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).
∫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.
∫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.
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.
∫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.
∫_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 0<L<∞ then ∫f and ∫g agree
Intuition: p-test ∫₁^∞ dx/xᵖ converges ⟺ p>1 · ∫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.