Finding the value of $\sum_{c|n}\mu(c)\log^2(\frac n c)$

Question

Given $\log^2(n)=\sum_{dd'|n}\Lambda(d)\Lambda(d')+\sum_{d|n}\Lambda(d)\log d$, it has been written (in the article "A discussion of the fundamental ideas behind Selberg’s “Elementary proof of the prime-number theorem” bySteve Balady, page 9) that,

Letting $c = dd'$ and replacing $d$ with $c$ in the second sum, $$\log^2(n)=\sum_{c|n}(\sum_{d|n}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c)$$ Now, applying $(4.1)$ to this sum with respect to $c$, we conclude that $$\sum_{c|n}\mu(c)\log^2(\frac n c)=\sum_{d|n}\Lambda(d)\Lambda(\frac n d)+\Lambda(n)\log n).$$

Here, $(4.1)$ tells $\sum_{d|n}\mu(d) = 1 $ if $n=1$, otherwise the sum is $0$.

QUESTION:

How we get $$\sum_{c|n}\mu(c)\log^2(\frac n c)=\sum_{d|n}\Lambda(d)\Lambda(\frac n d)+\Lambda(n)\log n)$$ from $$\log^2(n)=\sum_{c|n}(\sum_{d|n}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c)$$?

ATTEMPT:

If we replace $n$ by $\frac n c$

$$\log^2(\frac n c)=\sum_{c|\frac n c}(\sum_{d|\frac n c}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c) \cdots (1)$$

, then multiply by $\mu(c)$, we get,

$$\mu(c) \log^2(\frac n c)=\mu(c)\sum_{c|\frac n c}(\sum_{d|\frac n c}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c) \cdots (2)$$

$$=\sum_{c|\frac n c}\mu(c)(\sum_{d|\frac n c}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c)$$

, then we sum up -

$$\sum_{c|n}\mu(c) \log^2(\frac n c)=\sum_{c|n}\sum_{c|\frac n c} \mu(c) (\sum_{d|\frac n c}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c)\cdots (3)$$

But I have no idea how $\sum_{c|n}\sum_{c|\frac n c} \mu(c) (\sum_{d|\frac n c}\Lambda(d)\Lambda(\frac c d)+\Lambda(c)\log c)$ becomes $\sum_{d|n}\Lambda(d)\Lambda(\frac n d)+\Lambda(n)\log n)$.

EDIT:

Note that Mobius inversion formula directly gives the result but the author refers to a different way (multiply by $\mu(c)$, then sum up), I would like to see how that derivation is obtained.

Answer 1

We show the equivalence of the identities \begin{align*} \log^2(n)&=\sum_{c|n}\left(\sum_{d|c}\Lambda(d)\Lambda\left(\frac{c}{d}\right)+\Lambda(c)\log c\right)\tag{1}\\ \sum_{c|n}\mu(c)\log^2\frac{n}{c}&=\left(\sum_{d|n}\Lambda(d)\Lambda\left(\frac{n}{d}\right)\right)+\Lambda(n)\log n\tag{2} \end{align*} by starting with OP's approach. Since we want to substitute $n$ with $\frac{n}{c}$ in (1), we replace $c$ with $q$ at the right-hand side of (1) in order to avoid naming conflicts.

We obtain from (1): \begin{align*} \log^2(n)&=\sum_{q|n}\left(\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right)+\Lambda(q)\log q\right)\\ \log^2\left(\frac{n}{c}\right)&=\sum_{q|\frac{n}{c}}\left(\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right)+\Lambda(q)\log q\right)\tag{3}\\ \sum_{c|n}\mu(c)\log^2\frac{n}{c} &=\sum_{c|n}\mu(c)\sum_{q|\frac{n}{c}}\left(\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right)+\Lambda(q)\log q\right)\tag{4}\\ \end{align*} where we substituted in (3) $n$ with $\frac{n}{c}$ and get (4) after multiplication with $\mu(c)$ and summing up over the positive integral divisors $c$ of $n$.

