Integration · Improper Integrals

Improper Integrals — Integrating to Infinity and Beyond

What happens when an interval stretches to infinity, or the integrand blows up inside the limits? Standard integration breaks down, but improper integrals replace each problematic endpoint with a limit, letting us ask whether the infinite area converges to a finite number or escapes to infinity altogether.

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§ 01What Makes an Integral "Improper"?

Every integral you have evaluated so far has two finite limits and a bounded integrand. Improper integrals break at least one of those rules, and require a careful limit argument to handle correctly.

A definite integral ∫ₐᵇ f(x) dx is called improper when one or more of the following holds:

  • Type 1: One or both limits of integration are infinite (e.g., ∫₁^∞ or ∫_{−∞}^∞).
  • Type 2: The integrand f(x) has a vertical asymptote (blows up to ±∞) at one or both endpoints, or at some point c inside [a, b].

In both cases, we cannot directly apply the Fundamental Theorem of Calculus because one of its conditions, that f be continuous and bounded on a closed interval, is violated. Instead, we replace the bad endpoint with a variable t and take the limit as t approaches the problematic value.

Improper integrals arise throughout mathematics and physics: the Gamma function, the Laplace transform, probability density functions over all of ℝ, and the convergence of Fourier series all rely on improper integrals being finite.

Prerequisites You should be comfortable with: limits (including limits at infinity), basic antiderivatives, the Fundamental Theorem of Calculus, and L'Hôpital's Rule. Familiarity with ln, eˣ, and trig antiderivatives is assumed.

§ 02Type 1 — Infinite Limits of Integration

When one or both limits are ±∞, we replace the infinite limit with a finite variable and compute the limit of the resulting integral.

Type 1a
Infinite Upper Limit
∫ₐ^∞ f(x) dx = limt→∞ ∫ₐᵗ f(x) dx

Replace ∞ with t. Evaluate the definite integral, then take the limit as t → ∞.

Converges if the limit is a finite number. Diverges if the limit is ±∞ or does not exist.

Type 1b
Infinite Lower Limit
∫_{−∞}^b f(x) dx = limt→−∞ ∫ₜᵇ f(x) dx

Replace −∞ with t. Evaluate the definite integral, then take the limit as t → −∞.

Both Limits Infinite
∫_{−∞}^∞ f(x) dx = ∫_{−∞}^c f(x) dx + ∫_c^∞ f(x) dx

Split at any convenient constant c (usually 0). Both pieces must converge independently for the whole integral to converge. If either piece diverges, the whole integral diverges.

x y 1 ∞ finite area = 1 y = 1/x²
∫₁^∞ (1/x²) dx = 1. The curve approaches the x-axis but the total shaded area is finite — the integral converges.
The p-Integral — A Key Benchmark The integral ∫₁^∞ (1/xᵖ) dx converges if and only if p > 1. When p > 1 the value is 1/(p−1). When p ≤ 1 it diverges. This single fact is the foundation of the comparison test for improper integrals.
IntegralConditionResultStatus
∫₁^∞ 1/x² dxp = 2 > 11CONVERGES
∫₁^∞ 1/x dxp = 1∞DIVERGES
∫₁^∞ 1/√x dxp = ½ < 1∞DIVERGES
∫₁^∞ 1/x³ dxp = 3 > 11/2CONVERGES

§ 03Type 2 — Discontinuous Integrands

When the integrand has a vertical asymptote somewhere in [a, b], standard integration gives a nonsensical result. We must again replace the bad point with a variable and take a one-sided limit.

Type 2a — Asymptote at Right Endpoint
Blows up at x = b
∫ₐᵇ f(x) dx = limt→b⁻ ∫ₐᵗ f(x) dx

f is continuous on [a, b) but lim_{x→b⁻} f(x) = ±∞. Approach b from the left.

Type 2b — Asymptote at Left Endpoint
Blows up at x = a
∫ₐᵇ f(x) dx = limt→a⁺ ∫ₜᵇ f(x) dx

f is continuous on (a, b] but lim_{x→a⁺} f(x) = ±∞. Approach a from the right.

