Guides / Integrals

Improper integrals: replace infinity with a limit, and check for blow-ups inside

An integral is improper when a limit is infinite or the integrand blows up somewhere on the interval. You can't plug ∞ into an antiderivative, and you can't integrate straight across a vertical asymptote. Both are handled the same way: stop just short of the trouble, integrate, and take a limit.

Short answer

Infinite limit: ∫_a^∞ f(x) dx = lim_b→∞ ∫_a^b f(x) dx. Blow-up at an endpoint: stop at a + ε and let ε → 0⁺. Blow-up inside the interval, or both ends infinite: split into pieces, one problem per piece, and take each limit separately. The integral converges if every limit is a finite number and diverges if any one isn't. The benchmark is the p-test: ∫₁^∞ dx/x^p converges only for p > 1, and ∫₀¹ dx/x^p only for p < 1.

Why it works

The fundamental theorem, F(b) − F(a), needs a finite interval and an integrand that is continuous on all of it. An improper integral breaks one of those, so it is defined as a limit of ordinary integrals that don't. Integrate from 1 to b, which is a normal definite integral for every b, and then ask what happens as b grows.

An infinitely long region can still have a finite area, if it gets thin fast enough. 1/x² shrinks fast enough and the area under it from 1 to ∞ is exactly 1. 1/x shrinks too, but too slowly, and its area grows without limit. Near 0 it's the other way round: a small power such as 1/√x blows up slowly enough to leave a finite area.

Example 1: an infinite upper limit

Evaluate ∫₀^∞ e^−2x dx.

lim_b→∞ ∫₀^b e^−2x dxReplace ∞ with b. Now it's an ordinary integral.
= lim_b→∞ [−½e^−2x]₀^b = lim_b→∞ (½ − ½e^−2b)The antiderivative of e^−2x is −½e^−2x.
= ½e^−2b → 0 as b → ∞. The limit is finite, so the integral converges.

Example 2: one that diverges

Does ∫₁^∞ 1√x dx converge?

∫₁^b x^−1/2 dx = [2√x]₁^b = 2√b − 2The power rule, with b in place of ∞.
2√b − 2 → ∞, so the integral divergesThis is p = ½ in the p-test, and ½ isn't greater than 1. The integrand goes to 0, but not fast enough.

Example 3: the integrand blows up at an endpoint

Evaluate ∫₀⁸ x^−1/3 dx.

lim_ε→0⁺ ∫_ε⁸ x^−1/3 dxx^−1/3 = 1/∛x is infinite at x = 0, so start just to the right of 0.
= lim_ε→0⁺ [32x^2/3]_ε⁸ = lim_ε→0⁺ (6 − 32ε^2/3)8^2/3 = 4, and 32 · 4 = 6.
= 6It converges: this is p = ⅓ near 0, and ⅓ is less than 1.

Example 4: a blow-up in the middle

Evaluate ∫₀³ 1(x − 1)^2/3 dx.

∫₀³ = ∫₀¹ + ∫₁³The denominator is 0 at x = 1, inside the interval. Split there; each piece is improper at one end.
An antiderivative is 3∛(x − 1)The power rule on (x − 1)^−2/3. It is continuous at x = 1, so both one-sided limits exist.
∫₀¹ = 0 − 3∛(−1) = 3, ∫₁³ = 3∛2 − 0 = 3∛2Each piece converges on its own.
3 + 3∛2 ≈ 6.780Both pieces are finite, so the whole integral converges to their sum.

Common mistakes, and the exact line they happen on

1. Integrating straight through a vertical asymptote

Evaluate ∫₀² 1(x − 1)² dx.

An antiderivative of 1/(x − 1)² is −1/(x − 1)Right so far.
[−1x − 1]₀² = −1 − 1 = −2This is the mistake. The integrand is positive everywhere, so a negative answer is impossible. It blows up at x = 1, inside [0, 2], so the integral has to be split there. The right piece is ∫ from 1 + ε to 2, which equals 1/ε − 1 and goes to ∞: the integral diverges. Example 4 looked the same and converged; only splitting tells you which you have.

2. Deciding it converges because the integrand goes to 0

Does ∫₁^∞ 1x dx converge?

1x → 0 as x → ∞, so the area is finiteThis is the mistake. If the integrand doesn't go to 0 the integral certainly diverges, but going to 0 isn't enough. ∫ from 1 to b of 1/x is ln b, which is about 6.908 at b = 1000 and keeps growing without limit. The integral diverges; 1/x is the borderline case p = 1.

3. Taking a symmetric limit when both ends are infinite

Evaluate ∫_−∞^∞ x dx.

∫_−b^b x dx = 0 for every bTrue: x is odd, so the two sides cancel.
So ∫_−∞^∞ x dx = lim_b→∞ 0 = 0This is the mistake. Both ends have to go to infinity separately. Split at 0: ∫ from 0 to b of x is b²/2, which goes to ∞, so the integral diverges. Cutting off at −b and 2b instead gives 3b²/2, not 0. When different ways of approaching infinity give different answers, none of them is the value.

Practice

Evaluate ∫₁^∞ 1x⁴ dx.
Show answer
lim (13 − 13b³) = 13. It converges, as p = 4 > 1 says it should.
Evaluate ∫₀^∞ e^−x/3 dx.
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[−3e^−x/3]₀^b = 3 − 3e^−b/3 → 3.
Evaluate ∫₀⁴ 1√x dx.
Show answer
Improper at 0: [2√x]_ε⁴ = 4 − 2√ε → 4.
Does ∫₀¹ 1x³ dx converge?
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No. ∫_ε¹ x⁻³ dx = 12ε² − 12 → ∞. Near 0 the p-test needs p < 1, and here p = 3.
Evaluate ∫₂^∞ 1x(ln x)² dx.
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Let u = ln x, du = dx/x: [−1ln x]₂^b → 1ln 2 ≈ 1.443.