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The improper $p$ -test for rational exponents

Statement

For every rational $p$ , $\int_1^\infty x^{-p}\,dx\ \text{converges exactly when }p>1,$ and $\int_0^1 x^{-p}\,dx\ \text{converges exactly when }p<1.$ 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 $R^s\to\infty$ and $R^{-s}\to0$ as $R\to\infty$ , and $c^s\to0$ and $c^{-s}\to\infty$ as $c\downarrow0$ : in the finite-limit cases choose the thresholds $R>\varepsilon^{-1/s}$ and $0<c<\varepsilon^{1/s}$ , and use the analogous threshold for divergence (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Laws of rational exponents, Limits at $+\infty$ and $-\infty$ , and infinite limits at a point, The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ ).

[L3]

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

Proof

technique · cases

1.1

If $p>1$ , then $R^{1-p}\to0$ , so [L1] gives $\int_1^R x^{-p}dx\to1/(p-1)$ .

L1L2assume-case infinityhigh

1.2

If $p\le1$ , 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 $p\le0$ compare $x^{-p}\ge1$ with the constant one.

L1L2L3assume-case infinitylow

1.3

If $p<1$ , then $1-p>0$ , so [L1] and $c^{1-p}\to0$ from [L2] give convergence at zero with value $1/(1-p)$ .

L1L2assume-case zerolow

1.4

If $p\ge1$ , 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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 116 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

William F. Trench, Introduction to Real Analysis, Examples 3.4.1–3 (standard reference, not scraped)

The improper ppp-test for rational exponents