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)


1 Answer 1


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.


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