Key takeaways
- An improper integral has an infinite limit (or a curve that blows up).
- p-test: ∫₁^∞ 1/xᵖ dx = 1 ÷ (p − 1) when p > 1, and diverges when p ≤ 1.
- Never plug in ∞ like a number. Use a limit: stop at t, then let t grow.
- p = 1 diverges: the area under 1/x grows like ln t forever.
In this guide
What is an improper integral?
A normal definite integral has two finite limits, like 1 to 5. An improper integral has a limit of infinity (∞), or a curve that shoots up to infinity inside the interval. You can't stop at infinity, so you stop at a large number t, work out the area, and then watch what happens as t grows without end.
Picture the curve y = 1/x² from x = 1 onwards. It gets thin very fast. The tail is so thin that the total area adds up to exactly 1. But y = 1/x shrinks too slowly, and its area keeps growing forever.
Why do improper integrals matter?
Many important quantities add up something that never quite ends: a signal fading over time, a force weakening with distance, or the tail of a probability curve. Improper integrals tell you whether those endless sums give a sensible, finite answer.
The surprise is how thin the dividing line is. The curves 1/x and 1/x² both go to 0, yet one has infinite area and the other has area 1.
| Area from 1 to t | Under 1/x | Under 1/x² |
|---|---|---|
| t = 1,000 | about 6.91 | 0.999 |
| t = 1,000,000 | about 13.82 | 0.999999 |
| t → ∞ | grows forever | exactly 1 |
The area under 1/x grows so slowly that you'd need x beyond a million just to pass 14, but it never stops growing.
The formula (the p-test)
| Symbol | What it means | Example |
|---|---|---|
| p | The power in the bottom of the fraction (calculator needs p > 0) | 3 |
| 1 | The lower limit, where the area starts | x = 1 |
| ∞ | The upper limit: the region never ends | — |
| 1 ÷ (p − 1) | The exact area when p > 1 | 1 ÷ 2 = 0.5 |
If p ≤ 1 the integral diverges, and the calculator shows ∞.
How to evaluate an improper integral step by step
- 1Rewrite the fraction.
1/xᵖ = x⁻ᵖ. - 2Integrate with the power rule. You get
x¹⁻ᵖ ÷ (1 − p). If p = 1, the antiderivative isln xinstead. - 3Stop at t. Evaluate from 1 to t:
(t¹⁻ᵖ − 1) ÷ (1 − p). - 4Let t → ∞. If p > 1,
t¹⁻ᵖ → 0and the answer is1 ÷ (p − 1). If p ≤ 1,t¹⁻ᵖ(or ln t) grows forever, so the integral diverges.
Try a few powers yourself. Enter p below and see whether the area settles down or runs away.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Plugging in ∞ like a number. Always replace ∞ with t, then take the limit as t grows.
- Mixing up the direction of the test. Bigger p means a thinner tail, so p > 1 converges and p ≤ 1 diverges.
- Starting at 0. Near x = 0, 1/xᵖ blows up. That is a different kind of improper integral with the opposite rule (it converges when p < 1).
- Forgetting the sign.
1 ÷ (1 − p)is negative when p > 1, so write the answer as1 ÷ (p − 1).
Tips and tricks
- Compare with a known curve. If your function sits below a convergent 1/xᵖ, its area is finite too. This is the comparison test.
- Series link: the sum 1 + 1/2ᵖ + 1/3ᵖ + … converges for exactly the same p (p > 1). This is the integral test.
- Finite limits? For ordinary integrals, use the definite integral calculator.
Where you'll use it in real life
- Probability: the total area under the normal distribution curve, which stretches to infinity both ways, is exactly 1.
- Physics: the work to lift a rocket completely free of Earth's gravity is finite, because gravity weakens like 1/r².
- Engineering: the Laplace transform is an improper integral from 0 to ∞.
- Mathematics: testing whether infinite series add up to a finite number.
Quick summary
An improper integral handles an endless region by stopping at t and letting t grow. For ∫₁^∞ 1/xᵖ dx, the area is 1 ÷ (p − 1) when p > 1 and infinite when p ≤ 1. Use the calculator above to test any power and see the working.
Worked example
In physics, some forces weaken with distance like 1/x³. Does the total area under y = 1/x³ from x = 1 to infinity have a finite value, and what is it? Here p = 3.
- 01∫[1,∞] x^(−3) dx
- 02= [x^(−3+1) / (−3+1)]₁^∞
- 03= [x^-2 / -2]₁^∞
- 04As x→∞: x^-2 → 0 (since 2 > 0)
- 05= 0 − (1^-2 / -2)
- 06= 0 − (−1/2) = 1/2 = 0.5
- 07The integral CONVERGES to 0.5
These numbers are pre-filled in the calculator above.
Frequently asked questions
How do I know if an improper integral converges?
Replace ∞ with a variable t, integrate, and take the limit as t → ∞. If you get a finite number, it converges. For 1/xᵖ from 1 to ∞, it converges exactly when p > 1.
What is the integral of 1/x² from 1 to infinity?
It equals 1. The antiderivative is −1/x, which goes to 0 at infinity and equals −1 at x = 1, so the area is 0 − (−1) = 1.
Why does 1/x diverge but 1/x² converge?
Both go to 0, but 1/x² shrinks much faster. The area under 1/x from 1 to t is ln t, which grows forever, while the area under 1/x² is 1 − 1/t, which never passes 1.
What does the calculator show when the integral diverges?
For p between 0 and 1 (including p = 1) it shows ∞ and explains that the area grows without limit. It needs p > 0.
