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,



1 Answer 1


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.

  • $\begingroup$ Strong heuristic is still a big understatement. $\endgroup$
    – vitamin d
    May 17, 2021 at 19:59

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