# Questions on Riemann Hypothesis prediction and explicit formula for $\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}$

I've primarily been able to find information on the the summatory Euler totient function $$\Phi(x)$$ defined in (1) below, but I'm more interested in the $$\hat{\Phi}(x)$$ defined in (2) below.

(1) $$\quad\Phi(x)=\sum\limits_{n\le x}\phi(n)\,,\qquad\frac{\zeta(s-1)}{\zeta(s)}=\sum\limits_{n\le x}\frac{\phi(n)}{n^s}\quad\Re(s)>2$$

(2) $$\quad\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}\,,\qquad\frac{\zeta(s)}{\zeta(s+1)}=\sum\limits_{n\le x}\frac{\phi(n)}{n^{s+1}}\quad\Re(s)>1$$

The following figure illustrates the $$\hat{\Phi}(x)$$ function defined in (2) above seems to exhibit a near-linear growth analogous the the first and second Chebyshev functions $$\vartheta(x)$$ and $$\psi(x)$$.

Figure (1): Illustration of formula (2) for $$\hat{\Phi}(x)$$

Question (1): What does the Riemann hypothesis predict for the asymptotic growth and error bound with respect to the $$\hat{\Phi}(x)$$ function defined in (2) above?

I've been told the explicit formula for $$\Phi(x)$$ defined in (3) below doesn't converge, but I'm wondering if there's one that converges for $$\hat{\Phi}(x)$$.

(3) $$\quad\Phi_o(x)=\frac{3\,x^2}{\pi^2}+\sum\limits_\rho\frac{x^\rho\,\zeta(\rho-1)}{\rho\,\zeta'(\rho)}+\frac{1}{6}+\sum\limits_{n}\frac{x^{-2\,n}\,\zeta(-2\,n-1)}{(-2\,n)\,\zeta'(-2\,n)}$$

Question (2): Is there a convergent explicit formula for the $$\hat{\Phi}(x)$$ function defined in (2) above and if so, what is the definition?

This question is motivated in part by the relationship between $$f(x)$$ and $$\hat{\Phi}(x)$$ illustrated in (6) below which leads to the formula for the Dirichlet eta function $$\eta(s)$$ defined in (7) below which I believe converges for $$\Re(s)>1$$. Note $$a_n$$ is 2-periodic (1, -3, 1, -3, ...).

(4) $$\quad f(x)=\sum\limits_{n\le x}(-1)^{n-1}\qquad\eta(s)=\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\,,\quad\Re(s)>0$$

(5) $$\quad\hat{\Phi}(x)=\sum\limits_{n\le x}\frac{\phi(n)}{n}\qquad\quad\frac{\zeta(s)}{\zeta(s+1)}=\sum\limits_{n=1}^\infty\frac{\phi(n)}{n^{s+1}}\,,\quad\Re(s)>1$$

(6) $$\quad f(x)=\sum\limits_{n\le x}\frac{a_n}{n}\hat{\Phi}\left(\frac{x}{n}\right)\,,\qquad a_n=(-1)^{n-1}((2 n-1)\bmod 4)$$

(7) $$\quad\eta(s)=\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\sum\limits_{m=1}^\infty\frac{\phi(m)}{m}\,m^{-s}\,,\quad\Re(s)>1$$

Formula (7) above can be analytically continued as illustrated in formulas (8) and (9) below which confirm the correctness of relationship (6) above. In formulas (8) and (9) below the sum over $$m$$ in formula (7) above is analytically continued to $$\frac{\zeta(s)}{\zeta(s+1)}$$, and in formula (9) below the sum over $$n$$ in formula (7) above is also analytically continued to $$\left(1-2^{1-s}\right)\zeta(s+1)$$.

(8) $$\quad\eta(s)=\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\frac{\zeta(s)}{\zeta(s+1)}\,,\quad\Re(s)>0$$

(9) $$\quad\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$$

Note formula (8) above is equivalent to formula (10) below.

(10) $$\quad\zeta(s+1)=\frac{1}{1-2^{1-s}}\sum\limits_{n=1}^\infty\frac{a_n}{n^{s+1}}\,,\quad\Re(s)>0\land s\ne 2\,\pi\,i\,t\land t\in\mathbb{Z}$$

• No. Look at the residue theorem, at the estimates of $F(s)x^{-s}$ in each part, the problem of traversing the critical strip, the proofs of the explicit formula for $\sum_{p^k \le x} \log p$, look at those for $\sum_{n \le x} \mu(n)$ and their complication, see why it doesn't work for $\sum_{n \le x} \phi(n) n^{-k}$. – reuns Jul 1 at 3:11
• @reuns Ok, I'm disappointed there's no convergent explicit formula, but do these problems also rule out a Riemann hypothesis prediction? – Steven Clark Jul 1 at 3:48
• @StevenClark, the problem is that the error term for these functions is not dominated by the contribution from the zeroes, but rather an "arithmetic" part that is much bigger. As reuns mentioned, the issue is shifting the contour; one cannot bound the contours satisfactorily. See also this paper: sciencedirect.com/science/article/pii/S0022314X10001873 – Peter Humphries Jul 2 at 9:24
• @PeterHumphries The Nicolas criterion implies $\hat{\Phi}_n(N_k)-\hat{\Phi}_n(N_k-1)<\frac{\log(\log(N_k))}{e^{\gamma}}$ where $N_k$ is the product of the first $k$ primes is equivalent to the Riemann Hypothesis. I guess one can't do any better than that. – Steven Clark Jul 2 at 17:07