📐 Concept diagram

05-08 - Improper Integrals

Phase: 5 | Subject: 05-08 Prerequisites: 05-02-the-definite-integral.md (definite integrals), 05-03-the-fundamental-theorem-of-calculus.md Next subject: 05-09-applications-of-integration.md

Learning Objectives

By the end of this subject, you will be able to:

1. Recognise improper integrals with infinite limits
2. Recognise improper integrals with discontinuous integrands
3. Evaluate improper integrals as limits
4. Apply the comparison test for convergence

Core Content

What is an Improper Integral?

An improper integral has either: 1. Infinite limit of integration (∫[a to ∞] or ∫[-∞ to b] or ∫[-∞ to ∞]) 2. Discontinuous integrand at an endpoint or inside the interval

Type 1: Infinite Limits

$∫[a to ∞] f(x)dx = $lim(t→∞)$ ∫[a to t] f(x)dx
∫[-∞ to b] f(x)dx = $lim(t→-∞)$ ∫[t to b] f(x)dx
∫[-∞ to ∞] f(x)dx = ∫[-∞ to c] + ∫[c to ∞]
$

Convergent if the limit exists (finite). Divergent if the limit is ±∞ or doesn't exist.

Example: ∫[1 to ∞] (1/x²)dx = $lim(t→∞)$ [-1/x][1 to t] = $lim(t→∞)$ (-1/t + 1) = 1

Convergent (to 1).

Example: ∫[1 to ∞] (1/x)dx = $lim(t→∞)$ [ln(x)][1 to t] = $lim(t→∞)$ ln(t) = ∞

Divergent.

Type 2: Discontinuous Integrand

If f has a discontinuity at c where a < c < b:

$∫[a to b] f(x)dx = ∫[a to c] f(x)dx + ∫[c to b] f(x)dx
$

Each part is evaluated as a limit.

Example: ∫[0 to 2] (1/√x)dx f(x) = 1/√x has discontinuity at x = 0. = $lim(t→0⁺)$ ∫[t to 2] x^(-1/2)dx = $lim(t→0⁺)$ [2√x][t to 2] = $lim(t→0⁺)$ (2√2 - 2√t) = 2√2

Convergent.

Comparison Test

If 0 ≤ f(x) ≤ g(x) for x ≥ a: - If ∫[a to ∞] g(x)dx converges, then ∫[a to ∞] f(x)dx converges - If ∫[a to ∞] f(x)dx diverges, then ∫[a to ∞] g(x)dx diverges

Example: Show ∫[1 to ∞] (1/(x² + 1))dx converges. 0 < 1/(x² + 1) < 1/x² for x ≥ 1. Since ∫[1 to ∞] (1/x²)dx converges (to 1), so does ∫[1 to ∞] 1/(x²+1)dx.

Key Terms

• Convergent
• Divergent

Worked Examples

Example 1: Infinite limit, convergent

∫[0 to ∞] e^(-x)dx = $lim(t→∞)$ [-e^(-x)][0 to t] = $lim(t→∞)$ (-e^(-t) + 1) = 1

Example 2: Infinite limit, divergent

∫[1 to ∞] (1/√x)dx = $lim(t→∞)$ [2√x][1 to t] = $lim(t→∞)$ (2√t - 2) = ∞

Divergent.

Example 3: Discontinuous at endpoint

∫[0 to 1] (1/√(1 - x))dx Discontinuity at x = 1. = $lim(t→1⁻)$ ∫[0 to t] (1 - x)^(-1/2)dx = $lim(t→1⁻)$ [-2√(1 - x)][0 to t] = $lim(t→1⁻)$ (-2√(1 - t) + 2) = 2

Convergent.

Practice Problems

Problem 1: ∫[1 to ∞] (1/x³)dx

Problem 2: ∫[0 to ∞] e^(-2x)dx

Problem 3: ∫[-1 to 1] (1/x²)dx

Problem 4: ∫[0 to 1] ln(x)dx

Problem 5: Use comparison: does ∫[1 to ∞] e^(-x)/x dx converge?

Summary

Key takeaways:

• Type 1: infinite limits — evaluate as limit
• Type 2: discontinuous integrand — split at discontinuity, evaluate each limit
• Convergent if limit is finite, divergent otherwise
• Comparison test: compare with known convergent/divergent integral

Pitfalls

• Treating improper integrals as ordinary definite integrals. For ∫[1 to ∞] (1/x)dx, plugging ∞ directly into the antiderivative gives a meaningless expression. You must write the limit explicitly: lim_{t→∞} ∫[1 to t] (1/x)dx and evaluate the limit properly.
• Not checking for interior discontinuities. For ∫[−1 to 1] (1/x²)dx, the discontinuity at x = 0 lies inside the interval. Evaluating as a single integral without splitting at x = 0 misses the divergence — each piece must be checked independently.
• Misapplying the comparison test direction. To prove convergence, you need 0 ≤ f(x) ≤ g(x) with ∫g convergent — the larger function bounds from above. To prove divergence, you need 0 ≤ g(x) ≤ f(x) with ∫g divergent. Using the wrong inequality direction invalidates the conclusion.
• Assuming convergence implies the integrand must approach zero. While f(x) → 0 is necessary for convergence of a positive, eventually monotone function (like p-integrals and exponentials), it is not a general theorem. Some pathological oscillating functions can have convergent improper integrals without pointwise limits.
• Forgetting to handle both tails in doubly-infinite integrals. For ∫[−∞ to ∞] f(x)dx, you must split at some c: ∫[−∞ to c] + ∫[c to ∞]. Both limits must exist independently. Using a single symmetric limit lim_{t→∞} ∫[−t to t] may give a finite Cauchy principal value even when the improper integral actually diverges.

Quiz

Q1: ∫[1 to ∞] (1/x²)dx is:

A) Convergent, value = 1 B) Divergent C) Convergent, value = 0 D) Convergent, value = ∞

Q2: ∫[1 to ∞] (1/x)dx is:

A) Convergent B) Divergent C) Convergent to ln(2) D) Convergent to 1

Q3: ∫[0 to 1] (1/√x)dx is:

A) Divergent B) Convergent to 2 C) Convergent to 1 D) Convergent to 1/2

Q4: For ∫[a to ∞] f(x)dx, if f(x) > 0 and the integral converges, then:

A) f(x) must approach 0 as x → ∞ B) f(x) can approach any value C) f(x) must be bounded D) f(x) must be decreasing

Q5: ∫[0 to ∞] e^(-x²)dx:

A) Diverges B) Equals √π/2 C) Equals 1 D) Equals π

Next Steps

Next up: 05-09-applications-of-integration.md