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)


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.