Improper integrals push the boundaries of integration, tackling infinite limits and unbounded functions. They're crucial for understanding advanced calculus concepts and solving real-world problems involving unbounded domains or asymptotic behavior.

Evaluating improper integrals requires clever limit techniques and careful analysis of convergence. Mastering these skills opens doors to more complex mathematical modeling and deepens your understanding of infinite processes in calculus.

Improper Integrals

Integrals with infinite limits

• Improper integrals have infinite interval of integration ($\int_a^{\infty} f(x) dx$, $\int_{-\infty}^b f(x) dx$, or $\int_{-\infty}^{\infty} f(x) dx$) or unbounded integrand with vertical asymptote within the interval of integration ($\int_a^b f(x) dx$ where $f(x)$ is undefined at one or more points in $[a, b]$)
• Evaluate improper integrals with infinite limits by converting to a limit of a definite integral ($\int_a^{\infty} f(x) dx = \lim_{t \to \infty} \int_a^t f(x) dx$, $\int_{-\infty}^b f(x) dx = \lim_{t \to -\infty} \int_t^b f(x) dx$, $\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^{\infty} f(x) dx$ for any real number $c$) and evaluating the limit; if it exists, the improper integral converges, otherwise it diverges
• Evaluate improper integrals with unbounded integrands by splitting the integral at the point(s) where the integrand is undefined ($\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$ where $c$ is the point of discontinuity), evaluating each integral separately as a limit approaching the point of discontinuity ($\int_a^b f(x) dx = \lim_{t \to c^-} \int_a^t f(x) dx + \lim_{t \to c^+} \int_t^b f(x) dx$), and determining convergence if both limits exist

Types of improper integrals

• Type 1: Infinite interval of integration
  • Examples: $\int_1^{\infty} \frac{1}{x^2} dx$, $\int_{-\infty}^0 e^x dx$
  • Evaluate by converting to a limit and evaluating ($\int_1^{\infty} \frac{1}{x^2} dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2} dx = \lim_{t \to \infty} [-\frac{1}{x}]1^t = \lim{t \to \infty} (-\frac{1}{t} + 1) = 1$, $\int_{-\infty}^0 e^x dx = \lim_{t \to -\infty} \int_t^0 e^x dx = \lim_{t \to -\infty} [e^x]t^0 = \lim{t \to -\infty} (1 - e^t) = 1$)
• Type 2: Unbounded integrand with vertical asymptote
  • Examples: $\int_0^1 \frac{1}{\sqrt{x}} dx$, $\int_{-1}^1 \frac{1}{x^2} dx$
  • Evaluate by splitting the integral and evaluating limits ($\int_0^1 \frac{1}{\sqrt{x}} dx = \lim_{t \to 0^+} \int_t^1 \frac{1}{\sqrt{x}} dx = \lim_{t \to 0^+} [2\sqrt{x}]t^1 = \lim{t \to 0^+} (2 - 2\sqrt{t}) = 2$, $\int_{-1}^1 \frac{1}{x^2} dx = \lim_{t \to 0^-} \int_{-1}^t \frac{1}{x^2} dx + \lim_{t \to 0^+} \int_t^1 \frac{1}{x^2} dx = \lim_{t \to 0^-} [-\frac{1}{x}]{-1}^t + \lim{t \to 0^+} [-\frac{1}{x}]t^1 = \lim{t \to 0^-} (-\frac{1}{t} + 1) + \lim_{t \to 0^+} (-1 + \frac{1}{t}) = \infty$)

Convergence and Divergence

• Convergence occurs when the limit of integration exists and is finite
• Divergence happens when the limit does not exist or approaches infinity
• Vertical asymptotes often lead to divergence due to unbounded behavior
• Discontinuities in the integrand may affect convergence and require careful analysis
• The limit of integration is crucial in determining whether an improper integral converges or diverges

Comparison theorem for convergence

• Comparison Theorem: If $0 \leq f(x) \leq g(x)$ for all $x \geq a$, then if $\int_a^{\infty} g(x) dx$ converges, $\int_a^{\infty} f(x) dx$ converges, and if $\int_a^{\infty} f(x) dx$ diverges, $\int_a^{\infty} g(x) dx$ diverges
• Apply the Comparison Theorem:
  1. Find a suitable function $g(x)$ to compare with $f(x)$ that is simpler to integrate and has known convergence properties, ensuring $0 \leq f(x) \leq g(x)$ for all $x \geq a$ (or $x \leq b$ for $\int_{-\infty}^b f(x) dx$)
  2. Determine the convergence of $\int_a^{\infty} g(x) dx$
  3. Apply the Comparison Theorem to conclude the convergence of $\int_a^{\infty} f(x) dx$
• Example: Determine the convergence of $\int_1^{\infty} \frac{1}{x^2 + 1} dx$
  • Choose $g(x) = \frac{1}{x^2}$ since $\frac{1}{x^2 + 1} \leq \frac{1}{x^2}$ for all $x \geq 1$
  • $\int_1^{\infty} \frac{1}{x^2} dx$ converges (p-series with $p=2 > 1$)
  • By the Comparison Theorem, $\int_1^{\infty} \frac{1}{x^2 + 1} dx$ converges