for every number $t \ge a$, then $$\int_{a}^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_{a}^{t} f(x) \, dx$$ provided the limit exists (as a finite number).
  • If $\int_{t}^{b} f(x) \, dx$ exists for every number $t \le b$, then $$\int_{-\infty}^{b} f(x) \, dx = \lim_{t \to -\infty} \int_{t}^{b} f(x) \, dx$$ provided the limit exists (as a finite number).

If the limit exists, we say the improper integral converges. If the limit does not exist (or is infinite), we say the integral diverges.

Example

Let's determine if the following integral converges or diverges: $$\int_{1}^{\infty} \frac{1}{x} \, dx$$

Solution: $$\int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x} \, dx = \lim_{t \to \infty} [\ln|x|]_{1}^{t} = \lim_{t \to \infty} (\ln(t) - \ln(1)) = \lim_{t \to \infty} \ln(t) = \infty$$

Since the limit is infinite, the integral $$\int_{1}^{\infty} \frac{1}{x} \, dx$$ diverges.

Type 2: Improper Integrals - Discontinuous Integrands

When the integrand has a discontinuity at a point within the interval of integration, we again use limits to evaluate the integral. Suppose $f(x)$ is discontinuous at $c$, where $a < c < b$. Then, we split the integral into two parts and evaluate each part using limits:

$$\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx$$

We evaluate each integral on the right as follows:

$$\int_{a}^{c} f(x) \, dx = \lim_{t \to c^{-}} \int_{a}^{t} f(x) \, dx$$ $$\int_{c}^{b} f(x) \, dx = \lim_{t \to c^{+}} \int_{t}^{b} f(x) \, dx$$

If both limits exist, then the improper integral converges, and its value is the sum of the two limits. If either limit does not exist, then the improper integral diverges.

Comparison Test for Improper Integrals

Sometimes, it's difficult or impossible to find the exact value of an improper integral. In these cases, we can use the Comparison Test to determine whether an integral converges or diverges.

Suppose that $f(x)$ and $g(x)$ are continuous functions with $f(x) \ge g(x) \ge 0$ for $x \ge a$.

  • If $\int_{a}^{\infty} f(x) \, dx$ is convergent, then $\int_{a}^{\infty} g(x) \, dx$ is convergent.
  • If $\int_{a}^{\infty} g(x) \, dx$ is divergent, then $\int_{a}^{\infty} f(x) \, dx$ is divergent.

This test is particularly useful when dealing with integrals that do not have elementary antiderivatives.

Keep practicing, and you'll master the art of evaluating improper integrals! Good luck!