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Improper Integral

Improper Integral

A definite integral for which the integrand has a discontinuity between the bounds of integration, or which has ∞ and/or –∞ as a bound. Improper integrals are evaluated using limits as shown below. If the limit exists and is finite, we say the integral converges. If the limit does not exist or is infinite, we say the integral diverges.

 

Two examples of improper integrals evaluated with limits: ∫(1 to ∞) dx/x²=1 and ∫(-∞ to ∞) dx/(x²+1)=π

Evaluation of improper integral ∫₁³ dx/(x−2)^(2/3) using limits, splitting at discontinuity x=2, result equals 0.

 

 

See also

Integral test

Key Formula

Type 1 (infinite bound):∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx\text{Type 1 (infinite bound):} \quad \int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^{t} f(x)\,dx Type 2 (discontinuity at b):∫abf(x) dx=lim⁡t→b−∫atf(x) dx\text{Type 2 (discontinuity at } b\text{):} \quad \int_a^{b} f(x)\,dx = \lim_{t \to b^-} \int_a^{t} f(x)\,dx
Where:
  • f(x)f(x) = The integrand — the function being integrated
  • a,ba, b = The bounds of integration
  • tt = A temporary variable that approaches the problematic bound via a limit
  • ∞\infty = Indicates the interval extends without bound

Worked Example

Problem: Evaluate the improper integral ∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\,dx.
Step 1: Replace the infinite upper bound with a variable tt and write the integral as a limit.
∫1∞1x2 dx=lim⁡t→∞∫1t1x2 dx\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{t \to \infty} \int_1^{t} \frac{1}{x^2}\,dx
Step 2: Find the antiderivative of 1x2=x−2\frac{1}{x^2} = x^{-2}.
∫x−2 dx=−x−1=−1x\int x^{-2}\,dx = -x^{-1} = -\frac{1}{x}
Step 3: Evaluate the antiderivative from 11 to tt.
[−1x]1t=−1t−(−11)=−1t+1\left[-\frac{1}{x}\right]_1^{t} = -\frac{1}{t} - \left(-\frac{1}{1}\right) = -\frac{1}{t} + 1
Step 4: Take the limit as t→∞t \to \infty. Since 1t→0\frac{1}{t} \to 0, the expression approaches 11.
lim⁡t→∞(−1t+1)=0+1=1\lim_{t \to \infty}\left(-\frac{1}{t} + 1\right) = 0 + 1 = 1
Answer: The integral converges, and ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1.

Another Example

This example demonstrates a Type 2 improper integral, where the issue is a discontinuity at a bound rather than an infinite limit of integration. The limit approaches the problematic bound from inside the interval.

Problem: Evaluate the improper integral ∫011x dx\int_0^{1} \frac{1}{\sqrt{x}}\,dx.
Step 1: Identify the discontinuity. The integrand 1x\frac{1}{\sqrt{x}} is undefined at x=0x = 0 (the lower bound), so this is a Type 2 improper integral.
Step 2: Replace the problematic lower bound with a variable tt and write the integral as a limit.
∫011x dx=lim⁡t→0+∫t1x−1/2 dx\int_0^{1} \frac{1}{\sqrt{x}}\,dx = \lim_{t \to 0^+} \int_t^{1} x^{-1/2}\,dx
Step 3: Find the antiderivative of x−1/2x^{-1/2}.
∫x−1/2 dx=x1/21/2=2x\int x^{-1/2}\,dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x}
Step 4: Evaluate from tt to 11.
[2x]t1=21−2t=2−2t\left[2\sqrt{x}\right]_t^{1} = 2\sqrt{1} - 2\sqrt{t} = 2 - 2\sqrt{t}
Step 5: Take the limit as t→0+t \to 0^+. Since t→0\sqrt{t} \to 0, the result is 22.
lim⁡t→0+(2−2t)=2−0=2\lim_{t \to 0^+}(2 - 2\sqrt{t}) = 2 - 0 = 2
Answer: The integral converges, and ∫011x dx=2\int_0^{1} \frac{1}{\sqrt{x}}\,dx = 2.

