Is there an explicit formula connected to $(\log\zeta(s))^2$? Riemann used $\log\zeta(s)$ and, essentially, Perron's formula to find the explicit formula for his prime counting function, $\Pi(n)$:
$li(x)-\displaystyle\sum_{\rho}li(x^\rho)-\log 2-\int_{x}^{\infty}\frac{dt}{t(t^2-1)\log t}$
Is an explicit formula using $(\log\zeta(s))^2$ and Perron's formula to compute the partial sum $\Pi_2(n) = \displaystyle\sum_{j=2}^n\sum_{k=2}^{\lfloor \frac{n}{j} \rfloor}\frac{\Lambda(j)}{ \log j}\frac{\Lambda(k)}{\log k}$ known?
(and here, $\Lambda(n)$ is the Von Mangoldt function)
 A: Nathan - Maybe you are already aware that it is perhaps more natural to first try to get some kind of explicit formula for
$$\displaystyle\sum_{j=1}^n\sum_{k=1}^{\lfloor \frac{n}{j} \rfloor}\Lambda(j)\Lambda(k) = \sum_{j=1}^n (\Lambda \ast \Lambda)(j)$$
where $\Lambda \ast \Lambda$ is defined to be
$$(\Lambda \ast \Lambda)(n) = \sum_{d|n} \Lambda(d) \Lambda(n/d).$$
We have the Dirichlet series,
$$\left(\frac{\zeta'(s)}{\zeta(s)}\right)^2 = \sum_{n=1}^\infty (\Lambda \ast \Lambda)(n) n^{-s}.$$
Then for non-integer $x$, we have
$$\sum_{n<x} (\Lambda \ast \Lambda)(n) = \frac{1}{2 \pi i}  \int_{2-i\infty}^{2+i\infty} \left(\frac{\zeta'(s)}{\zeta(s)}\right)^2 \frac{x^s}{s}\,dx.$$
If we move the line of integration left towards $-\infty$, we should pick up all the residues and get some kind of explicit formula.  I think we get something like:
$$\sum_{n<x} (\Lambda \ast \Lambda)(n) = x \log{x} - (1+2\gamma)x + (\log 2\pi)^2 + \sum_{\rho} \operatorname{Res}_{s=\rho}{\left(\left(\frac{\zeta'(s)}{\zeta(s)}\right)^2 \frac{x^s}{s}\right)},$$
where the sum is over both the trivial and non-trivial zeros of $\zeta$.  This is an explicit formula, although unfortunately I think there is no easy way to explicitly write down the residues that occur at the zeta zeros.