Asymptote Inside the Interval — Split It
∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx

If f has a vertical asymptote at c ∈ (a, b), split the integral there. Each piece is then a Type 2a or 2b improper integral and must be evaluated separately. Both must converge for the whole integral to converge.

x y 1 0 asymptote at x = 0 finite area = 2 y = 1/√x
∫₀¹ (1/√x) dx = 2. Despite the vertical asymptote at x = 0, the shaded area is finite — this Type 2 improper integral converges.
The Hidden Asymptote Trap The most dangerous mistake with Type 2 integrals is failing to notice the discontinuity. For example, ∫_{−1}^1 (1/x²) dx looks like a normal integral, but 1/x² blows up at x = 0. Applying FTC blindly gives [−1/x]_{−1}^1 = −1 − 1 = −2, which is absurd — areas cannot be negative for a non-negative function. The correct answer is that this integral diverges.

§ 04Convergence and Divergence

The language of convergence and divergence is borrowed from sequences and series, and it works the same way here.

✓ Converges — the limit exists and is a finite number
✗ Diverges — the limit is ±∞, or does not exist

To determine convergence, always write out the limit explicitly. Do not shortcut this step: the limit is the definition of an improper integral, and skipping it leads to errors.

The Comparison Test for Improper Integrals

Sometimes finding the exact antiderivative is impossible, but we can still determine whether the integral converges by comparing it to a simpler function whose behaviour we already know.

Comparison Test
0 ≤ f(x) ≤ g(x) for all x ≥ a

If ∫ₐ^∞ g(x) dx converges, then ∫ₐ^∞ f(x) dx also converges.

If ∫ₐ^∞ f(x) dx diverges, then ∫ₐ^∞ g(x) dx also diverges.

In other words: smaller than a convergent integral → converges. Bigger than a divergent integral → diverges.

The Limit Comparison Test

When direct comparison is awkward, compare the growth rates using a limit. If f(x) and g(x) are positive and

Limit Comparison Test
limx→∞ f(x)/g(x) = L, where 0 < L < ∞

then ∫ₐ^∞ f(x) dx and ∫ₐ^∞ g(x) dx either both converge or both diverge.

Choosing the Right Comparison Function Use the p-integral ∫₁^∞ 1/xᵖ dx as your benchmark. For a function like 1/(x²+3x+1), observe that for large x it behaves like 1/x², so compare with 1/x² (which converges, p=2). For 1/(x+1) it behaves like 1/x (which diverges, p=1).

§ 05Worked Examples

Type 1 Example 1 — ∫₁^∞ 1/x² dx

The benchmark convergent integral.

  • Replace ∞ with t: limt→∞ ∫₁ᵗ x⁻² dx.

  • Antiderivative: −x⁻¹ = −1/x.

    [−1/x]₁ᵗ = −1/t − (−1) = 1 − 1/t

  • Take the limit: limt→∞ (1 − 1/t) = 1 − 0 = 1.

∫₁^∞ 1/x² dx = 1  (Converges)
Type 1 Example 2 — ∫₁^∞ 1/x dx (Harmonic — diverges)

  • limt→∞ ∫₁ᵗ (1/x) dx = limt→∞ [ln x]₁ᵗ.

  • = limt→∞ (ln t − ln 1) = limt→∞ ln t = ∞.
∫₁^∞ 1/x dx = ∞  (Diverges) — the harmonic integral
Type 1 Example 3 — ∫₀^∞ e⁻ˣ dx

  • limt→∞ ∫₀ᵗ e⁻ˣ dx = limt→∞ [−e⁻ˣ]₀ᵗ.

  • = limt→∞ (−e⁻ᵗ + e⁰) = limt→∞ (1 − e⁻ᵗ) = 1 − 0 = 1.
∫₀^∞ e⁻ˣ dx = 1  (Converges)
Type 1 Example 4 — ∫_{−∞}^∞ 1/(1+x²) dx

Both limits are infinite: split at 0.

  • Left piece: limt→−∞ [arctan x]ₜ⁰ = arctan(0) − arctan(−∞) = 0 − (−π/2) = π/2.

