Improper Integrals

An improper integral is a definite integral where either the interval of integration is unbounded, or the integrand has an infinite discontinuity within (or at the boundary of) the interval.

Overview

A proper definite integral $\int_a^b f(x),dx$ requires two conditions:

  1. The interval $[a, b]$ is finite.
  2. The integrand $f(x)$ is continuous on $[a, b]$.

An improper integral arises when at least one of these conditions fails. Such integrals are evaluated by replacing the problematic point with a variable and taking a limit.

Type 1: Infinite Limits of Integration

Occurs when one or both limits of integration are $\pm\infty$.

Definition

If $\int_a^t f(x),dx$ exists for every $t \ge a$:

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

If $\int_t^b f(x),dx$ exists for every $t \le b$:

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

For both limits infinite (choose any $c \in \mathbb{R}$):

$$\int_{-\infty}^\infty f(x),dx = \int_{-\infty}^c f(x),dx + \int_c^\infty f(x),dx$$

The integral converges if the limit exists as a finite number; otherwise it diverges.

Type 1 p-Test

For $a > 0$:

$$\int_a^\infty \frac{1}{x^p},dx \quad \text{converges} \iff p > 1$$

$$\int_a^\infty \frac{1}{x^p},dx \quad \text{diverges if} \quad p \le 1$$

Key Examples (Type 1)

Example — Convergent ($p = 2$): $$\int_1^\infty \frac{1}{x^2},dx = \lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^t = 1$$

Example — Divergent ($p = 1$): $$\int_1^\infty \frac{1}{x},dx = \lim_{t \to \infty} \bigl[\ln x\bigr]_1^t = \infty$$

Example — Exponential Decay: $$\int_0^\infty e^{-x},dx = \lim_{t \to \infty} \bigl[-e^{-x}\bigr]_0^t = 1$$

Type 2: Discontinuous Integrand

Occurs when $f(x)$ has a vertical asymptote (infinite discontinuity) at a point in the integration interval.

Definition

Discontinuity at $a$ (lower limit): $$\int_a^b f(x),dx = \lim_{t \to a^+} \int_t^b f(x),dx$$

Discontinuity at $b$ (upper limit): $$\int_a^b f(x),dx = \lim_{t \to b^-} \int_a^t f(x),dx$$

Discontinuity at $c \in (a, b)$ (interior): $$\int_a^b f(x),dx = \int_a^c f(x),dx + \int_c^b f(x),dx$$

The integral converges only if both one-sided improper integrals converge independently.

Type 2 p-Test

For the integral $\int_0^1 \frac{1}{x^p},dx$:

$$\text{Converges} \iff p < 1$$

$$\text{Diverges if} \quad p \ge 1$$

Key Examples (Type 2)

Example — Convergent ($p = \frac{1}{2} < 1$): $$\int_0^1 \frac{1}{\sqrt{x}},dx = \lim_{t \to 0^+} \bigl[2\sqrt{x}\bigr]_t^1 = 2$$

Example — Divergent ($p = 1$): $$\int_0^1 \frac{1}{x},dx = \lim_{t \to 0^+} \bigl[\ln x\bigr]_t^1 = \infty$$

Example — Discontinuity in Interior: $$\int_{-1}^1 \frac{1}{x^2},dx = \int_{-1}^0 \frac{1}{x^2},dx + \int_0^1 \frac{1}{x^2},dx$$

Both one-sided integrals diverge (each behaves like $\lim_{t \to 0^+} [-\frac{1}{x}]_t^1 = \infty$), so the original integral diverges.

The Comparison Test

Used when direct evaluation of an improper integral is difficult. The idea is to compare the integrand with a simpler function whose convergence is already known.

Theorem Statement

[!theorem] Comparison Test Let $f$ and $g$ be continuous on $[a, \infty)$ with $0 \le g(x) \le f(x)$ for all $x \ge a$.

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

Logical Structure

If... and... Then...
$g(x) \le f(x)$ $\int f$ converges $\int g$ converges
$g(x) \le f(x)$ $\int g$ diverges $\int f$ diverges

[!warning] Important If $\int f$ diverges, the test gives no information about $\int g$. If $\int g$ converges, the test gives no information about $\int f$.

Common Comparison Functions

Function Behavior at $\infty$
$\displaystyle\frac{1}{x^p}$ Converges iff $p > 1$
$e^{-kx}$ ($k > 0$) Always converges
$\displaystyle\frac{1}{x\ln x}$ Diverges (compare with $1/x$)

Worked Example — Comparison Test

Determine convergence of $\displaystyle\int_1^\infty \frac{\sin^2 x}{x^2},dx$.

Solution:

For all $x \ge 1$, $0 \le \sin^2 x \le 1$, so:

$$0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2}$$

From the p-test, $\displaystyle\int_1^\infty \frac{1}{x^2},dx$ converges ($p = 2 > 1$). By the comparison test:

$$\int_1^\infty \frac{\sin^2 x}{x^2},dx \quad \text{converges}$$

Summary Table

Criterion Type 1 $\int_1^\infty \frac{1}{x^p},dx$ Type 2 $\int_0^1 \frac{1}{x^p},dx$
Converges when $p > 1$ $p < 1$
Diverges when $p \le 1$ $p \ge 1$
Convergent example $\frac{1}{x^2}$ $\frac{1}{\sqrt{x}}$
Divergent example $\frac{1}{x}$ $\frac{1}{x}$

Strategy for Testing Improper Integrals

  1. Identify the type: infinite limit (Type 1), discontinuity (Type 2), or both.
  2. Rewrite as a limit of a proper integral.
  3. Evaluate the proper integral, then take the limit.
  4. If direct evaluation is impossible, apply the comparison test using a known function ($1/x^p$, $e^{-kx}$, etc.).
  5. Declare convergence or divergence based on the limit result.

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