2.4 Improper Integrals and Convergence Criteria

2.4 Improper Integrals and Convergence Criteria

Definite Riemann integration requires a bounded integrand on a closed, bounded interval $[a, b]$. When either the domain of integration becomes infinite (Type I) or the integrand becomes unbounded at one or more points (Type II), the integral is classified as improper and must be defined as a limit of proper Riemann integrals. On the GRE Mathematics Subject Test, questions frequently test whether an improper integral converges or diverges, the values of parameters that guarantee convergence, and evaluations utilizing special functions such as the Euler Gamma function.

Classification of Improper Integrals

Type I: Infinite Intervals

1. Upper limit infinite: $\int_a^\infty f(x),dx = \lim_{R \to \infty} \int_a^R f(x),dx$.
2. Lower limit infinite: $\int_{-\infty}^b f(x),dx = \lim_{L \to -\infty} \int_L^b f(x),dx$.
3. Both limits infinite: For any real number $c$: $\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^\infty f(x)\,dx = \lim_{L \to -\infty} \int_L^c f(x)\,dx + \lim_{R \to \infty} \int_c^R f(x)\,dx$ The integral converges if and only if both limits exist independently. If either limit diverges, the entire integral diverges.

Type II: Discontinuous or Unbounded Integrands

1. Singularity at upper limit: If $f$ is continuous on $[a, b)$ and unbounded as $x \to b^-$: $\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_a^{b - \varepsilon} f(x)\,dx$
2. Singularity at lower limit: If $f$ is continuous on $(a, b]$ and unbounded as $x \to a^+$: $\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_{a + \varepsilon}^b f(x)\,dx$
3. Interior singularity: If $f$ is unbounded at an interior point $c \in (a, b)$: $\int_a^b f(x)\,dx = \lim_{\varepsilon_1 \to 0^+} \int_a^{c - \varepsilon_1} f(x)\,dx + \lim_{\varepsilon_2 \to 0^+} \int_{c + \varepsilon_2}^b f(x)\,dx$ Both limits must converge independently for the improper integral to converge.

The Canonical $p$-Integral Benchmarks

The behavior of power functions forms the basis of all integral comparison tests:

Notice the inversion: near infinity, larger powers ($p > 1$) accelerate decay to zero and ensure convergence; near zero, smaller powers ($p < 1$) tame the vertical asymptote and ensure convergence.

Convergence Tests for Non-Negative Functions

When antiderivatives cannot be expressed in elementary terms, comparison tests determine convergence.

Direct Comparison Test (DCT)

Let $0 \le f(x) \le g(x)$ for all $x \ge a$:

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

Limit Comparison Test (LCT)

Let $f(x) > 0$ and $g(x) > 0$ on $[a, \infty)$, with $L = \lim_{x \to \infty} \frac{f(x)}{g(x)}$:

1. If $0 < L < \infty$, then $\int_a^\infty f(x),dx$ and $\int_a^\infty g(x),dx$ either both converge or both diverge.
2. If $L = 0$ and $\int_a^\infty g(x),dx$ converges, then $\int_a^\infty f(x),dx$ converges.
3. If $L = \infty$ and $\int_a^\infty g(x),dx$ diverges, then $\int_a^\infty f(x),dx$ diverges.

Absolute and Conditional Convergence

An improper integral $\int_a^\infty f(x),dx$ converges absolutely if $\int_a^\infty |f(x)|,dx$ converges. Absolute convergence guarantees ordinary convergence: $\int_a^\infty |f(x)|\,dx < \infty \implies \int_a^\infty f(x)\,dx \text{ converges}$ If $\int_a^\infty f(x),dx$ converges but $\int_a^\infty |f(x)|,dx$ diverges, the integral converges conditionally.

• For example, $\int_1^\infty \frac{\sin x}{x},dx$ converges conditionally by Dirichlet's Test for integrals, but $\int_1^\infty \frac{|\sin x|}{x},dx$ diverges to $\infty$.

