Improper integrals Knowledge-Based Study

Study notes

1. Understanding Improper Integrals

   The Fundamental Theorem of Calculus requires an integrand to be continuous on a closed, bounded interval $[a, b]$. When one or both limits of integration become infinite, or when the integrand has an infinite discontinuity on the interval of integration, the integral is called an improper integral.

   Improper integrals are classified into two primary categories:

   • Type 1: Integrals over infinite intervals.
   • Type 2: Integrals with discontinuous (unbounded) integrands.

   An improper integral is said to converge if the corresponding limit exists as a finite real number. If the limit fails to exist or approaches $\infty$ or $-\infty$, the improper integral is said to diverge.

   Integrals with Infinite Limits

   Integrals over unbounded intervals are evaluated by replacing the infinite limit with a finite variable and computing a limit.

   Definitions

   • If $f$ is continuous on $[a, \infty)$, then: $$\int_{a}^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_{a}^{t} f(x)\,dx$$
   • If $f$ is continuous on $(-\infty, b]$, then: $$\int_{-\infty}^{b} f(x)\,dx = \lim_{t \to -\infty} \int_{t}^{b} f(x)\,dx$$
   • If $f$ is continuous on $(-\infty, \infty)$, we split the integral at any real number $c$ (frequently $c = 0$): $$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_{c}^{\infty} f(x)\,dx$$ Both integrals on the right must converge independently for the original integral to converge.

   Example

   Evaluate $\int_{1}^{\infty} \frac{1}{x^2}\,dx$:

   $$\lim_{t \to \infty} \int_{1}^{t} x^{-2}\,dx = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_{1}^{t} = \lim_{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 0 + 1 = 1$$

   Because the limit is finite, the integral converges to 1.

   The p-Integral Test Over Infinite Intervals

   A fundamental reference family for convergence over infinite intervals is the class of $p$-integrals of the form:

   $$\int_{1}^{\infty} \frac{1}{x^p}\,dx$$

   Convergence Criteria

   • If $p > 1$, the integral converges to $\frac{1}{p - 1}$.
   • If $p \le 1$, the integral diverges.

   Demonstration for $p = 1$

   When $p = 1$, the integrand is $\frac{1}{x}$:

   $$\int_{1}^{\infty} \frac{1}{x}\,dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x}\,dx = \lim_{t \to \infty} \ln(t) = \infty$$

   Thus, $\int_{1}^{\infty} \frac{1}{x}\,dx$ diverges. Knowing this threshold helps evaluate more complex integrands by comparison.

   Integrals with Infinite Discontinuities

   When the integrand $f(x)$ has a vertical asymptote at one of the endpoints or at an interior point of $[a, b]$, the integral must be set up with a one-sided limit.

   Definitions

   • If $f$ is continuous on $[a, b)$ and discontinuous at $b$: $$\int_{a}^{b} f(x)\,dx = \lim_{t \to b^{-}} \int_{a}^{t} f(x)\,dx$$
   • If $f$ is continuous on $(a, b]$ and discontinuous at $a$: $$\int_{a}^{b} f(x)\,dx = \lim_{t \to a^{+}} \int_{t}^{b} f(x)\,dx$$
   • If $f$ has an infinite discontinuity at an interior point $c \in (a, b)$: $$\int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx$$ Both individual limits must exist for the overall integral to converge.

   Example with an Endpoint Discontinuity

   Evaluate $\int_{0}^{1} \frac{1}{\sqrt{x}}\,dx$:

   $$\lim_{t \to 0^{+}} \int_{t}^{1} x^{-1/2}\,dx = \lim_{t \to 0^{+}} \left[ 2\sqrt{x} \right]_{t}^{1} = \lim_{t \to 0^{+}} (2 - 2\sqrt{t}) = 2$$

   The integral converges to 2. Notice that for $\int_{0}^{1} \frac{1}{x^p}\,dx$, the convergence condition is reversed: it converges when $p < 1$ and diverges when $p \ge 1$.

   Direct and Limit Comparison Tests

   When an antiderivative cannot be expressed in elementary functions, comparison tests allow us to determine convergence or divergence by referencing a known benchmark integral.

   Direct Comparison Test

   Let $f$ and $g$ be continuous functions with $0 \le f(x) \le g(x)$ for all $x \ge a$:

   • If $\int_{a}^{\infty} g(x)\,dx$ converges, then $\int_{a}^{\infty} f(x)\,dx$ also converges.
   • If $\int_{a}^{\infty} f(x)\,dx$ diverges, then $\int_{a}^{\infty} g(x)\,dx$ also diverges.

   Limit Comparison Test

   If $f(x) > 0$ and $g(x) > 0$ on $[a, \infty)$, and $$L = \lim_{x \to \infty} \frac{f(x)}{g(x)}$$ where $0 < L < \infty$, then both $\int_{a}^{\infty} f(x)\,dx$ and $\int_{a}^{\infty} g(x)\,dx$ either both converge or both diverge.