Is this a valid application of the Möbius Inversion Formula:
Define: $$\psi\left(x\right) = \sum\limits_{p^k \le x} \log p$$
So that: $$\log x! = \sum\limits_{k=1}^{\infty}\psi\left(\frac{x}{k}\right)$$
Then, applying the Möbius Inversion Formula gives:
$$\psi\left(x\right) = \sum_{k=1}^{\infty}\mu\left(k\right)\log\left(\left\lfloor\frac{x}{k}\right\rfloor!\right)$$
I'm thinking that the logic used here applies so that:
$$ \begin{align} \sum_{k \geq 1} \mu\left(k\right) \log\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) &= \sum_{k \geq 1} \mu\left(k\right) \sum_{l \geq 1} \psi \left(\frac{x}{kl}\right) \\ &= \sum_{n \geq 1} \sum_{k \mid n}\mu\left(k\right)\psi\left(\frac{x}{n}\right) \\ &= \sum_{n \geq 1} \left( \delta_{1,n}\right)\psi\left(\frac{x}{n}\right) \\ &= \psi(x) \end{align}$$
where $\delta_{i,j} = \begin{cases}1&\mbox{ if } i=j\\0&\mbox{ if } i \ne j\end{cases}$
Is that right?
Thanks very much,
-Larry
Yes this is correct, also note that
$$\psi(x)=\sum_{n\leq x}\Lambda(n)=\sum_{n\leq x}\mu(n)*\ln(n)=\sum_{n\leq x}\mu(n)\sum_{k\leq \frac{x}{n}}\ln(k)=\sum_{n\leq x}\mu(n)\log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right)$$
You can also re-write this as, $$\psi(x)=\sum_{n\leq x}-\mu(n)\ln(n)\left\lfloor\frac{x}{n}\right\rfloor$$
And also note that $$\sum_{n=1}^\infty-\mu(n)\ln(n)\frac{1}{n}=\lim_{s\to1}-\frac{\zeta'(s)}{\zeta(s)^2}=1$$
Which gives a strong heuristic for the fact that,
$\psi(x)\sim x$
Which is equivalent to the prime number theorem.