LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Logarithmic integral asymptotic expansion

Statement

For each fixed integer $m \geq 1$ , as $x \to \infty$ , $Li (x) = \sum_{j = 0}^{m - 1} \frac{j! x}{\log^{j + 1} x} + O_{m} (\frac{x}{\log^{m + 1} x}) .$

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Logarithmic integral: For real $x \geq 2$ , define $Li (x) = \int_{2}^{x} \frac{d t}{\log t} .$ In particular $Li (2) = 0$ . The integral never crosses the singularity at one.

Proof

1.1

Let $J_{k} (x) = \int_{2}^{x} (\log t)^{- k} d t$ . Integration by parts gives $J_{k} = x / \log^{k} x - 2 / \log^{k} 2 + k J_{k + 1}$ . Starting with $Li = J_{1}$ , apply this identity m times: the remainder is $m! J_{m + 1}$ and the lower-end constant is $- \sum_{j = 0}^{m - 1} 2 j! / \log^{j + 1} 2$ .

F1algebra

2.1

For $x \geq 4$ , split $J_{m + 1}$ at $\sqrt{x}$ . Its first part is at most $\sqrt{x} / (\log 2)^{m + 1}$ and its second at most $2^{m + 1} x / \log^{m + 1} x$ . The first bound and the fixed lower-end constant are also $O_{m} (x / \log^{m + 1} x)$ . This proves the expansion for each fixed m, including m=1.

step 1.1algebra∎

Depends on

Logarithmic integral

Used by

Prime number theorem Corollary
Prime number theorem arithmetic progressions Theorem

Dependency tree · one level

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Sources

§6.2, equation (6.15), pp.179–180 (standard reference, not scraped)