Is $\frac1\pi \arctan \frac\pi{\ln x}- \frac1{\ln x}$ related to the trivial solutions $\zeta(-2n)$?

The Prime Counting Function $\pi(x)$ is given $$\pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x} ,$$ with $\operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})$.

(At least to me) It's interesting to see, that $\operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho})$ runs over the non-trivial roots of $\zeta$ and its pole at $1$. So my question is:

Is $\displaystyle \frac1\pi \arctan \frac\pi{\ln x}- \frac1{\ln x}$ related to the trivial solutions $\zeta(-2n)$?

I'm just guessing, but could this be related to $\displaystyle \sum_{n}R(x^{-2n})$, the sum over the trivial soultions?

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EDIT

Shortly : Yes, the 'remaining term' at the right is minus the sum on the trivial zeros : $-\sum_{n=1}^\infty \operatorname{R}(x^{-2n})$ so that the complete formula may be rewritten : $$\pi_0(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho})$$ with $\rho$ running over all the zeros of $\zeta$ (you may find affirmations that the non-trivial zeros only should be considered but I think they originated from a wrong interpretation of an approximate expression neglecting the trivial zeros...).

(for more about $\pi_0$, Gram series and $\zeta$ zeros see Matthew Watkins' entertaining page on the 'encoding' of the distribution of prime numbers)

The non-trivial zeros will be considered grouped by symmetrical pairs extending away from the real line (but even in this case I didn't see a proof of convergence) while the real zeros contribution is very regular and small for large $x$ so that your 'remaining term' is indeed often forgotten...

To obtain the expression for $\pi_0(x)$ let's start with Riemann's famous 'explicit formula' (the $\rho'$ are non-trivial zeros) : $$f(x)= \operatorname{li}(x)-\sum_{\rho'} \operatorname{li}(x^{\rho'}) -\log(2)+\int_x^\infty \frac{dt}{t\left(t^2-1\right)\log(t)}$$

$\displaystyle f(x)=\sum_{n=1}^\infty \pi_0\left(x^{\frac 1n}\right)$ so that we may use a special case of Möbius inversion formula to get the converse expression ($\mu$ is the Möbius function) :

$$\pi_0(x)=\sum_n \frac{\mu(n)}n f\left(x^{\frac 1n}\right)$$ and apply this transformation on $f$ to each of the terms at the right :

of course $\operatorname{li}(x)$ becomes the 'Gram series' $\operatorname{R}(x)$ given by : $$\operatorname{R}(x)=1+\sum_{m=1}^\infty \frac{\log(x)^m}{m!m \zeta(m+1)}= \sum_{n=1}^\infty \frac{\mu(n)}n \operatorname{li}\left(x^{\frac 1n}\right)$$

the same way $\displaystyle \sum_{\rho'} \operatorname{li}(x^{\rho'})$ becomes $\displaystyle \sum_{\rho'} \operatorname{R}(x^{\rho'})$

this identity of Möbius functions $\displaystyle \sum_{n=1}^\infty \frac{\mu(n)}n=0$ allows to eliminate the constant term.

At this point we should have : $$\pi_0(x)= \operatorname{R}(x)-\sum_{\rho'} \operatorname{R}(x^{\rho'}) + \sum_{n=1}^\infty \frac{\mu(n)}n\cdot\int_{x^{\frac 1n}}^\infty \frac{dt}{(t^2-1)t\log(t)}$$

we may now develop $\displaystyle \frac 1{t^2-1}$ as $\displaystyle \frac 1{t^2}+\frac 1{t^4}+\frac 1{t^6}+\cdots$ and get :

$$\int_x^\infty \frac{dt}{(t^2-1)t\log(t)}= -\left[\operatorname{li}(x^{-2})+\operatorname{li}(x^{-4})+\operatorname{li}(x^{-6})+\cdots\right]$$ (using the integral expression of the logarithmic integral or the exponential integral...)

so that the 'remaining term' of interest to you becomes (as you supposed) : $$\sum_{n=1}^\infty \frac{\mu(n)}n\cdot\int_{x^{\frac 1n}}^\infty \frac{dt}{(t^2-1)t\log(t)}= -\sum_{n=1}^\infty \operatorname{R}(x^{-2n})$$

Concerning the proposed closed form (which appears too in the rather good article 'Prime-counting function' from Wikipedia) you may consult the fine article of Riesel and Göhl from 1970 'Some calculations related to Riemann's prime number formula' (p976s).

All this should show that for $\rho$ running over all the zeros of $\zeta$ we have indeed : $$\pi_0(x)= \operatorname{R}(x)-\sum_{\rho} \operatorname{R}(x^{\rho})$$

(of course there should be a more direct derivation of this for example starting with Riemann's expression $f(x)$ written with trivial zeros...)

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+1 thanks, so $\operatorname{R}(x^{\rho_t}) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{\rho_t/n})$ with trvial roots $\rho_t=-2n$ is kind of a representation of $\frac1\pi \arctan \frac\pi{\ln x}- \frac1{\ln x}$? – draks ... Mar 14 '12 at 21:42
@draks: shortly yes. I'll update my answer a little later with a more detailed derivation. – Raymond Manzoni Mar 14 '12 at 21:58
Thanks again Raymond: I would be glad if you could have a look at this question on Two Representations of the Prime Counting Function as well. Cheers, draks... – draks ... Jan 4 '13 at 19:09
First many thanks for the very generous bounty @draks !! Of course this was not necessary since the subject itself is really beautiful (I had noticed your other question and wanted to investigate it this weekend...). Cheers ! – Raymond Manzoni Jan 4 '13 at 21:38