The Cauchy Principal Value (P.V.)

For an integral with dual infinite limits, the Cauchy Principal Value is defined with a single symmetric limit: $\text{P.V.} \int_{-\infty}^\infty f(x)\,dx = \lim_{R \to \infty} \int_{-R}^R f(x)\,dx$ For an interior singularity at $c \in (a, b)$: $\text{P.V.} \int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \left( \int_a^{c - \varepsilon} f(x)\,dx + \int_{c + \varepsilon}^b f(x)\,dx \right)$ Critical Distinction: If an improper integral converges in the standard Riemann sense, its Cauchy Principal Value exists and equals the improper integral. However, the converse is false. For instance, $\text{P.V.} \int_{-\infty}^\infty x,dx = \lim_{R \to \infty} \left[\frac{x^2}{2}\right]{-R}^R = 0$, but the improper integral $\int{-\infty}^\infty x,dx$ diverges because $\lim_{R \to \infty} \int_0^R x,dx = \infty$.

The Euler Gamma Function and Gaussian Integrals

The Euler Gamma function is defined for all $\text{Re}(z) > 0$ by the improper integral: $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$ Key properties tested on the GRE:

1. Functional equation: $\Gamma(z + 1) = z \Gamma(z)$.
2. Factorial correspondence: For any $n \in \mathbb{N}^+$, $\Gamma(n) = (n - 1)!$, with $\Gamma(1) = 0! = 1$.
3. Half-integer value: $\Gamma(1/2) = \sqrt{\pi}$.
4. Gaussian connection: Substituting $t = x^2$ yields $\int_0^\infty e^{-x^2},dx = \frac{1}{2} \Gamma(1/2) = \frac{\sqrt{\pi}}{2}$, which implies $\int_{-\infty}^\infty e^{-x^2},dx = \sqrt{\pi}$.

Worked Problem: Parameterized Improper Integral

Problem: Determine all values of $p$ for which $\int_0^\infty \frac{1}{x^p (1 + x)},dx$ converges. Solution: The integral is improper at both $x = 0$ (Type II) and $x = \infty$ (Type I). Split at $x = 1$: $I = \int_0^1 \frac{1}{x^p (1 + x)}\,dx + \int_1^\infty \frac{1}{x^p (1 + x)}\,dx = I_1 + I_2$

• Near $x \to 0^+$: $1 + x \approx 1$, so $\frac{1}{x^p (1 + x)} \sim \frac{1}{x^p}$. By the Type II $p$-test, $I_1$ converges if and only if $p < 1$.
• Near $x \to \infty$: $1 + x \approx x$, so $\frac{1}{x^p (1 + x)} \sim \frac{1}{x^{p+1}}$. By the Type I $p$-test, $I_2$ converges if and only if $p + 1 > 1 \iff p > 0$. Combining both constraints, the integral converges if and only if $0 < p < 1$.

Common GRE Traps & Pitfalls

• Applying FTC across interior vertical asymptotes: Calculating $\int_{-1}^2 \frac{1}{x^3},dx = \left[-\frac{1}{2x^2}\right]_{-1}^2 = -\frac{1}{8} - \left(-\frac{1}{2}\right) = \frac{3}{8}$ fails because the function blows up at $x = 0$; each one-sided limit diverges, so the integral diverges.
• Confusing $p$-rules between $0$ and $\infty$: Assuming $p > 1$ implies convergence at $x = 0$ reverses the correct criterion ($p < 1$ converges at $0$, while $p > 1$ converges at $\infty$).
• Equating Principal Value with convergence: Concluding that $\int_{-\infty}^\infty \sin x,dx = 0$ because the integrand is odd overlooks that $\lim_{R \to \infty} \int_0^R \sin x,dx = \lim_{R \to \infty} (1 - \cos R)$ does not exist.