Mertens function has, by residues, an explicit formula of
$M(x)=\displaystyle\sum_{\rho}\frac{x^\rho}{\rho\zeta'(\rho)}-2+\sum_{n=1}^\infty\frac{(-1)^{2 n}(2\pi)^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}$
where $\rho$ are the zeros of $\zeta(s)$, as usual.
Meanwhile, if we use this generalized identity for the number of divisors function, $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$, it's not much work to see that the Moebius function $\mu(n)$ is equal to $d_{-1}(n)$, and with $D_z(n) = \sum_{j=1}^n d_z(j)$, that $M(n) = D_{-1}(n)$.
Is there an explicit formula, similar to that of $M(n)$, above, for the more general case of $D_z(n)$ that the formula for $M(n)$ is a specialization of?
Some more detail, in response to Eric N.:
I understand that we can't use residues to get an explicit formula, for the reasons mentioned. But does that lead naturally to the idea that there isn't / couldn't be explicit formulas for $D_k(n), k>0$ that use the Zeta Zeros?
I want to make a visual, intuitive argument here. Here's an identity for $D_z(n)$ for complex z.
$\displaystyle D_z(n) = \frac{z^0}{0!}1+\frac{z^1}{1!}\sum_{j=2}^n \kappa(j)+\frac{z^2}{2!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor} \kappa(j) \kappa(k)+\frac{z^3}{3!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor}\sum_{l=2}^{\lfloor \frac{n}{j k} \rfloor} \kappa(j) \kappa(k) \kappa(l)+\frac{z^4}{4!}…$
where $\kappa(n) = \frac{\Lambda(n)}{\log n}$.
Define $\displaystyle P_k(n)=\sum_{j=2}^{n}\kappa(j) P_{k-1}(\lfloor \frac{n}{j} \rfloor)$ with $P_0(n)=1$, and restate that as
$\displaystyle D_z(n) = \frac{z^0}{0!}P_0(n)+\frac{z^1}{1!}P_1(n)+\frac{z^2}{2!}P_2(n)+\frac{z^3}{3!}P_3(n)+\frac{z^4}{4!}P_4(n)+…$
$P_k(n) = 0$ if $n < 2^k$, so only $\log_2 n$ terms are non-zero. This means, if you've computed those non-zero $P_k(n)$ terms, it's trivial to compute $D_z(n)$ for any z in $\log_2 n$ operations.
Now, use this identity to animate, in Mathematica, $\displaystyle\frac{(D_z(n)-1)}{z}$ over the range $z = 1$ to $z = -1$.
K[n_] := FullSimplify[MangoldtLambda[n]/Log[n]]
P[n_, k_] := P[n, k] = Sum[ K[j] P[Floor[n/j], k – 1], {j, 2, n}];P[n_, 0] := 1
DD[n_, k_] := Sum[ k^j/j! P[n, j], {j, 0, Log[2, n]}]
Animate[DiscretePlot[ (DD[n, z = Cos[k] ] – 1)/z, {n, 1, 100}], {k, 0, 2 Pi, .0001}]
What you'll see, if you watch this animation, is an animating line that starts as f(x)=(x-1), races down and at its fastest is the Riemann Prime Counting function right when z=0, and then finally comes to a halt at (1-Mertens Function), before it cycles back up – all in all, a nice gradual transformation between those three important functions.
I know it's only an appeal to visuals, but I feel like what's going on at $D_{-.2}(n)/-.2$ looks continuous with what's going on at $D_{.2}(n)/.2$. Here's a closer look at that.
Animate[DiscretePlot[(DD[n, z = Cos[k]*.2] – 1)/z, {n, 1, 400}], {k, 0, 2 Pi, .0001}]
I guess anything's possible, but it deeply offends my senses of symmetry to think the Zeta Zeros are accounting for the high frequency part of the line there, from -.2 to 0, and then in the blink of an eye, something else is accounting for what is almost the exact same high frequency information. Are my instincts wrong?
A Few More Notes About All This
That identity for $D_z(n)$ stems from Linnik's identity summed, inverted, and generalized a bit. There's a corresponding identity for $d_z(n)$ as well.
In the notation above, $P_1(n) = \Pi(n)$, the Riemann Prime Counting Function.
As was casually demonstrated above, $\displaystyle \lim_{z \to 0}\frac{D_z(n)-1}{z} = \Pi(n)$.
You can get this last result more easily by taking $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$ and noting that $\displaystyle \lim_{z \to 0}\frac{d_z(n)}{z} = \frac{\Lambda(n)}{\log n}$ except at 1, where the limit is infinity.
Best Answer
I can give a partial answer to this. My apologies if I am telling you what you already know.
First, let's review how the explicit formula for the Merten's function is derived. We have that $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}.$$
We have $$M(x) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^s}{s} dx,$$ and the residue theorem gives you the explicit formula you mentioned above. Note that the zeros of $\zeta$ become poles of $\zeta^{-1}$, so the two summations in your formula correspond to the nontrivial zeros and the trivial zeros of Riemann zeta function respectively. Also the $1/s$ term gives us a pole at $s=0$, giving us a residue of $1/\zeta(0) = -2$.
More generally, we can define a function $d_k(n)$ by the Dirichlet series given by $\zeta^k$:
$$\zeta(s)^k = \sum_{n=1}^\infty d_k(n) n^{-s}.$$
Then, for $k\geq 1$, $d_k(n)$ gives the number of representations of $n$ as a product of $k$ positive integers, given explicitly by your product formula above. In particular, $d_2(n)$ gives the number of positive divisors of $n$. Also, as you mentioned, $d_{-1}(n) = \mu(n)$, the Möbius function.
(For $k<-1$, I am have not checked whether my definition using the Dirichlet series will coincide with your product formula.)
Now you can attempt to find an explicit formula by carrying out the contour integral $$D_k(x) = \sum_{n<x} d_k(n) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \zeta(s)^k \frac{x^s}{s} dx.$$
Now we can see that the explicit formula you will get will differ greatly depending on whether $k$ is positive or negative. If $k$ is positive, the main term will come from the residue of the pole at $s=1$, and this main term will be of the form $xP(\log x)$, where $P$ is a polynomial of degree $k-1$. If $k$ is negative, you'll pick up residues at all the zeros of $\zeta$, as in the case of $k=-1$.