Chapter 21: Improper Integrals: How to Solve Problems

Why this mattersAn improper integral asks one bold question: does an endless area add up to a finite number, or run away to infinity? Every test in this chapter answers exactly that, and the toolkit returns for infinite series.

1 · The two ways an integral goes improper

An integral is improper when the interval never ends (Type 1) or the integrand blows up somewhere inside (Type 2). Both are tamed the same way: replace the suspicious spot with a limit, then ask if it is finite.

Type 1: ∫_a^∞ f(x) dx means lim_t→∞ ∫_a^t f(x) dx. Finite limit ⇒ converges; otherwise diverges.
Type 2: blow-up at an endpoint c — take lim_t→c of the partial integral. Blow-up in the middle — split into left and right pieces, one one-sided limit each.
Middle blow-up at c means ∫_a^b = ∫_a^c + ∫_c^b; both pieces must converge. One bad piece sinks the whole integral.

2 · The systematic checklist

Identify the type. Infinite limit? A point where the integrand is undefined or unbounded?
Split at every blow-up. Interior singularity at c? Cut the integral there and test each piece separately.
Handle negative values first. Compare on |f(x)|; if ∫|f| converges, the original converges absolutely — the right move whenever the sign changes.
Compare, don't compute. You usually need only the fate, not the value — reaching for antiderivatives early is a trap.

3 · The comparison toolbox at a glance

Direct comparison: 0 ≤ f ≤ g and ∫g converges ⇒ ∫f converges.
Limit comparison: if lim f/g = L with 0 < L < ∞, then ∫f and ∫g share the same fate. The workhorse for rational functions.
p-test: ∫₁^∞ dx/x^p converges iff p > 1; ∫₀¹ dx/x^p converges iff p < 1. Same family, opposite conditions — memorize both.
Absolute convergence: ∫|f| converges ⇒ ∫f converges. Cancellation-only survival is merely conditional.

4 · Behavior near ±∞

At infinity, growth rate is destiny.

Polynomials: the degree gap rules — a rational function converges when the bottom outgrows the top by at least 2. (x² + 3)/(x⁵ + x) behaves like 1/x³.
Exponentials beat powers: e^−x crushes any polynomial, so ∫₁^∞ x^k e^−x dx always converges.
Logarithms lose: ln x grows slower than any x^ε. The integral ∫_e^∞ dx / (x (ln x)^p) converges iff p > 1 — substitute u = ln x and it is a p-test in disguise.
Trig functions: ∫₁^∞ sin x dx diverges (partial integrals oscillate, never settle). But ∫₁^∞ |sin x| / x² dx converges by direct comparison with 1/x². Judge |sin x| by its average 2/π, not its peaks.

5 · Behavior near 0

Near zero the cast changes, but the method does not: find what the integrand looks like infinitesimally.

Polynomials: harmless — the danger is a denominator power. 1/x^p near 0 diverges for p ≥ 1, converges for p < 1.
Trig: sin x ~ x near 0, so 1/sin x ~ 1/x — and ∫₀^a dx / sin x diverges.
Exponentials: e^x − 1 ~ x, so ∫₀^a dx / (e^x − 1) diverges — a disguised 1/x.
Logs: ln x has a vertical asymptote at 0 yet is integrable: ∫₀¹ ln x dx = [x ln x − x] = −1. Log blow-ups converge; p-power blow-ups with p ≥ 1 do not.

6 · Trouble spots that are neither 0 nor ∞

Singularities hide anywhere: inside a root, inside a log, in denominator zeros, in fractional powers. The rule never changes: find every point where the integrand is undefined or unbounded, split there, and test each piece separately. A convergent tail never rescues a divergent piece on the other side.

Worked example Example 1 — Classify ∫₁^∞ (3x + 1)/(x³ + x) dx.
Step 1: Type 1 (infinite upper limit); no interior blow-up.
Step 2: For large x, top ~ 3x, bottom ~ x³, so the integrand ~ 3/x².
Step 3: Limit comparison with g(x) = 1/x²: lim x²(3x + 1)/(x³ + x) = 3.
Step 4: 0 < 3 < ∞ and ∫₁^∞ dx/x² converges (p = 2 > 1), so the integral converges.

Example 2 — Split at the blow-up: ∫₀^∞ dx/(x − 2)².
Step 1: The integrand explodes at x = 2, an interior point — split into ∫₀² + ∫₂^∞.
Step 2: ∫₀² dx/(x − 2)² = lim_t→2⁻ [−1/(x − 2)]₀^t = ∞. The first piece alone diverges, so the whole integral diverges — the second piece is never tested.

Watch outComparing against 1/x and then claiming convergence proves nothing — ∫₁^∞ dx/x diverges. Your integrand must sit below a convergent partner (or above a divergent one). And never integrate straight through an interior blow-up: the two sides may not cancel each other's infinity.

Chapter formula sheet ∫_a^∞ f(x) dx = lim_t→∞ ∫_a^t f(x) dx · blow-up at c ⇒ split, use one-sided limits
p-test at ∞: ∫₁^∞ dx/x^p converges ⇔ p > 1 · p-test at 0: ∫₀¹ dx/x^p converges ⇔ p < 1
Direct comparison: 0 ≤ f ≤ g and ∫g converges ⇒ ∫f converges
Limit comparison: lim f/g = L ∈ (0, ∞) ⇒ ∫f and ∫g share fate
Absolute convergence: ∫|f| converges ⇒ ∫f converges
Log p-test: ∫_e^∞ dx / (x (ln x)^p) converges ⇔ p > 1
∫₀¹ ln x dx = −1 · e^x outruns any polynomial · ln x trails any x^ε

Common mistakes

Claiming convergence by beating 1/x — the harmonic integral diverges; only p > 1 partners work.
Integrating straight through an interior blow-up instead of splitting there.
Judging a sign-changing integrand without testing |f| first.
Concluding ∫₁^∞ sin x dx converges because |sin x| ≤ 1 — bounded oscillation is not decay.
Swapping the two p-tests: at ∞ you need p > 1, at 0 you need p < 1.
Assuming a finite answer survives just because some pieces converge — every piece must.