# Chapter 21: Improper Integrals: How to Solve Problems

> A field guide to improper integrals: split at every blow-up, compare against p-integrals, and read off convergence — without ever finishing the integral.

**Canonical:** https://calculus.kitzo.work/chapter/improper-integrals-solving  
**Level:** Advanced  
**Estimated time:** 50 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Splitting at blow-up points
- Summary of tests
- Behavior near ±∞
- Behavior near 0
- Comparison vs p-test
- Absolute convergence

**Why this matters**

An 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 limt→∞ ∫at f(x) dx. Finite limit ⇒ **converges**; otherwise **diverges**.

- **Type 2**: blow-up at an endpoint c — take limt→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 ∫ab = ∫ac + ∫cb; **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/xp converges iff p > 1; ∫01 dx/xp 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∞ xk e−x dx always converges.

- **Logarithms lose**: ln x grows slower than any xε. 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/xp near 0 diverges for p ≥ 1, converges for p < 1.

- **Trig**: sin x ~ x near 0, so 1/sin x ~ 1/x — and ∫0a dx / sin x **diverges**.

- **Exponentials**: ex − 1 ~ x, so ∫0a dx / (ex − 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 = limt→2⁻ [−1/(x − 2)]0t = ∞. The first piece alone diverges, so the whole integral diverges — the second piece is never tested.

**Watch out**

Comparing 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 = limt→∞ ∫at f(x) dx · blow-up at c ⇒ split, use one-sided limits

p-test at ∞: ∫1∞ dx/xp converges ⇔ p > 1 · p-test at 0: ∫01 dx/xp 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 · ex outruns any polynomial · ln x trails any xε

## 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.

## Practice exercises

1. The integral ∫ 1 ∞ dx / x³ is:

- A. convergent, by the p-test with p = 3 > 1
- B. divergent, since the area is unbounded
- C. convergent, but only conditionally
- D. impossible to classify without computing it

2. Compute ∫ 1 ∞ dx / x².

3. Classify ∫ 1 ∞ dx / x — converge or diverge?

4. ∫ 0 1 dx / x p converges exactly when:

- A. p < 1
- B. p > 1
- C. p ≥ 1
- D. for every real p

5. Compute ∫ 0 1 dx / √x.

6. Why does ∫ 1 ∞ sin x / x² dx converge?

- A. |sin x| ≤ 1, so the integrand is ≤ 1/x² in absolute value — direct comparison wins
- B. because sin x is always positive on [1, ∞)
- C. because 1/x² is smaller than sin x for large x
- D. no comparison test can settle it

7. For ∫ 1 ∞ (x + 1)/(x³ + 5) dx, limit comparison uses g(x) = 1/x p . Give the value of p.

8. The integral ∫ −1 1 dx / x is:

- A. divergent — split at x = 0 and each piece diverges
- B. equal to 0 by odd symmetry
- C. convergent, because ln 1 = 0
- D. divergent, but only at the right endpoint

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