6.13 Evaluating Improper Integrals

BC only. An improper integral either runs over an infinite interval or has an integrand that blows up somewhere in the interval. You evaluate it as a limit of ordinary definite integrals: if the limit is a finite number the integral converges, and otherwise it diverges.

Key terms

improper integral
converge
diverge
infinite limit of integration
unbounded integrand

Two kinds of improper integrals

This topic is tested only on the BC exam.

Type 1, an infinite limit: ∫ from a to ∞ of f(x) dx, or ∫ from −∞ to b, or from −∞ to ∞. Type 2, an unbounded integrand: f has a vertical asymptote at an endpoint or somewhere inside [a, b], like 1/√x on [0, 1].

How to evaluate them

Replace the trouble spot with a variable, do an ordinary definite integral, then take the limit.

∫ from a to ∞ of f(x) dx = lim (b → ∞) ∫ₐᵇ f(x) dx.
If f blows up at the left end a: ∫ₐᵇ f(x) dx = lim (t → a⁺) ∫ₜᵇ f(x) dx.
If f blows up at an inside point c: split it into ∫ from a to c plus ∫ from c to b. Each piece must converge on its own for the whole integral to converge.
For ∫ from −∞ to ∞, split at any number, such as 0. Both halves must converge.

Converge or diverge

If the limit is a finite number, the integral converges to that number. If the limit is infinite or doesn't exist, the integral diverges. An infinitely long region can have finite area: the region under 1/x² from 1 to ∞ has area exactly 1.

A benchmark to remember: ∫ from 1 to ∞ of 1/xᵖ dx converges when p > 1 (to 1/(p − 1)) and diverges when p ≤ 1. So 1/x² converges and 1/x diverges. On [0, 1] it's the reverse: ∫₀¹ 1/xᵖ dx converges when p < 1 and diverges when p ≥ 1. This connects directly to p-series and the integral test (10.4, 10.5).

Writing it the way scorers want

Use limit notation. Writing ∫ from 1 to ∞ of 1/x³ dx = [−1/(2x²)] from 1 to ∞ = 0 + ½ treats ∞ like a number and can lose points. Write lim (b → ∞) [−1/(2b²) + ½] = ½.

Limits you'll use often: as b → ∞, 1/b → 0, e^(−b) → 0, ln b → ∞, and arctan b → π/2. To evaluate a limit like b·e^(−b), you may need L'Hospital's Rule (4.7).

Worked examples

Try each one yourself first, then open the solution.

Example 1
Infinite interval
Evaluate ∫ from 1 to ∞ of 1/x³ dx, or show it diverges.
Show the solutionHide the solution
1. Step 1: Write it as a limit: lim (b → ∞) ∫₁ᵇ x⁻³ dx.
2. Step 2: Antiderivative: x⁻²/(−2) = −1/(2x²).
3. Step 3: ∫₁ᵇ x⁻³ dx = −1/(2b²) + 1/2.
4. Step 4: As b → ∞, −1/(2b²) → 0, so the limit is 1/2.
Answer: Converges to 1/2
Example 2
Infinite interval with substitution
Evaluate ∫ from 0 to ∞ of x e^(−x²) dx.
Show the solutionHide the solution
1. Step 1: Limit form: lim (b → ∞) ∫₀ᵇ x e^(−x²) dx.
2. Step 2: Substitute u = −x², du = −2x dx: an antiderivative is −½ e^(−x²).
3. Step 3: ∫₀ᵇ x e^(−x²) dx = −½ e^(−b²) + ½.
4. Step 4: As b → ∞, e^(−b²) → 0, so the limit is ½.
Answer: Converges to 1/2
Example 3
Trap: a hidden asymptote
Evaluate ∫₀³ 1/(x − 1)² dx.
Show the solutionHide the solution
1. Step 1: Careless work: [−1/(x − 1)] from 0 to 3 = −½ − 1 = −3/2. That can't be right: the integrand is positive, so the integral can't be negative.
2. Step 2: The problem: 1/(x − 1)² has a vertical asymptote at x = 1, inside [0, 3]. Split there.
3. Step 3: ∫₀¹ 1/(x − 1)² dx = lim (t → 1⁻) [−1/(x − 1)] from 0 to t = lim (t → 1⁻) [−1/(t − 1) − 1].
4. Step 4: As t → 1⁻, t − 1 → 0⁻, so −1/(t − 1) → +∞. This piece diverges.
5. Step 5: Since one piece diverges, the whole integral diverges.
Answer: Diverges

Common mistakes

Not noticing a vertical asymptote inside the interval and just plugging in the endpoints.
Saying an integral converges because the integrand approaches 0. 1/x → 0 as x → ∞, but ∫ from 1 to ∞ of 1/x dx diverges.
Splitting at an inside asymptote and letting the two pieces “cancel.” Each piece must converge separately.
Writing ∞ as a limit of integration and plugging it in without limit notation.

On the exam

BC multiple-choice questions often ask which improper integrals converge, or for the value of one. Know the p-benchmark for 1/xᵖ cold.
In free response, improper integrals sometimes appear in an area question (area of an unbounded region) or with the integral test. Show the limit explicitly.

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