# Chapter 20: Improper Integrals: Basic Concepts > When the interval is infinite or the integrand blows up, do we still get a finite answer? This chapter answers it: convergence versus divergence, and the four tests that decide. **Canonical:** https://calculus.kitzo.work/chapter/improper-integrals-concepts **Level:** Advanced **Estimated time:** 45 minutes **Exercises:** 8 (auto-marked; answers stay server-side) ## Topics - Convergence and divergence - Type 1: infinite intervals - Type 2: vertical asymptotes - The comparison test - The limit comparison test - The p-test - Absolute convergence **Why this matters** Some areas go on forever — yet their total can still be a small, finite number. Learning to tell those apart is the seed of everything from series to probability, and it teaches you to ask the calculus question that matters most: "is the total finite — yes or no?" ## 1 · Two ways an integral goes improper An integral is **improper** when it fails to look like the usual ∫ₐᵇ f(x)dx: - **Type 1:** one or both limits are infinite, like ∫₁^∞ f(x)dx. - **Type 2:** the integrand has a blow-up somewhere in the interval, like ∫₀¹ dx/x. In both cases we cheat honestly: replace the bad spot with a real number or a real limit and ask whether the ordinary integral survives. ## 2 · Type 1 — infinite intervals Define ∫₁^∞ f(x)dx = lim_{t→∞} ∫₁ᵗ f(x)dx. If that limit exists and is finite, the integral **converges**; otherwise it **diverges** (the total area runs away). **Worked example 1** ∫₁^∞ dx/x² = lim_{t→∞} (1 − 1/t) = 1. The infinite area is exactly 1 — finite. Convergence. But ∫₁^∞ dx/x = lim_{t→∞} ln t = ∞. This area genuinely runs away. Divergence. Despite 1/x² being much smaller than 1/x, size alone doesn't decide — rate does. ## 3 · Type 2 — blow-ups inside the interval If f is undefined or unbounded at an endpoint c, write ∫_c^b f(x)dx = lim_{t↓c} ∫_t^b f(x)dx. If the blow-up is *inside* (at x = c with a < c < b), split: the integral converges only when **both** ∫_a^c and ∫_c^b converge. **Worked example 2** ∫₀¹ dx/√x = lim_{a↓0} 2(1 − √a) = 2. Converges — an infinite height squeezed into 0 width can still have finite area. Compare ∫₀¹ dx/x = lim_{a↓0} −ln a = ∞. Diverges. Same shape, opposite fate. ## 4 · The comparison test — sandwiching areas If 0 ≤ f(x) ≤ g(x) on [1, ∞) and ∫₁^∞ g converges, then ∫₁^∞ f converges (even smaller area). If ∫₁^∞ f diverges, so does ∫₁^∞ g (even more area to add up). For Type 2 the same idea works near the blow-up, as long as the inequality holds near 0. **Watch out** The inequality must point the right way *and* apply on the tail you care about. Comparing against the wrong known integral (divergent vs divergent) proves nothing — you need a bird's-eye comparison with a *known* one. ## 5 · The limit comparison test — proportional tails When two functions behave like a constant multiple of each other at ∞ (or near a blow-up at 0), they converge or diverge together. Compute L = lim f(x)/g(x). If 0 < L < ∞, then ∫f and ∫g share their fate. **Worked example 3** ∫₁^∞ (x + 1)/(x³ + 5) dx. As x→∞, the integrand behaves like x/x³ = 1/x². The ratio (x+1)/(x³+5) ÷ (1/x²) = x²(x+1)/(x³+5) → 1. So it converges, exactly like ∫₁^∞ dx/x². ## 6 · The p-test — the integral world’s golden rule ∫₁^∞ dx/xᵖ converges if and only if p > 1. ∫₀¹ dx/xᵖ converges if and only if p < 1. Why: the antiderivative is x^{1−p}/(1−p), and its limit at ∞ (or 0) is finite exactly under those conditions. This one test settles most power-law integrals at once. ## 7 · Absolute convergence — smaller than small If ∫|f(x)|dx converges, ∫f(x)dx does too (any wiggle left cancels or adds, but total variation is bounded). The converse fails: ∫₁^∞ sin x/x² dx converges conditionally (it shrinks) while nothing interesting happens either way — absolute convergence is a sufficient, not necessary, luxury. **Chapter formula sheet** ∫₁^∞ f(x)dx = lim_{t→∞} ∫₁ᵗ f(x)dx ∫_c^b f(x)dx (Type 2) = lim_{t↓c} ∫_t^b f(x)dx 0 ≤ f ≤ g, ∫g conv. ⇒ ∫f conv. · ∫f div. ⇒ ∫g div. Limit comparison: if lim f/g = L with 01 · ∫₀¹ dx/xᵖ converges ⟺ p<1 ∫|f| converges ⇒ ∫f converges; area of signed pieces can cancel, area of |f| cannot ## Common mistakes - Trying to integrate across a vertical asymptote without splitting the integral. - Treating “∞ as an answer” instead of a limit — if the limit fails, the integral diverges. - Applying comparison in the wrong direction (larger integral must be the known one). - Forgetting the p-test flips at 0 versus at ∞: p > 1 converges at ∞, p < 1 converges at 0. - Using “the integrand → 0” to claim convergence — it’s necessary, not sufficient (remember 1/x). - Sign: for improper integrals with sign changes, compare |f| before comparing f. ## Practice exercises 1. ∫₁^∞ dx/x³ is: - A. convergent (value 1/2) - B. divergent - C. equal to 0 - D. equal to ∞ 2. Compute ∫₁^∞ dx/x². 3. ∫₁^∞ dx/x — converge or diverge? 4. ∫₀¹ dx/xᵖ converges exactly when: - A. p < 1 - B. p > 1 - C. p = 1 - D. always 5. Compute ∫₀¹ dx/√x. 6. Why does ∫₁^∞ sin x / x² dx converge? - A. |sin x| ≤ 1 ≤ x² tail, comparison with 1/x² - B. sin x → 0 at ∞ - C. its antiderivative is bounded - D. it is absolutely convergent by limit comparison alone 7. For ∫₁^∞ (x + 1)/(x³ + 5) dx, limit comparison with 1/x² gives a ratio limit equal to: 8. ∫₋₁¹ dx/x (with a blow-up at 0) is: - A. divergent — it must be split at 0 - B. 0 by symmetry - C. converges to 0 - D. log 2 --- Practice interactively: https://calculus.kitzo.work/chapter/improper-integrals-concepts Citation: CalculusKitzo (https://calculus.kitzo.work/). Independent study notes inspired by the standard calculus syllabus; not affiliated with Princeton University Press. See https://calculus.kitzo.work/disclaimer