Overview

Let's begin by asking ourselves why $$\int_{-1}^1 \frac{1}{x^2} \ dx \neq -2.$$

The Fundamental Theorem of Calculus

Recall that the Fundamental Theorem of Calculus (FTC) says if $$f$$ is a continuous function on the closed interval $$[a,b]$$, then:

$$\int_a^b f(x) \, dx = \left[F(x)\right]_a^b = F(b)-F(a),$$

where $$F$$ is any antiderivative of $$f$$.

This explains why $$\int_{-1}^1 \frac{1}{x^2} \ dx \neq -2$$. We see that $$\frac{1}{x^2}$$ has a discontinuity at $$x=0$$ and is therefore not continuous on $$[-1,1]$$.

There are two ways to extend the Fundamental Theorem of Calculus, leading to the concept of an improper integral:

• Use an infinite interval (e.g., $$(-\infty, \infty)$$, $$[a, \infty)$$, or $$(-\infty, b]$$).
• Allow the interval $$[a,b]$$ to contain an infinite discontinuity of $$f(x)$$.

Motivating Example

This example illustrates the following:

Type I: Infinite Intervals

Integrals with infinite limits of integration are called improper integrals of Type I:

• If $$f(x)$$ is continuous on $$[a,\infty)$$, then:

  $$\int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx$$

• If $$f(x)$$ is continuous on $$(-\infty,b]$$, then:

  $$\int_{-\infty}^b f(x) \, dx = \lim_{a \to -\infty} \int_a^b f(x) \, dx.$$

• If $$f(x)$$ is continuous on $$(-\infty,\infty)$$, then:

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

Convergence and Divergence

• If the limit is finite, the improper integral converges to that value.
• If the limit is infinite (or fails to exist), the improper integral diverges.

Type II: Infinite Discontinuity

Integrals with an infinite discontinuity at a limit of integration or within the interval are called improper integrals of Type II:

• If $$f(x)$$ is continuous on $$(a,b]$$ and discontinuous at $$a$$, then:

  $$\int_a^b f(x) \, dx = \lim_{c \to a^+} \int_c^b f(x) \, dx.$$

• If $$f(x)$$ is continuous on $$[a,b)$$ and discontinuous at $$b$$, then:

  $$\int_a^b f(x) \, dx = \lim_{c \to b^-} \int_a^c f(x) \, dx.$$

• If $$f(x)$$ is discontinuous at $$c$$ (where $$a < c < b$$) and continuous on $$[a,c) \cup (c,b]$$, then:

  $$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx.$$

Convergence and Divergence

• If the limit is finite, the improper integral converges to that value.
• If the limit is infinite (or fails to exist), the improper integral diverges.

Type III: Infinite Intervals and Discontinuity

When we encounter integrals with infinite limits of integration and an infinite discontinuity, we apply concepts from both Type I and Type II integrals.

Comparison Tests for Integrals

When we cannot evaluate an improper integral directly, we compare it to a known improper integral:

Direct Comparison Test

If $$0 \leq f(x) \leq g(x)$$ on $$(a,\infty)$$:

1. If $$\int_a^\infty g(x) \, dx$$ converges, then $$\int_a^\infty f(x) \, dx$$ also converges.

2. If $$\int_a^\infty f(x) \, dx$$ diverges, then $$\int_a^\infty g(x) \, dx$$ also diverges.

Limit Comparison Test

If $$f(x)$$ and $$g(x)$$ are positive and continuous on $$[a,\infty)$$ and $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = L \ \ (\text{given } 0 < L < \infty)$$, then $$\int_a^\infty f(x) \, dx$$ and $$\int_a^\infty g(x) \, dx$$ both converge or diverge.