  • Right piece: limt→∞ [arctan x]₀ᵗ = arctan(∞) − arctan(0) = π/2 − 0 = π/2.

  • Total = π/2 + π/2 = π.
∫_{−∞}^∞ 1/(1+x²) dx = π  (Converges)
Type 1 Example 5 — ∫₁^∞ xe⁻ˣ dx (Integration by Parts)

  • Use IBP: u = x, dv = e⁻ˣ dx → du = dx, v = −e⁻ˣ.

    ∫xe⁻ˣ dx = −xe⁻ˣ − ∫(−e⁻ˣ)dx = −xe⁻ˣ − e⁻ˣ = −e⁻ˣ(x+1)

  • Evaluate: limt→∞ [−e⁻ˣ(x+1)]₁ᵗ.

    = limt→∞ [−e⁻ᵗ(t+1)] − [−e⁻¹(2)]

  • As t → ∞, e⁻ᵗ(t+1) → 0 (exponential beats polynomial). So: 0 + 2e⁻¹ = 2/e.

∫₁^∞ xe⁻ˣ dx = 2/e ≈ 0.736  (Converges)
Type 1 Example 6 — ∫₀^∞ cos x dx (Diverges)

  • limt→∞ [sin x]₀ᵗ = limt→∞ sin t.

  • sin t oscillates between −1 and 1 and has no limit as t → ∞.

∫₀^∞ cos x dx — Diverges (limit does not exist due to oscillation)
Type 2 Example 7 — ∫₀¹ 1/√x dx (asymptote at x = 0)

  • Replace 0 with t (approach from the right): limt→0⁺ ∫ₜ¹ x⁻¹/² dx.

  • [2x^(1/2)]ₜ¹ = 2 − 2√t

  • limt→0⁺ (2 − 2√t) = 2 − 0 = 2.

∫₀¹ 1/√x dx = 2  (Converges)
Type 2 Example 8 — ∫₀¹ ln x dx (asymptote at x = 0)

  • limt→0⁺ ∫ₜ¹ ln x dx. Use IBP: ∫ln x dx = x ln x − x + C.

  • [x ln x − x]ₜ¹ = (1·0 − 1) − (t ln t − t) = −1 − t ln t + t

  • Need limt→0⁺ t ln t. Use L'Hôpital: lim t ln t = lim (ln t)/(1/t) = lim (1/t)/(−1/t²) = lim(−t) = 0.

  • So lim (−1 − t ln t + t) = −1 − 0 + 0 = −1.

∫₀¹ ln x dx = −1  (Converges — the area below the x-axis has magnitude 1)
Type 2 Example 9 — ∫₀² 1/(x−1)² dx (asymptote inside — diverges)

The integrand blows up at x = 1, which is inside [0, 2]. We must split there.

  • Split: ∫₀¹ 1/(x−1)² dx + ∫₁² 1/(x−1)² dx.

  • First piece: limt→1⁻ [−1/(x−1)]₀ᵗ = limt→1⁻ [−1/(t−1) + 1/(−1)] = limt→1⁻ [−1/(t−1)] − 1.

  • As t → 1⁻, (t−1) → 0⁻, so −1/(t−1) → +∞. The first piece diverges.

∫₀² 1/(x−1)² dx — Diverges (first piece → ∞). Blindly applying FTC gives the wrong answer −2.
Type 1 Example 10 — Comparison Test: ∫₁^∞ e^(−x²) dx

No elementary antiderivative exists. Use comparison.

  • For x ≥ 1: x² ≥ x, so −x² ≤ −x, so e^(−x²) ≤ e^(−x).

  • We know ∫₁^∞ e⁻ˣ dx = [−e⁻ˣ]₁^∞ = 0 + e⁻¹ = 1/e < ∞ (converges).

  • Since 0 ≤ e^(−x²) ≤ e^(−x) and the larger integral converges, by the Comparison Test:

∫₁^∞ e^(−x²) dx converges (value ≈ 0.1394, related to the Gaussian integral)
Type 1 Example 11 — Limit Comparison: ∫₁^∞ (2x+1)/(x²+x+1) dx

  • For large x, (2x+1)/(x²+x+1) ≈ 2x/x² = 2/x. Compare with g(x) = 1/x.

