Improper Integrals — Infinity and Discontinuity

A standard definite integral ∫ₐᵇ f(x)dx requires a finite interval and a bounded function. When either condition fails — the interval extends to infinity, or the function blows up at some point — you have an improper integral. It might still have a finite value. Whether it does is a question about limits, not just antiderivatives.

Two Types of Improper Integrals

• Type 1 (Infinite Bounds): ∫₁^∞, ∫₋∞^b, or ∫₋∞^∞. Replace ∞ with a parameter t and take the limit as t→∞.
• Type 2 (Unbounded Integrand): the function blows up within the interval. Split at the discontinuity and take one-sided limits.

Type 1 Examples

• ∫₁^∞ e⁻ˣ dx = lim(t→∞) [−e⁻ˣ]₁ᵗ = lim(t→∞)(−e⁻ᵗ+e⁻¹) = 0+e⁻¹ = 1/e. Converges.
• ∫₁^∞ 1/x dx = lim(t→∞) [ln x]₁ᵗ = lim ln(t) → ∞. Diverges.

The p-Integral Test

• ∫₁^∞ 1/xᵖ dx converges if p>1 (equals 1/(p−1)), diverges if p≤1.
• ∫₀¹ 1/xᵖ dx converges if p<1 (equals 1/(1−p)), diverges if p≥1.
• This test is used constantly to determine convergence without computing the full integral.

Comparison Test for Convergence

• If 0 ≤ f(x) ≤ g(x) for all x≥a:
• If ∫ₐ^∞ g converges → ∫ₐ^∞ f converges.
• If ∫ₐ^∞ f diverges → ∫ₐ^∞ g diverges.
• Useful when f can't be integrated directly but can be bounded by a known function.

Why Standard Riemann Integration Fails

The Riemann integral is defined for functions that are bounded on a closed, bounded interval [a,b]. When either condition fails — when a bound is infinite, or when the function blows up somewhere in the interval — the definition breaks down. Improper integrals are the extension that handles these cases by introducing limits. The idea is elegant: approximate the improper integral by a proper one, then let the approximation approach the problematic limit.

Type 1 — Infinite Bounds

For ∫₁^∞ f(x)dx, replace ∞ with a parameter t: ∫₁^∞ f(x)dx = lim(t→∞) ∫₁ᵗ f(x)dx. The inner integral is now proper for each finite t, and we take the limit of the result. If this limit is a finite number L, the improper integral converges to L. If the limit is ±∞ or does not exist, the integral diverges.

Type 2 — Unbounded Integrands

For ∫₀¹ 1/√x dx, the function blows up at x = 0. Replace the problematic endpoint with ε: ∫₀¹ 1/√x dx = lim(ε→0⁺) ∫ε¹ x^(−1/2) dx = lim(ε→0⁺) [2√x]ε¹ = lim(ε→0⁺)(2 − 2√ε) = 2. This improper integral converges to 2, even though the integrand is unbounded at x = 0.

Four Detailed Examples

The P-Integral — The Most Important Test

The p-integral test is the single most useful tool for quickly determining convergence without computing the full integral:

• ∫₁^∞ 1/xᵖ dx converges if p > 1 (equals 1/(p−1)), diverges if p ≤ 1.
• ∫₀¹ 1/xᵖ dx converges if p < 1 (equals 1/(1−p)), diverges if p ≥ 1.

Note the asymmetry: the boundary condition flips. At infinity, you need p > 1 for convergence (the function must shrink fast enough). Near zero, you need p < 1 (the singularity must be weak enough). For p = 1 (which gives 1/x), the integral diverges in both cases.

Comparison Test for Improper Integrals

If you cannot compute an improper integral directly, bound it above or below by one you know. If 0 ≤ f(x) ≤ g(x) for all x ≥ a: ∫ₐ^∞ g converges → ∫ₐ^∞ f converges (f is smaller, so its area is finite too). ∫ₐ^∞ f diverges → ∫ₐ^∞ g diverges (g is larger, so its area is also infinite).

Example: does ∫₁^∞ e^(−x²)dx converge? Note e^(−x²) ≤ e^(−x) for x ≥ 1. And ∫₁^∞ e^(−x)dx = [−e^(−x)]₁^∞ = e^(−1) < ∞. So ∫₁^∞ e^(−x²)dx converges by comparison. (The full ∫₋∞^∞ e^(−x²)dx = √π requires a more sophisticated proof.)

Splitting at Interior Discontinuities

If f has a vertical asymptote at c inside [a,b], you must split the integral at that point: ∫ₐᵇ f dx = ∫ₐᶜ f dx + ∫ᶜᵇ f dx, and each piece must be evaluated as a separate improper integral. Both pieces must converge for the total integral to converge. A critical error: treating an integral like ∫₋₁¹ 1/x dx as if it were proper (it has a singularity at x=0). If you naively apply the FTC: [ln|x|]₋₁¹ = 0−0 = 0, which is wrong. The correct answer is that each piece diverges, so the integral diverges (though the Cauchy principal value is 0).