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The cubic moment of $Z(t)$ is also discussed, and a mean value result is presented which supports the author's conjecture that $$ \\int_1^TZ^3(t)\\,dt \\;=\\;O_\\varepsilon(T^{3/4+\\varepsilon}). $$"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1511.07140","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.NT","submitted_at":"2015-11-23T09:04:47Z","cross_cats_sorted":[],"title_canon_sha256":"a64b1be1a58a3c5789ce8255aaff1b3fc25a49b57bb9e63bb9b96402e2c834ec","abstract_canon_sha256":"ed12035208ff2989b6292f5faccfedd789d6a32bb1449e57ac2b25bd5d1dd2de"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:27:22.492707Z","signature_b64":"8bNpbalyQ4YLBX1WtWDsMYoEgCONbAQM105rsOuEMtkldGjSwyIf7pdWFb2EyncGN4HGYsMeGj5OwD4g2F9fAA==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"c977f2e08b8a63832a5c0a0c37f7870a6a520c3afa6507e244bc24bd932edb34","last_reissued_at":"2026-05-18T00:27:22.492111Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:27:22.492111Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"On a cubic moment of Hardy's function with a shift","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.NT","authors_text":"Aleksandar Ivi\\'c","submitted_at":"2015-11-23T09:04:47Z","abstract_excerpt":"An asymptotic formula for $$ \\int_{T/2}^{T}Z^2(t)Z(t+U)\\,dt\\qquad(0< U = U(T) \\le T^{1/2-\\varepsilon}) $$ is derived, where $$ Z(t) := \\zeta(1/2+it){\\bigl(\\chi(1/2+it)\\bigr)}^{-1/2}\\quad(t\\in\\Bbb R), \\quad \\zeta(s) = \\chi(s)\\zeta(1-s) $$ is Hardy's function. 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