  • limx→∞ f/g = lim (2x+1)/(x²+x+1) · x = lim (2x²+x)/(x²+x+1) = 2 (finite and positive).

  • Since ∫₁^∞ 1/x dx diverges (p=1) and the limit is finite and positive, by LCT:

∫₁^∞ (2x+1)/(x²+x+1) dx — Diverges
Type 1 Example 12 — The Gamma Function: ∫₀^∞ x^(n−1)e⁻ˣ dx

The Gamma function Γ(n) = ∫₀^∞ x^(n−1)e⁻ˣ dx is one of the most important improper integrals in mathematics. It extends the factorial to non-integers: Γ(n) = (n−1)! for positive integers n.

  • For n = 1: Γ(1) = ∫₀^∞ e⁻ˣ dx = 1 = 0! ✓

  • For n = 2: Γ(2) = ∫₀^∞ xe⁻ˣ dx. IBP gives [−xe⁻ˣ − e⁻ˣ]₀^∞ = 0 − (0−1) = 1 = 1! ✓

  • The recurrence Γ(n+1) = n·Γ(n) follows from integration by parts, extending the factorial to all positive reals.

Γ(n) = (n−1)! for positive integers; converges for all n > 0

§ 06Common Mistakes

Mistake 1 — Forgetting to write the limit An improper integral is defined as a limit. Writing ∫₁^∞ (1/x²) dx = [−1/x]₁^∞ = 0 − (−1) = 1 is sloppy — the notation [F(x)]₁^∞ hides the limit that makes this rigorous. Always write limt→∞ [F(x)]₁ᵗ first.
Mistake 2 — Missing the interior discontinuity Always scan the integrand for vertical asymptotes inside [a, b], not just at the endpoints. ∫_{−1}^1 (1/x) dx is improper and diverges, despite looking like a routine integral. Plugging in directly gives 0 (by symmetry), which is wrong — each half diverges.
Mistake 3 — Splitting ∫_{−∞}^∞ as a single limit Writing ∫_{−∞}^∞ f dx = limt→∞ ∫_{−t}^t f dx is the Cauchy principal value, which is not the same as the improper integral. The correct definition requires splitting at a point and taking two independent limits. For ∫_{−∞}^∞ x dx, the Cauchy PV is 0, but the integral properly diverges.
Mistake 4 — Using the wrong direction for Type 2 If the asymptote is at the left endpoint a, the limit is t → a⁺ (from the right). If at the right endpoint b, it is t → b⁻ (from the left). Using the wrong direction gives a nonsensical or infinite result.
Mistake 5 — Comparison Test direction error The comparison works as follows: f ≤ g and ∫g converges → ∫f converges. But f ≤ g and ∫g diverges tells you nothing about ∫f. You need f ≥ g and ∫g diverges → ∫f diverges. Be precise about which function is larger.

§ 07Quick Reference — Key Improper Integrals

IntegralValueStatusNotes
∫₁^∞ x⁻ᵖ dx1/(p−1)CONV if p>1p-integral benchmark
∫₁^∞ x⁻ᵖ dx∞DIV if p≤1Includes harmonic (p=1)
∫₀^∞ e⁻ˣ dx1CONVExponential decay
∫₀^∞ e^(ax) dx, a>0∞DIVExponential growth
∫_{−∞}^∞ 1/(1+x²) dxπCONVArctan antiderivative
∫₀¹ x⁻ᵖ dx1/(1−p)CONV if p<1Type 2 p-integral
∫₀¹ x⁻ᵖ dx∞DIV if p≥1Note: opposite rule to Type 1!
∫₀¹ ln x dx−1CONVL'Hôpital needed for limit
∫₀^∞ x^(n−1)e⁻ˣ dx(n−1)!CONV, n>0Gamma function Γ(n)
Type 1 vs Type 2 p-Integral — The Reversed Rule For ∫₁^∞ 1/xᵖ dx: converges when p > 1. For ∫₀¹ 1/xᵖ dx: converges when p < 1. The rules are mirror images of each other. A common exam trap is applying the wrong rule for the wrong type.

