TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-24 (gpt-6-sol)

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Prime number theorem logarithmic integral

Statement

Assume countable choice.

For some absolute $c > 0$ and every $x \geq 2$ , $π (x) = Li (x) + O (x e^{- c \sqrt{\log x}}) .$

Facts & Assumptions

Given: Countable choice and the data and hypotheses of the statement.

[F1]

Chebyshev theta prime number theorem error: For some absolute $c > 0$ and all $x \geq 2$ , $θ (x) = x + O (x e^{- c \sqrt{\log x}}) .$

[F2]

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.

[F3]

Abel summation recovers the prime-counting function from theta: For every real $x \geq 2$ , $π (x) = \frac{θ (x)}{\log x} + \int_{2}^{x} \frac{θ (t)}{t \log^{2} t} d t .$

Proof

1.1F2F3

Set $E (t) = θ (t) - t$ . The exact partial-summation identity gives $π (x) = x / \log x + \int_{2}^{x} d t / \log^{2} t + E (x) / \log x + \int_{2}^{x} E (t) / (t \log^{2} t) d t$ . Integration by parts in the definition of Li makes its main term $Li (x) + 2 / \log 2$ .

2.1F1step 1.1∎

Under countable choice, use the supplied theta error and, for $x \geq 4$ , split the error integral at $\sqrt{x}$ . The initial part is $O (\sqrt{x})$ because $E (t) = O (t)$ . The second is $O (x e^{- (c_{0} / \sqrt{2}) \sqrt{\log x}})$ using the theta error and $\log t \geq \log 2$ . The endpoint error has the same form. Decrease the positive exponent constant and absorb $2 / \log 2$ and the compact range $2 \leq x \leq 4$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Theorem 6.9, equation (6.14) and proof, pp.179–181 (standard reference, not scraped)