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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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The improper p-test for rational exponents

Statement

For every rational p, ∫1∞x−p dx converges exactly when p>1, and ∫01x−p dx converges exactly when p<1. When they converge, their values are respectively 1/(p−1) and 1/(1−p).

Facts & Assumptions

Given: A rational exponent p.

[L1]

The compact-truncation formula and the dyadic p=1 bounds are in Truncated integrals of rational powers.

[L2]

For rational s>0, monotonicity and the rational-power laws give Rs→∞ and R−s→0 as R→∞, and cs→0 and c−s→∞ as c↓0: in the finite-limit cases choose the thresholds R>ε−1/s and 0<c<ε1/s, and use the analogous threshold for divergence (Monotonicity of r↦ar and of a↦ar, Laws of rational exponents, Limits at +∞ and −∞, and infinite limits at a point, The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)).

[L3]

Nonnegative comparison transfers improper convergence (Comparison tests for improper integrals).

Proof

technique · cases
1.1

If p>1, then R1−p→0, so [L1] gives ∫1Rx−pdx→1/(p−1).

L1L2assume-case infinityhigh
1.2

If p≤1, the integral at infinity diverges: for 0<p<1 the formula in [L1] is unbounded by [L2], for p=1 use the first dyadic bound, and for p≤0 compare x−p≥1 with the constant one.

L1L2L3assume-case infinitylow
1.3

If p<1, then 1−p>0, so [L1] and c1−p→0 from [L2] give convergence at zero with value 1/(1−p).

L1L2assume-case zerolow
1.4

If p≥1, the zero-endpoint integral diverges: use the unbounded formula and [L2] for p>1, and the second dyadic bound for p=1.

L1L2assume-case zerohigh
2.1

The alternatives in steps 1.1–1.4 exhaust all rational exponents and establish both thresholds and values.

step 1.1step 1.2step 1.3step 1.4cases-exhaustive∎

Depends on

Used by

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Sources