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Let $$ \theta(x) = \sum\limits_{p \leq x} \log{p} \quad \ ; \ \psi(x)=\sum\limits_{n=1}^{\infty} \theta(x^{1/n})$$ then how does one prove $$e^{\psi(x)}= \text{lcm}[ 1,2,\cdots , \lfloor{x\rfloor}]$$

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up vote 4 down vote accepted

$e^{\psi(x)} = \exp({\displaystyle \sum_{n=1}^{\infty} \theta(x^{1/n})}) = \displaystyle \Pi_{n=1}^{\infty} \exp(\theta (x^{1/n})) = \displaystyle \Pi_{n=1}^{\infty} \exp(\displaystyle \sum_{p \leq x^{1/n}} \log(p))$

$ = \Pi_{n=1}^{\infty} (\Pi_{p \leq x^{1/n}} p) = (\Pi_{p \leq x} p) \times (\Pi_{p \leq x^{1/2}} p) \times (\Pi_{p \leq x^{1/3}} p) ... = lcm[1,2,3,...,[x]]$

$\textbf{EDIT:}$ To make the last equality a bit more clear, the highest power of $p$ dividing $lcm[1,2,3,...,[x]]$ is given by $[\log_p([x])]$

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