The challenge is now to show the equality of the RHS of (4) and (2): \begin{align*} \color{blue}{\sum_{c|n}\mu(c)\sum_{q|\frac{n}{c}}\left(\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right)+\Lambda(q)\log q\right) =\left(\sum_{d|n}\Lambda(d)\Lambda\left(\frac{n}{d}\right)\right)+\Lambda(n)\log n} \end{align*} which boils down to show the validity of \begin{align*} \sum_{c|n}\mu(c)\sum_{q|\frac{n}{c}}\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right) &=\sum_{d|n}\Lambda(d)\Lambda\left(\frac{n}{d}\right)\tag{5}\\ \sum_{c|n}\mu(c)\sum_{q|\frac{n}{c}}\Lambda(q)\log q&=\Lambda(n)\log n\tag{6} \end{align*}

In order to do so it is convenient to use the Dirichlet convolution operator $\ast$: \begin{align*} \left(f\ast g\right)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right) \end{align*} We also use the arithmetical functions \begin{align*} \zeta_{0}(n)&=1\qquad\qquad n\in\mathbb{N}\\ \delta(n)&= \begin{cases} 1\qquad&\ n=1\\ 0\qquad &\ \text{otherwise} \end{cases} \end{align*} where the notation is from Introduction to Arithmetical Functions by P.J. McCarthy.

We start with the easier part. We obtain \begin{align*} \color{blue}{\sum_{c|n}}&\color{blue}{\mu(c)\sum_{q|\frac{n}{c}}\Lambda(q)\log q}\\ &=\sum_{c|n}\mu(c)\left(\left(\Lambda\cdot\log\right) \ast\zeta_0\right)\left(\frac{n}{c}\right)\tag{7}\\ &=\left(\mu\ast\left(\left(\Lambda\cdot\log\right) \ast\zeta_0\right)\right)(n)\\ &=\left(\mu\ast\left(\zeta_{0}\ast\left(\Lambda\cdot\log\right) \right)\right)(n)\tag{8}\\ &=\left(\left(\mu\ast\zeta_{0}\right)\ast\left(\Lambda\cdot\log\right) \right)(n)\tag{9}\\ &=\left(\delta\ast\left(\Lambda\cdot\log\right) \right)(n)\tag{10}\\ &\,\,\color{blue}{=\left(\Lambda\cdot\log\right) (n)}\tag{11}\\ \end{align*} and the claim (6) follows.

Comment:

• In (7) we use the identity $\sum_{d|n}f(d)=\sum_{d|n}f(d)\zeta_0\left(\frac{n}{d}\right)=\left(f\ast \zeta_0\right)(n)$.

• In (8) we use the commutativity $f\ast g = g\ast f$ of the convolution operator.

• In (9) we use the associativity $(f\ast g)\ast h=f\ast(g\ast h)$.

• In (10) we recall that $\mu$ and $\zeta_0$ are inverse arithmetical functions: $\mu\ast\zeta_0=\delta=\zeta_0\ast\mu$.

• In (11) we use the identity $f\ast\delta = f = \delta\ast f$.

In the same way we can show the validity of (5):

We obtain \begin{align*} \color{blue}{\sum_{c|n}}&\color{blue}{\mu(c)\sum_{q|\frac{n}{c}}\left(\sum_{d|q}\Lambda(d)\Lambda\left(\frac{q}{d}\right)\right)}\\ &=\sum_{c|n}\mu(c)\sum_{q|\frac{n}{c}}\left(\Lambda\ast\Lambda\right)(q)\\ &=\sum_{c|n}\mu(c)\left(\left(\Lambda\ast\Lambda\right)\ast\zeta_0\right)\left(\frac{n}{c}\right)\\ &=\left(\mu\ast\left(\left(\Lambda\ast\Lambda\right)\ast\zeta_0\right)\right)(n)\\ &=\left(\mu\ast\left(\zeta_{0}\ast\left(\Lambda\ast\Lambda\right)\right)\right)(n)\\ &=\left(\left(\mu\ast\zeta_{0}\right)\ast\left(\Lambda\ast\Lambda\right)\right)(n)\\ &=\left(\delta\ast\left(\Lambda\ast\Lambda\right)\right)(n)\\ &=\left(\Lambda\ast\Lambda\right)(n)\\ &\,\,\color{blue}{=\sum_{d|n}\Lambda(d)\Lambda\left(\frac{n}{d}\right)} \end{align*} and the claim (5) follows.

Note that when using the Dirichlet convolution operator $\ast$ the equivalence of (1) and (2) can be written as \begin{align*} \log^2(n)&=\left(\left(\Lambda\ast\Lambda+\Lambda\cdot\log\right)\ast\zeta_{0}\right)(n)\tag{1'}\\ \left(\mu\ast\log^2\right)(n)&=\left(\Lambda\ast\Lambda+\Lambda\cdot\log\right)(n)\tag{2'} \end{align*}