Lesson 14: Improper Integrals

An improper integral extends the definite integral to an infinite interval, or to an integrand that blows up inside the interval. In both cases the definition is a limit: replace the infinite or singular end by a finite parameter, integrate, then let the parameter approach its target. Sometimes the limit is finite — an infinitely long region with finite area — and sometimes it is not. Deciding which, and computing the value when finite, is this lesson. The concepts are pure Lesson 1 and 2 (limits) pointed at Lesson 7 (integrals).

Type 1: infinite interval

∫ from 1 to ∞ of (1/x^2) dx means lim(b→∞) ∫ from 1 to b of (1/x^2) dx = lim(b→∞) (1 − 1/b) = 1. The region under 1/x^2 from 1 rightward is unbounded in extent yet has area exactly 1 — infinity of length, finitude of content. Its neighbour behaves oppositely: ∫ from 1 to ∞ of (1/x) dx = lim ln b = ∞ — diverges. The p-test summarises the family: ∫ from 1 to ∞ of dx/x^p converges exactly when p > 1. The boundary case p = 1 diverges, logarithmically slowly. Worked Example 1. ∫ from 0 to ∞ of e^(−x) dx = lim(b→∞) (1 − e^(−b)) = 1.

Type 2: infinite discontinuity

∫ from 0 to 1 of dx/sqrt(x): the integrand explodes at 0. Define it as lim(a→0+) ∫ from a to 1 of x^(−1/2) dx = lim(a→0+) (2 − 2 sqrt(a)) = 2 — convergent, despite the blow-up. But ∫ from 0 to 1 of dx/x diverges (lim −ln a = ∞). The p-test mirrors: at a singularity, ∫ dx/x^p converges exactly when p < 1. Note the reversed inequality — near ∞, big p decays fast enough; near 0, small p explodes slowly enough.

Worked Example 2 (parts returns). ∫ from 0 to ∞ of x e^(−x) dx. Integrate by parts on [0, b]: [−(x + 1) e^(−x)] from 0 to b = −(b + 1) e^(−b) + 1. As b→∞, (b + 1) e^(−b) → 0 (exponential beats polynomial — Lesson 12's hierarchy), so the integral equals 1.

Comparison: deciding without computing

Often convergence is all that matters. If 0 ≤ f ≤ g and ∫ g converges, then ∫ f converges; if ∫ f diverges, so does ∫ g. Worked Example 3. ∫ from 1 to ∞ of (1 + sin^2 x)/x^2 dx: the integrand lies between 1/x^2 and 2/x^2, both of whose integrals converge, so the integral converges — its exact value is a harder, separate question. And ∫ from 1 to ∞ of dx/(x + sqrt(x)) diverges because the integrand exceeds 1/(2x) for x ≥ 1... check: x + sqrt(x) ≤ 2x when sqrt(x) ≤ x, i.e. x ≥ 1, so 1/(x + sqrt(x)) ≥ 1/(2x), and ∫ dx/(2x) diverges, forcing divergence. Comparison arguments are inequality management plus one known benchmark.

Application: probability densities and Gabriel's horn

The exponential distribution with mean 5 has density f(x) = (1/5) e^(−x/5) on [0, ∞). Total probability: ∫ from 0 to ∞ of f = 1 ✓ (a density must integrate to 1 — improper integrals are where this is checked). Its mean is ∫ from 0 to ∞ of x f(x) dx = 5, computed by parts as in Worked Example 2 — the parameter 5 in the density really is the average wait, and the improper integral is the proof. Then the paradox to argue about: rotate y = 1/x (x ≥ 1) about the x-axis. The volume is π∫ from 1 to ∞ of dx/x^2 = π — finite! The surface area is 2π∫ from 1 to ∞ of (1/x) sqrt(1 + 1/x^4) dx ≥ 2π∫ dx/x = ∞ — infinite. Gabriel's horn can be filled with π cubic units of paint but can never be painted. The resolution (the paint would need to be infinitely thin in the narrowing tail) is a fine discussion; the mathematics is two improper integrals with different verdicts.

Common mistakes

• Writing ∫ from 1 to ∞ as if ∞ were a number. "F(∞) − F(1)" is shorthand, not a computation. Expand the limit, evaluate the finite integral, then pass to the limit — the steps where divergence actually shows itself.
• Splitting blindness. ∫ from −1 to 1 of dx/x^2 is improper at 0, in the middle. Naively applying FTC gives [−1/x] from −1 to 1 = −1 − 1 = −2 — a negative "area" for a positive function, which should set off alarms. Split at 0: both halves diverge, so the integral diverges. (Similarly ∫ from −1 to 1 of dx/x diverges as an improper integral, despite symmetric cancellation — the "principal value" is a different definition.)
• Applying the ∞ p-test at 0. ∫ from 0 to 1 of dx/x^2 diverges (p = 2 > 1 hurts at 0, helps at ∞). The two p-tests have opposite inequalities; mixing them up inverts every answer.
• Concluding convergence from "the integrand tends to 0." 1/x → 0 and its integral still diverges. Tending to 0 is necessary, never sufficient — the rate is everything, which is why the p-test boundary sits where it does.

