Math · Calculus II · Worked example
Decide whether an improper integral converges
Decide whether the integrals of 1/x and of 1/x³ from 1 to ∞ converge, and evaluate any that do.
The integral of 1/x
The integral from 1 to b of dx/x is ln b, which grows without bound as b → ∞: the integral diverges, even though 1/x → 0.
The integral of 1/x³
An antiderivative of 1/x³ is −1/(2x²), so the integral from 1 to b is 1/2 − 1/(2b²), which tends to 1/2. At b = 10 it is already 0.495.
Compare with the p rule
Both are p-integrals: p = 1 diverges, and p = 3 > 1 converges to 1/(p − 1).
Use comparison
For x ≥ 1, 1/(x³ + 1) ≤ 1/x³, so the integral of 1/(x³ + 1) from 1 to ∞ also converges, to a value below 1/2, with no antiderivative needed.
Result
The integral of 1/x from 1 to ∞ diverges; the integral of 1/x³ from 1 to ∞ converges to 1/2.
Your turn
Does the integral of 1/√x from 1 to ∞ converge?
Show the answer and explanation
No: it diverges.
It is a p-integral with p = 1/2 ≤ 1. Indeed the integral from 1 to b is 2√b − 2, which grows without bound.
Keep exploring
Limits & one-sided behavior verifies that 1/2 − 1/(2b²) tends to 1/2; change the function to ln b to see it grow without bound.
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