Improper Integrals: Convergence, p-Integrals and Examples

Integrals That Go On Forever

An ordinary definite integral has finite limits and a function that stays finite in between. But what is the area under y = 1/x² from x = 1 all the way to infinity? Or under y = 1/√x from 0 to 1, where the function shoots up to infinity at x = 0? These are improper integrals. Surprisingly, some of them have finite values: an infinitely long region can have a finite area. This tutorial explains the two types of improper integral, how to evaluate them using limits, how to decide whether they converge or diverge, the important p-integrals, and applications in probability and physics.

Two Types of Improper Integrals

Type 1: infinite limits. One or both limits of integration are infinite, such as the integral from 1 to ∞ of e^(−x) dx.
Type 2: infinite integrand. The function becomes infinite somewhere in the interval, usually at an endpoint, such as the integral from 0 to 1 of 1/√x dx.

In both cases, replace the problem point with a variable, integrate normally, and then take a limit. If the limit is a finite number, the integral converges to that value. If the limit is infinite or does not exist, the integral diverges.

Type 1: Infinite Limits

Definition: the integral from a to ∞ of f(x) dx = the limit as R → ∞ of the integral from a to R of f(x) dx.

Example 1. Evaluate the integral from 1 to ∞ of 1/x² dx.

The integral from 1 to R is [−1/x] from 1 to R = −1/R + 1. As R → ∞, 1/R → 0, so the integral converges to 1. The region under 1/x² to the right of x = 1 is infinitely long, but its area is exactly 1.

Example 2. Evaluate the integral from 1 to ∞ of 1/x dx.

The integral from 1 to R is [ln x] = ln R. As R → ∞, ln R → ∞, so the integral diverges. Even though 1/x tends to 0, it doesn't do so quickly enough.

Example 3. Evaluate the integral from 0 to ∞ of e^(−2x) dx.

The integral from 0 to R is [−½e^(−2x)] = −½e^(−2R) + ½. As R → ∞, e^(−2R) → 0, so the value is ½.

Example 4 (both limits infinite). Evaluate the integral from −∞ to ∞ of 1/(1 + x²) dx.

Split at any point, say 0, and evaluate each half separately. The integral from 0 to R is tan⁻¹ R, which tends to π/2. By symmetry the other half is also π/2. The total is π. Both halves must converge for the whole integral to converge.

Type 2: An Infinite Integrand

If f(x) becomes infinite at x = a, define the integral from a to b of f(x) dx as the limit as ε → 0⁺ of the integral from a + ε to b.

Example 5. Evaluate the integral from 0 to 1 of 1/√x dx.

The integral from ε to 1 of x^(−1/2) dx = [2√x] from ε to 1 = 2 − 2√ε. As ε → 0, √ε → 0, so the integral converges to 2.

Example 6. Evaluate the integral from 0 to 1 of 1/x² dx.

The integral from ε to 1 is [−1/x] = −1 + 1/ε, which tends to ∞. The integral diverges.

Example 7 (a problem point inside the interval). Evaluate the integral from −1 to 1 of 1/x² dx.

A careless calculation gives [−1/x] from −1 to 1 = −1 − 1 = −2, which is impossible: the function is positive everywhere, so the area can't be negative. The problem is that 1/x² is infinite at x = 0, inside the interval. Split it: the integral from 0 to 1 diverges (Example 6), so the whole integral diverges. Always check for points where the integrand blows up before integrating.

The p-Integrals

These two results are worth memorising, because many other integrals are compared with them.

The integral from 1 to ∞ of 1/xᵖ dx converges if p > 1 (to 1/(p − 1)) and diverges if p ≤ 1.
The integral from 0 to 1 of 1/xᵖ dx converges if p < 1 (to 1/(1 − p)) and diverges if p ≥ 1.

So 1/x² and 1/x^1.5 have finite area out to infinity, while 1/x and 1/√x do not. Near zero it is the other way round: 1/√x has finite area, 1/x and 1/x² do not. The borderline case p = 1 diverges in both.

Comparison Tests for Improper Integrals

Often you only need to know whether an integral converges, not its exact value. If 0 ≤ f(x) ≤ g(x) for all large x:

if the integral of g converges, so does the integral of f (a smaller positive function has smaller area);
if the integral of f diverges, so does the integral of g.

Example 8. Does the integral from 1 to ∞ of 1/(x² + 5) dx converge?

For x ≥ 1, 1/(x² + 5) < 1/x², and the integral of 1/x² from 1 to ∞ converges. So this integral converges too (its value is less than 1).

Example 9. Does the integral from 2 to ∞ of 1/(√x − 1) dx converge?

For x ≥ 2, √x − 1 < √x, so 1/(√x − 1) > 1/√x. The integral of 1/√x from 2 to ∞ diverges (p = ½ ≤ 1), so this one diverges.

Example 10. Does the integral from 1 to ∞ of e^(−x²) dx converge?

For x ≥ 1, x² ≥ x, so e^(−x²) ≤ e^(−x), whose integral from 1 to ∞ is 1/e. So it converges. (The full integral from −∞ to ∞ of e^(−x²) dx equals √π, a famous result behind the normal distribution.)

Applications of Improper Integrals

Probability. For a continuous random variable, the total probability is the integral of the probability density over all values, which is often an improper integral. For the exponential distribution with density λe^(−λx) for x ≥ 0, the total is the integral from 0 to ∞ of λe^(−λx) dx = 1, as it must be. The mean, the integral of xλe^(−λx) from 0 to ∞, works out by integration by parts to 1/λ.

Escape velocity. The work needed to move a mass m from the Earth's surface (radius R) to infinity against gravity is the integral from R to ∞ of GMm/r² dr = GMm/R. This finite value is why a finite launch speed, about 11.2 km/s, is enough to escape the Earth.

Electric potential. The potential at a distance r from a point charge is defined as the work done bringing a unit charge from infinity, again an improper integral of an inverse-square force.

Summary

Improper integrals have an infinite limit or an integrand that becomes infinite.
Replace the problem point with a variable, integrate, then take the limit; a finite limit means convergence.
From 1 to ∞, 1/xᵖ converges only for p > 1; from 0 to 1, only for p < 1.
Always check for points inside the interval where the function blows up.
Comparison with a known integral settles convergence without evaluating.