Exam Experience

1. ∫ from 1 to ∞ of dx/x^(3/2) equals

• (A) 1
• (B) 2
• (C) 3/2
• (D) Diverges

Answer: (B). lim(b→∞) [ −2 x^(−1/2) ] from 1 to b = 0 − (−2) = 2. p = 3/2 > 1, so convergence was guaranteed by the p-test; the value is 2.

2. Which of the following integrals converges?

• (A) ∫ from 0 to 1 of dx/x
• (B) ∫ from 0 to 1 of dx/x^(1/2)
• (C) ∫ from 1 to ∞ of dx/x
• (D) ∫ from 1 to ∞ of dx/sqrt(x)

Answer: (B). At 0 the p-test needs p < 1: 1/2 qualifies. (A) is p = 1 at 0 — diverges; (C) is p = 1 at ∞ — diverges; (D) is p = 1/2 < 1 at ∞ — diverges. Only (B) sits on the convergent side of its boundary.

3. ∫ from 0 to ∞ of 3 e^(−3x) dx equals

• (A) 1
• (B) 3
• (C) 1/3
• (D) Diverges

Answer: (A). lim(b→∞) [ −e^(−3x) ] from 0 to b = 0 − (−1) = 1. (It is a probability density — total probability 1 — which is a useful reality check.)

Free-response question. (a) Evaluate ∫ from 1 to ∞ of (1/x^4) dx. (b) Use comparison to determine whether ∫ from 1 to ∞ of (x + 1)/(x^3) dx converges or diverges; do not evaluate it. (c) A classmate claims ∫ from 0 to ∞ of (1/(1 + x)) dx converges "because the integrand goes to 0." Refute the claim with a computation.

Scoring guideline.

• (a) 1 point. lim(b→∞) [ −1/(3x^3) ] from 1 to b = 0 + 1/3 = 1/3. Writing the limit explicitly is required for the point.
• (b) 1 point. For x ≥ 1, (x + 1)/x^3 ≤ 2x/x^3 = 2/x^2, and ∫ from 1 to ∞ of 2/x^2 converges (= 2), so the given integral converges by comparison. Any valid dominating function earns the point.
• (c) 1 point. ∫ from 0 to b of dx/(1 + x) = ln(1 + b) → ∞ as b→∞, so the integral diverges despite the integrand tending to 0. Tending to 0 is necessary but not sufficient — the refutation must include the divergent computation, not just the slogan.

Going deeper: tails, and what convergence buys you numerically

A convergent improper integral can be used — approximated by a finite integral with a controlled error. For ∫ from 1 to ∞ of dx/x^2 = 1, the tail beyond b is ∫ from b to ∞ of dx/x^2 = 1/b. Want the total to within 0.01? Capture everything up to b = 100 and the missing tail is exactly 0.01. The rate at which tails shrink is the practical meaning of convergence speed, and it varies enormously: the tail of ∫ e^(−x) dx beyond b is e^(−b), so b = 10 already leaves less than 0.00005 behind, while the p = 2 tail needs b = 100 for a mere 0.01. Divergent integrals have their own diagnostic value: ∫ from 2 to ∞ of dx/(x ln x) diverges (substitute u = ln x: it becomes ∫ du/u from ln 2 onward) — logarithmically slowly, which is why the series Σ 1/(n ln n) fools intuition; it grows like ln(ln N), reaching barely 2 after N = e^(e^2) ≈ 1600 terms. Slow divergence is still divergence. One more benchmark to own: ∫ from 0 to ∞ of e^(−x^2) dx = sqrt(π)/2 ≈ 0.886 (proved by a famous two-dimensional trick beyond this course, but the value powers the normal distribution's normalisation — Lesson 16 met its finite-interval cousin). Improper integrals are not pathology; they are where probability, physics, and signal processing keep their most-used constants.

One more application: the mean of an exponential wait

The density f(x) = λ e^(−λx) has mean ∫ from 0 to ∞ of x·λe^(−λx) dx = 1/λ (parts, using the Worked Example 3 survival computation pattern: the boundary term vanishes and ∫ λe^(−λx) dx = 1 leaves 1/λ). For λ = 2 per hour the mean wait is 1/2 hour. This single evaluation is the engine behind queueing theory's basic models, radioactive-decay mean lifetimes (half-life's quieter sibling: mean life = 1/λ), and reliability engineering. The improper integral converges for every λ > 0 — exponential tails always beat the linear factor x, which is the comparison theorem's hierarchy (Lesson 12's growth ladder) appearing as a modelling guarantee: exponential waits have finite means, always.