Improper Integrals of the First and Second Kind: Convergence Tests

An improper integral extends the definite integral to an infinite interval (first kind) or to an unbounded integrand (second kind) by taking a limit: ∫_a^∞ f(x) dx = lim_b→∞ ∫_a^b f(x) dx. In brief: ∫₁^∞ dx/x² = 1 converges, while ∫₁^∞ dx/x diverges. For ∫₀¹ dx/x^p the situation is reversed: it converges exactly for p < 1. Convergence tests (comparison and limit comparison) decide convergence without computing the value.

Many important integrals in probability and physics extend over an infinite range or have a singularity at an endpoint: the total of a distribution, the energy of a field, the Gaussian integral. The Riemann integral does not cover these cases, so they are defined as limits. This guide defines both kinds, gives the key benchmark integrals, explains the convergence tests with examples and presents the Gaussian and gamma integrals, with a table of the p-integrals.

What are improper integrals of the first kind?

The interval is infinite. Define ∫_a^∞ f(x) dx = lim_b→∞ ∫_a^b f(x) dx, and ∫_−∞^b f = lim_a→−∞ ∫_a^b f. For ∫_−∞^∞ f, split at any point c: both ∫_−∞^c f and ∫_c^∞ f must converge separately. If the limit exists and is finite, the integral converges; otherwise it diverges.

Example 1

∫₀^∞ e^−x dx = lim_b→∞ [−e^−x]₀^b = lim (1 − e^−b) = 1. The area under the exponential curve is finite.

Example 2

∫_−∞^∞ dx/(1 + x²) = lim_a→−∞ lim_b→∞ [arctan x]_a^b = π/2 − (−π/2) = π.

Example 3: divergence

∫₁^∞ dx/x = lim [ln x]₁^b = lim ln b = ∞. The area under 1/x is infinite, although the height tends to 0.

What are improper integrals of the second kind?

The integrand is unbounded near an endpoint (or an interior point). If f is unbounded at b, ∫_a^b f = lim_c→b⁻ ∫_a^c f. If f is unbounded at an interior point c, split the integral at c and require both parts to converge.

Example 4

∫₀¹ dx/√x = lim_a→0⁺ [2√x]_a¹ = 2 − 0 = 2. The area is finite despite the vertical asymptote.

Example 5

∫₀¹ ln x dx = lim_a→0⁺ [x ln x − x]_a¹ = −1 − lim (a ln a − a) = −1, since a ln a → 0.

Example 6: an interior singularity

∫₋₁¹ dx/x²: the integrand is unbounded at 0. ∫₀¹ dx/x² = lim [−1/x]_a¹ = ∞. So the integral diverges. (The formal use of the Newton–Leibniz formula, [−1/x]₋₁¹ = −2, is wrong, since the function has a pole inside.) Similarly ∫₋₁¹ dx/x diverges, though by symmetry its principal value lim_ε→0 (∫₋₁^−ε + ∫_ε¹) dx/x = 0.

What are the benchmark integrals?

• ∫₁^∞ dx/x^p converges if and only if p > 1, and then equals 1/(p − 1).
• ∫₀¹ dx/x^p converges if and only if p < 1, and then equals 1/(1 − p).
• ∫₀^∞ e^−kx dx = 1/k for k > 0.
• ∫_a^b dx/(x − a)^p converges if and only if p < 1 (singularity at a).

These match the p-series Σ 1/n^p, which converges for p > 1, as: the integral test connects series and improper integrals.

What are the convergence tests?

Comparison test. If 0 ≤ f(x) ≤ g(x) for x ≥ a: if ∫ g converges, then ∫ f converges; if ∫ f diverges, then ∫ g diverges. Limit comparison test. If f, g > 0 and f(x)/g(x) → L with 0 < L < ∞ as x → ∞ (or to the singularity), then both integrals converge or both diverge. Absolute convergence. If ∫ |f| converges, then ∫ f converges. The same ideas hold for the second kind of integral, comparing near the singularity.

Example 7

∫₁^∞ dx/(x³ + 1): 0 < 1/(x³ + 1) < 1/x³, and ∫₁^∞ x⁻³ dx = 1/2 converges. So the integral converges (and is less than 1/2).

Example 8

∫₂^∞ dx/√(x² − 1): compare with 1/x, since √(x² − 1) < x, so 1/√(x² − 1) > 1/x, and ∫ dx/x diverges. So the integral diverges.

Example 9: limit comparison

∫₁^∞ (2x + 1)/(x³ + x + 5) dx: for large x, the integrand behaves like 2x/x³ = 2/x². The ratio with 1/x² tends to 2, and ∫ x⁻² converges. So the integral converges.

Example 10: a second kind comparison

∫₀¹ dx/(√x (1 + x)): near 0, 1/(√x(1 + x)) ≤ 1/√x, and ∫₀¹ x^−1/2 dx = 2 converges. So it converges. (The exact value: with x = u², ∫₀¹ 2 du/(1 + u²) = π/2.)

Example 11: an oscillating integrand

∫₁^∞ sin x/x² dx converges absolutely, since |sin x/x²| ≤ 1/x². In contrast ∫₁^∞ sin x/x dx converges (by parts: [−cos x/x]₁^∞ − ∫ cos x/x² dx), but not absolutely, since ∫ |sin x|/x diverges. Its value over [0, ∞) is π/2 (the Dirichlet integral).

What are important improper integrals?

• Gaussian integral: ∫_−∞^∞ e^−x2 dx = √π ≈ 1.7725. It has no elementary antiderivative, but the value is found by a clever computation in polar coordinates. It is the basis of the normal distribution.
• Gamma function: Γ(s) = ∫₀^∞ x^s−1e^−x dx converges for s > 0, and Γ(n) = (n − 1)! for positive integers n. Also Γ(1/2) = √π. By parts, Γ(s + 1) = sΓ(s).
• Exponential distribution: ∫₀^∞ λe^−λx dx = 1, a valid probability density, and its mean ∫₀^∞ x λe^−λx dx = 1/λ.
• Laplace transform: ∫₀^∞ e^−st f(t) dt.

Example 12

Γ(4) = 3! = 6: ∫₀^∞ x³e^−x dx. By parts three times, with the result 3 · 2 · 1 = 6.

Table of p-integrals

Where do improper integrals come from?

Evangelista Torricelli found in the 1640s that the solid obtained by rotating the curve y = 1/x for x ≥ 1 about the x-axis (Gabriel’s horn) has a finite volume (π) but an infinite surface area, a result that astonished his contemporaries. Euler studied the gamma function in the 1730s, Cauchy introduced the principal value in the 1820s and the systematic theory of convergence was developed in the nineteenth century.

Where are they used?

Probability densities must integrate to 1 over infinite ranges (normal, exponential distributions). In physics, the energy needed to move a charge to infinity, the gravitational potential energy and the total radiation emitted are improper integrals: the work to escape from the Earth’s gravity, ∫_R^∞ GMm/r² dr = GMm/R, is finite, which gives the escape velocity. Integral transforms (Laplace, Fourier) are improper integrals.

Which mistakes are common?

Ignoring a singularity inside the interval

∫₋₁¹ dx/x² is not −2. Always check where the integrand is unbounded.

Thinking that a function tending to 0 has a finite integral

1/x → 0, but ∫₁^∞ dx/x diverges. The decay must be faster than 1/x.

Writing ∞ − ∞ = 0

For ∫_−∞^∞ x dx, each half diverges. Splitting symmetrically (principal value) gives 0, but the integral does not converge in the usual sense.

Mixing up the p-conditions

At infinity convergence needs p > 1. At 0 it needs p < 1. The conditions are opposite.