LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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Truncated integrals of rational powers

Statement

For rational $p \neq 1$ and $0 < A < B$ , $\int_{A}^{B} x^{- p} d x = \frac{B^{1 - p} - A^{1 - p}}{1 - p} .$ For $p = 1$ and every positive integer $N$ , $\int_{1}^{2^{N}} \frac{d x}{x} \geq \frac{N}{2}, \int_{2^{- N}}^{1} \frac{d x}{x} \geq \frac{N}{2} .$

Facts & Assumptions

Given: Positive endpoints $A < B$ and the stated rational exponent.

[L1]

Positive rational powers satisfy the exponent, product, and monotonicity laws (Rational powers $a^{r}$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L2]

Proper substitution and the FTC evaluate integrals from verified primitives (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[L3]

Integer-power differentiation, the product rule, and the reciprocal rule follow from the derivative algebra (Integer powers $a^{m}$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · computation

1.1

Write $p = m / q$ with an integer $m$ and a positive integer $q$ . Substituting $x = t^{q}$ on the positive interval and using [L1]–[L3] reduces the integrand to $q t^{q - m - 1}$ . The integer-power rule gives primitive $q t^{q - m} / (q - m)$ when $m \neq q$ . Substituting $t = x^{1 / q}$ back gives $x^{1 - p} / (1 - p)$ . The FTC proves the displayed formula.

L1L2L3

2.1

On $[2^{k}, 2^{k + 1}]$ , $1 / x \geq 2^{- (k + 1)}$ , so its integral is at least $1 / 2$ . Adding the first $N$ dyadic blocks proves the first lower bound. The intervals $[2^{- (k + 1)}, 2^{- k}]$ give the second in the same way.

given∎

Depends on

Used by

1/x² has no finite Cauchy principal value at zero Counterexample
|x-c|^-1/2 has a convergent improper integral across an interior singularity Example
∫₀¹ x^-1/2 dx=2 Example
The rational p-threshold reverses between zero and infinity Example
The improper p-test for rational exponents Theorem

Dependency tree · two levels

71 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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