I've been reading this paper for Merten's second theorem. Every thing is just fine but I think that theorem 11 (3.5.2) can have an easier proof. Is there any better approach or an explanation that why that $\ln([x]!)$ can be written in the form of those thetas.
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You can also use some insights from here. – rtybase Oct 08 '20 at 16:16
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It was well-known that the divisor sum of von Mangoldt function $\Lambda(n)=[n=p^k]\log p$ is logarithm: $$ \sum_{d|n}\Lambda(d)=\log n $$
Using identities of logarithm, we have
$$ \begin{aligned} \log\lfloor x\rfloor! &=\sum_{n\le x}\log n=\sum_{n\le x}\sum_{d|n}\Lambda(d) \\ &=\sum_{qd\le x}\Lambda(q)=\sum_{n\le x}\Lambda(n)\sum_{m\le x/n}1 \\ &=\sum_{n\le x}\Lambda(n)\left\lfloor\frac xn\right\rfloor \end{aligned} $$ Now, if we gather all the prime powers, then $$ \log\lfloor x\rfloor!=\sum_{p\le x}\left(\sum_{k\ge1}\left\lfloor x\over p^k\right\rfloor\right)\log p $$ By exchanging summation order, we have
$$ \begin{aligned} \log\lfloor x\rfloor! &=\sum_{k\ge1}\sum_{p\le x}\left\lfloor x\over p^k\right\rfloor\log p=\sum_{k\ge1}\sum_{p\le x}\sum_{m\le x/p^k}\log p \\ &=\sum_{mp^k\le x}\log p=\sum_{m,k\ge1}\sum_{p\le\sqrt[k]{x/m}}\log p \\ &=\sum_{m,k\ge1}\vartheta\left(\sqrt[k]\frac xm\right) \end{aligned} $$
where $\vartheta(x)=\sum_{p\le x}\log p$.
TravorLZH
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1Seems like that your first line has a little problem.There should $d$ as the input of the Mangoldt function. Can I you please correct it so others can go through reading it more confidently ? – Parsa Noori Oct 08 '20 at 13:22
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1You're welcome. Forgot to say that it's on the second mathematical line too. – Parsa Noori Oct 08 '20 at 14:17
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