The formula this page uses
∫ₐ^∞ f(x) dx = lim[b→∞] ∫ₐᵇ f(x) dx ∫₀¹ f(x) dx = lim[a→0⁺] ∫ₐ¹ f(x) dx when f blows up at 0
← Back to calculators
Improper Integral Calculator
Solve improper integral problems with clear steps, notation, and a final check.
∫
Calculate without using AI.
Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
Choose an example or enter a formula.
Show your work Step-by-step method
- No account, AI request, or paid token is used.
Recent calculations
No calculations yet.
Calculus
Improper Integral Calculator explained
The short version
- An improper integral has an infinite limit, or an integrand that blows up somewhere in the interval.
- You do not integrate to infinity. You integrate to b and then let b run away.
- The answer is either a finite number, which is called converging, or infinite, which is called diverging.
What each part means
SymbolWhat it means
a, b — The finite stand-in limitA real number that replaces the infinity or the bad point while you integrate.
f(x) — The integrandCheck where it is undefined before choosing which endpoint to replace.
p — The exponent in x^(−p)The single number that decides convergence for the whole family of power integrals.
limit — The final limitA finite value means convergence. Infinity, or no limit at all, means divergence.
Show your work: a full example
- The integral∫₁^∞ 1/x² dx
- Replace infinity with blim[b→∞] ∫₁ᵇ x⁻² dx
- Antiderivative−1/x
- Evaluate from 1 to b(−1/b) − (−1/1) = 1 − 1/b
- Let b growas b → ∞, 1/b → 0
- The limit1 − 0 = 1, so the integral converges to 1
- Sanity checkthe region is infinitely long but gets thin fast enough that its total area is only 1
A second, different case
- A different case: the integrand blows up at an endpoint∫₀¹ 1/√x dx, which is undefined at x = 0
- Replace the bad endpoint with alim[a→0⁺] ∫ₐ¹ x^(−1/2) dx
- Antiderivative2√x
- Evaluate from a to 12√1 − 2√a = 2 − 2√a
- Let a shrink to 02 − 0 = 2, so it converges even though the curve is infinitely tall
- Now change the exponent∫₀¹ 1/x dx has antiderivative ln x, giving 0 − ln a
- Let a shrink againln a → −∞, so 0 − ln a → ∞ and this one diverges
Copy-ready example
∫₁^∞ 1/x² dx
The integral
Questions about the Improper Integral Calculator
Why does 1/x diverge when 1/x² converges?
Both shrink toward zero, but 1/x² shrinks fast enough that the leftover tail areas add to a finite total. 1/x is the exact borderline: its tail area from 1 to b is ln b, which grows without limit, just very slowly.
What is the p-test in one sentence?
For ∫₁^∞ x^(−p) dx you need p > 1 to converge, and for ∫₀¹ x^(−p) dx you need p < 1. Notice that the two conditions are opposite, which is why 1/√x converges near 0 but diverges out to infinity.
What if both endpoints are improper?
Split the integral at any convenient interior point and handle each half with its own limit. The whole thing converges only if both halves do; one divergent half makes the entire integral divergent.
Does diverging mean the calculation failed?
No, it is a real answer. It says the accumulated area grows without bound, which in a physics problem might mean an infinite total charge or an escape velocity that cannot be reached. Report it as divergent rather than as an error.