Asymptotic formula in Analytic Number Theory

Hi,

Could anyone tell me in what sense the following is an "asymptotic formula":

Theorem 1 from

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.mmj/1029000373

(this is open access, so I think I'm allowed to link it)

At the moment, I'm just trying to understand what the bit with $Q\leq x$ means. I assume the Theorem says something like: the LHS varies like the main term, in that LHS/[main term] tends to one as x tends to infinity. But it's not clear to me that [error terms]/[main terms] tend to zero. There seems to be definitely something I'm not understanding with the relationship between $Q$ and $x$.

Can anyone clarify?

Thanks very much.

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 I answered your question, then I edited it and realized that half of it was missing. Next time please use TeX and post your question only when it appears in full as you intended it. Also, your questions are not of research level I am afraid. – GH Dec 26 at 2:37 I added more detail in a comment below. – GH Dec 26 at 19:18

There is no Theorem 1 in this paper. The statement called "Theorem" contains an asymptotic formula: e.g. (2) has a main term $Qx\log x$ and two error terms $O(\dots)$. The error terms are $o(Qx\log x)$ for $x (\log x)^{-A} < Q < x$ and $x\to\infty$, say, hence in this case the left hand side is asymptotic to the main term.

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 Oh ye it's just "Theorem", sorry. I will use TeX in the future. As for the level of the question, I didn't realise it had to be "research level". I thought it was ok, but maybe not then. (I see threads such as "What should be learned in an Analytic Number Theory course", which, whilst more useful than my thread, is not "research level".) Thanks for your answer. Can I ask what exactly is the limit taken over? I know we say $\rightarrow \infty$ but what exactly happens with $Q$ when we do that. I don't really understand $x(\log x)^{-A}Q\leq x$. – Tomos Parry Dec 26 at 11:25 Actually, it's ok, your answer has been enough. Thank you. – Tomos Parry Dec 26 at 11:31 @Tomos: I meant that the error terms are $o(Qx\log x)$ as $x\to\infty$, uniformly in $Q$ satisfying $x(\log x)^{-A}0$ there is $x_0=x_0(\epsilon,A)$ such that for $x>x_0$ and \$x(\log x)^{-A}