§ 08Test Yourself — 10-Question Quiz

Questions test convergence decisions, exact values, and comparison reasoning. Some fill-in answers should be entered as decimals; convergence/divergence questions use multiple choice.

Score: 0 / 0
Question 1 of 10

Does ∫₁^∞ 1/x³ dx converge or diverge?

p = 3 > 1, so by the p-integral rule, ∫₁^∞ 1/x³ dx converges. Its value is 1/(3−1) = 1/2.

Question 2 of 10

Evaluate ∫₀^∞ e⁻³ˣ dx. Enter your answer as a fraction (decimal form).

  1. limt→∞ [−(1/3)e⁻³ˣ]₀ᵗ = lim(−(1/3)e⁻³ᵗ + 1/3) = 0 + 1/3 = 1/3 ≈ 0.333.
Question 3 of 10

Which statement is correct about ∫_{−1}^1 1/x dx?

1/x has a vertical asymptote at x = 0 inside [−1, 1]. We must split there. Both ∫_{−1}^0 (1/x)dx and ∫_0^1 (1/x)dx diverge (each gives ±∞). The Cauchy principal value is 0 by symmetry, but the proper improper integral diverges. Options (b) and (c) are the classic wrong answers.

Question 4 of 10

Evaluate ∫₀¹ 1/√x dx. Enter your answer.

  1. limt→0⁺ [2√x]ₜ¹ = 2(1) − 2(0) = 2.
Question 5 of 10

By the Comparison Test, what can we conclude about ∫₁^∞ sin²(x)/x² dx?

Since 0 ≤ sin²(x) ≤ 1, we have 0 ≤ sin²(x)/x² ≤ 1/x². The integral ∫₁^∞ 1/x² dx = 1 converges. By the Comparison Test, ∫₁^∞ sin²(x)/x² dx also converges. Note: (c) is wrong because ∫₁^∞ 1 dx = ∞ diverges: bounding by a divergent function is useless.

Question 6 of 10

What is the value of ∫_{−∞}^∞ 1/(1+x²) dx? Enter the exact value as a multiple of π (enter the coefficient of π).

As shown in Example 4: ∫_{−∞}^∞ 1/(1+x²) dx = π. The coefficient of π is 1.

Question 7 of 10

Does ∫₀¹ 1/x dx converge or diverge?

limt→0⁺ ∫ₜ¹ (1/x) dx = limt→0⁺ [ln x]ₜ¹ = ln(1) − ln(t) = 0 − (−∞) = ∞. This is the Type 2 analogue of the harmonic integral: it diverges for the same reason (p = 1).

Question 8 of 10

Evaluate ∫₂^∞ 1/x² dx. Enter the exact decimal value.

  1. limt→∞ [−1/x]₂ᵗ = lim(−1/t + 1/2) = 0 + 1/2 = 0.5.
Question 9 of 10

For the Type 2 p-integral ∫₀¹ x⁻ᵖ dx, when does it converge?

The Type 2 p-integral ∫₀¹ x⁻ᵖ dx converges when p < 1 (value = 1/(1−p)) and diverges when p ≥ 1. This is the opposite of the Type 1 rule, a common exam trap. When p < 1, the function doesn't blow up fast enough near 0 for the area to be infinite.

Question 10 of 10

By the Limit Comparison Test with 1/x, what can we say about ∫₁^∞ 1/(√(x²+1)) dx?

For large x: √(x²+1) ≈ x, so 1/√(x²+1) ≈ 1/x. Formally: limx→∞ [x/√(x²+1)] = lim 1/√(1+1/x²) = 1. Since the limit is 1 (finite and positive) and ∫₁^∞ 1/x diverges, by the Limit Comparison Test ∫₁^∞ 1/√(x²+1) also diverges. Note: (b) is wrong, knowing f < g when g diverges tells us nothing useful.

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