Frequently Asked Questions

How do you know if an improper integral converges or diverges?
You rewrite the integral using a limit and then evaluate that limit. If the limit exists and equals a finite number, the integral converges. If the limit is infinite or does not exist, the integral diverges. For example, ∫1∞1x dx\int_1^{\infty} \frac{1}{x}\,dx diverges because its limit grows without bound, while ∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\,dx converges to 11.
What is the difference between Type 1 and Type 2 improper integrals?
A Type 1 improper integral has an infinite bound of integration, such as ∫1∞f(x) dx\int_1^{\infty} f(x)\,dx. A Type 2 improper integral has a finite interval but the integrand is discontinuous at or between the bounds, such as ∫011x dx\int_0^{1} \frac{1}{\sqrt{x}}\,dx. Both types require limits to evaluate, but the limit targets different things: an infinite bound versus a point of discontinuity.
What happens when both bounds are infinite?
When both bounds are infinite, as in ∫−∞∞f(x) dx\int_{-\infty}^{\infty} f(x)\,dx, you split the integral at any convenient point cc (often c=0c = 0) into two separate improper integrals: ∫−∞cf(x) dx+∫c∞f(x) dx\int_{-\infty}^{c} f(x)\,dx + \int_{c}^{\infty} f(x)\,dx. Each piece must converge independently. If either piece diverges, the entire integral diverges.

Improper Integral vs. Definite Integral (proper)

Improper IntegralDefinite Integral (proper)
IntervalInfinite interval or discontinuity in the integrandFinite interval with a continuous integrand
Evaluation methodRequires limits to handle infinite bounds or discontinuitiesDirectly evaluated using the Fundamental Theorem of Calculus
ResultMay converge (finite value) or diverge (no finite value)Always produces a finite number
Example∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1∫13x2 dx=263\int_1^{3} x^2\,dx = \frac{26}{3}

Why It Matters

Improper integrals appear frequently in calculus courses when you study areas under curves that extend infinitely, probability distributions (the entire normal distribution curve is defined by an improper integral), and the integral test for series convergence. They also arise in physics when computing quantities like gravitational potential or electric fields over unbounded regions. Mastering improper integrals is essential for any course beyond Calculus I.

Common Mistakes

Mistake: Evaluating the integral directly without using a limit, such as plugging ∞\infty into the antiderivative.
Correction: Always introduce a finite variable tt and take a limit. Write lim⁡t→∞∫atf(x) dx\lim_{t \to \infty} \int_a^t f(x)\,dx, evaluate the antiderivative at tt, and then compute the limit. Skipping the limit is not rigorous and can lead to errors, especially when determining divergence.
Mistake: Forgetting to check for discontinuities inside the interval of integration.
Correction: Before evaluating any definite integral, examine the integrand for discontinuities on [a,b][a, b]. For example, ∫−111x2 dx\int_{-1}^{1} \frac{1}{x^2}\,dx has a discontinuity at x=0x = 0. You must split this into two improper integrals at the discontinuity: ∫−101x2 dx+∫011x2 dx\int_{-1}^{0} \frac{1}{x^2}\,dx + \int_{0}^{1} \frac{1}{x^2}\,dx. (Both diverge, so the original integral diverges.)

Related Terms

  • Definite Integral — The proper version with finite, continuous bounds
  • Limit — The core tool used to evaluate improper integrals
  • Converge — Describes an improper integral with a finite value
  • Diverge — Describes an improper integral without a finite value
  • Discontinuity — A break in the integrand that creates a Type 2 case
  • Integrand — The function being integrated inside the integral
  • Bounds of Integration — The limits of the interval; may be infinite
  • Integral Test — Uses improper integrals to test series convergence