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Lemma 15.1 in Montgomery-Vaughan’s analytic number theory book is Landau’s theorem for integrals. My question is, why is it necessary to have $A(x)$ bounded on every interval $[1,X]$? Doesn’t the requirement of $A(x)$ being Riemann integrable on those intervals suffice? The same theorem appears in Ingham’s book (The Distribution of Prime Numbers), but the requirement is made part of the definition of the Dirichlet integrals themselves. However, in Apostol’s analytic number theory book, this theorem appears as an exercise in Chapter 11, and the boundedness requirement is not there. The theorem also appears as exercise 2.5.19 in Murty’s book (Problems in Analytic Number Theory), and again, the boundedness requirement is missing here. I would like to understand why two of these four authors have this boundedness requirement as one of the conditions on the integrand for the theorem. Is it, or is it not necessary to have the boundedness condition?

Edit 18:16 UTC In the case of Ingham, local boundedness is used in the proof of theorem that is given. However, presumably, the proof can be carried out without this requirement? The full proof is not given in any of the other sources cited.

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    $\begingroup$ It's not necessary; it's merely sufficient. For all applications that the authors have in mind, $A(x)$ is bounded on every interval. $\endgroup$
    – Peter Humphries
    Commented Jan 26 at 15:00
  • $\begingroup$ @PeterHumphries Thank yo. But, regardless of the applications, if it is not necessary, can one omit this requirement entirely from the statement of the theorem or is it necessary to have something else in its place? And if so, what might that be, and why would the theorem not be viable with some extra condition? (I gather this might be the case given your comment about sufficiency?) I am still puzzled by the fact that, some authors write the statement of the theorem entirely without any requirement additional to Riemann integrability, while others add some requirement. $\endgroup$
    – EGME
    Commented Jan 27 at 11:11
  • $\begingroup$ @PeterHumphries I was thinking that, if A is non-negative from 1 on, then maybe you do not need the local boundedness requirement, and that this might only be needed if the constancy of sign comes later than 1 (the integrals in the theorem being from 1 to infinity). $\endgroup$
    – EGME
    Commented Jan 27 at 15:50
  • $\begingroup$ Riemann integrability on an interval implies boundness on that interval and Apostol's exercise specifically mentions Riemann integrability hence implicitly local boundness too; in Murty there are other conditions like finite variation which implies local boundness too; this being said I do not see any reason why one cannot substitute the monotone convergence theorem and just assume that $c$ satisfies the required integrability in the sense of Lebesgue - since the theorem is generally used for summatory functions coming from Dirichlet series which are locally bounded not sure why one would care $\endgroup$
    – Conrad
    Commented Feb 16 at 15:22

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