Improper integrals, explained with examples

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

An improper integral has an infinite limit or a function that blows up somewhere in the interval. Surprisingly, the area of an infinitely long region can still be finite. Improper integrals are an AP Calculus BC topic and the foundation for the integral test for series.

Two kinds of improper integral

• Infinite limits: one or both limits of integration are ∞ or −∞, as in ∫₁^∞ 1/x² dx.
• Infinite discontinuities: the integrand becomes infinite somewhere in the interval, such as ∫₀¹ 1/√x dx, where 1/√x is undefined at x = 0.

In both cases, you cannot just substitute the limits. Instead, replace the problem point with a variable, integrate, and take a limit.

The method

For an infinite upper limit:

∫_a^∞ f(x) dx = lim_b→∞ ∫_a^b f(x) dx

For a discontinuity at the lower limit a:

∫_a^c f(x) dx = lim_t→a⁺ ∫_t^c f(x) dx

If the limit exists and is a finite number, the integral converges to that value. If the limit is infinite or does not exist, the integral diverges. On AP free-response questions, you must show the limit notation; writing “∞” straight into an antiderivative loses points. Write the limit on every line until you actually evaluate it, then state clearly whether the integral converges or diverges.

Worked examples

Example 1: a convergent integral

Evaluate ∫₁^∞ 1/x² dx.

• = lim_b→∞ [−1/x]₁^b = lim_b→∞ (−1/b + 1).
• As b → ∞, −1/b → 0, so the integral converges to 1. An infinitely long region has an area of exactly 1.

Example 2: a divergent integral

Evaluate ∫₁^∞ 1/x dx.

• = lim_b→∞ [ln x]₁^b = lim_b→∞ ln b.
• ln b grows without bound, so the integral diverges. Although 1/x approaches 0, it does not do so fast enough for the area to be finite.

Example 3: an exponential

Evaluate ∫₀^∞ e^−x dx.

• = lim_b→∞ [−e^−x]₀^b = lim_b→∞ (1 − e^−b).
• Converges to 1.

Example 4: an infinite discontinuity

Evaluate ∫₀¹ 1/√x dx.

• The integrand is undefined at x = 0. = lim_t→0⁺ [2√x]_t¹ = lim_t→0⁺ (2 − 2√t).
• Converges to 2.

Example 5: integration by parts

Evaluate ∫₀^∞ x e^−x dx.

• By parts, ∫ x e^−x dx = −x e^−x − e^−x = −e^−x(x + 1).
• lim_b→∞ [−e^−b(b + 1) + 1] = 0 + 1, since exponentials beat polynomials (L'Hôpital's rule).
• Converges to 1.

Example 6: an inverse tangent

Evaluate ∫₀^∞ 1/(1 + x²) dx.

• = lim_b→∞ [arctan x]₀^b = lim_b→∞ arctan b.
• arctan b → π/2, so the integral converges to π/2.

Example 7: a hidden discontinuity

Evaluate ∫₋₁¹ 1/x² dx.

• The integrand is infinite at x = 0, inside the interval, so split it: ∫₋₁⁰ + ∫₀¹.
• ∫₀¹ 1/x² dx = lim_t→0⁺ (1/t − 1), which is infinite, so the whole integral diverges.
• Ignoring the discontinuity and computing [−1/x] from −1 to 1 gives −2, a negative “area” for a positive function, which is clearly wrong.

Both limits infinite

For an integral from −∞ to ∞, split it at any convenient point, such as 0, and evaluate both pieces separately: ∫_−∞^∞ f(x) dx = ∫_−∞⁰ f(x) dx + ∫₀^∞ f(x) dx. The whole integral converges only if both pieces converge. For 1/(1 + x²), each half gives π/2, so the total is π. You must not take a single symmetric limit, because divergent pieces can falsely cancel.

A famous paradox: Gabriel's horn

Rotate y = 1/x, for x ≥ 1, about the x-axis to make an infinitely long horn. Its volume is π∫₁^∞ 1/x² dx = π, which is finite. But its surface area involves an integral at least as large as 2π∫₁^∞ 1/x dx, which diverges. So the horn could be filled with a finite amount of paint, yet its inside surface could never be painted. Improper integrals produce results like this that defy everyday intuition, which is part of their appeal.

The p-integral rule

Integrals of 1/x^p appear constantly, so it pays to know their behaviour:

• ∫₁^∞ 1/x^p dx converges if p > 1 and diverges if p ≤ 1.
• ∫₀¹ 1/x^p dx converges if p < 1 and diverges if p ≥ 1.

Examples 1, 2 and 4 fit the pattern: p = 2 converges on [1, ∞), p = 1 diverges, and p = ½ converges on [0, 1]. The borderline case p = 1 diverges in both situations.

Comparison reasoning

Sometimes you cannot find an antiderivative, but you can still decide convergence by comparing with a known integral. If 0 ≤ f(x) ≤ g(x) and ∫g converges, then ∫f converges too. If f(x) ≥ g(x) ≥ 0 and ∫g diverges, then ∫f diverges. For example, 1/(x² + 5) < 1/x² for x ≥ 1, and ∫₁^∞ 1/x² dx converges, so ∫₁^∞ 1/(x² + 5) dx also converges.

The link to series

The integral test for infinite series uses improper integrals: if f is positive and decreasing, the series Σf(n) and the integral ∫₁^∞f(x) dx either both converge or both diverge. That is why the p-series Σ1/n^p converges exactly when p > 1, matching the p-integral rule.

Common mistakes

• Substituting ∞ directly instead of using a limit.
• Missing a discontinuity inside the interval.
• Assuming an integral converges because the function tends to 0.
• Getting the p-integral conditions the wrong way round for [0, 1] and [1, ∞).

Practice questions

Try each one before opening the answer.

Frequently asked questions

Are improper integrals on AP Calculus AB?

No. Improper integrals are part of AP Calculus BC, in Unit 6, and they reappear with the integral test in Unit 10.

Are improper integrals in A level Maths?

Not in A level Maths. They appear in A level Further Maths, where improper integrals with infinite limits and discontinuities are studied.

Key takeaways

• Improper integrals have infinite limits or infinite discontinuities.
• Replace the problem point with a variable and take a limit.
• A finite limit means convergence; otherwise the integral diverges.
• ∫₁^∞ 1/x^p converges only for p > 1.
• Check for discontinuities inside the interval before integrating.

Related guides

• AP Calculus BC series
• Integration by parts
• L'Hôpital's rule