# Questions on Convergence of Explicit Formulas for $f(x)=\sum\limits_{n=1}^x a(n)$ where $a(n)\in\{\left|\mu(n)\right|,\mu(n),\phi(n),\lambda(n)\}$

This question is a follow-on to my earlier question at the following link.

What is the explicit formula for $\Phi(x)=\sum\limits_{n=1}^x\phi(n)$?

This question pertains to the explicit formulas for the following four functions where $\mu(n)$ is the Möbius function, $\phi(n)$ is the Euler totient function, and $\lambda(n)$ is the Liouville function. Also note $M(x)$ is the Mertens function.

(1) $\quad Q(x)=\sum\limits_{n=1}^x\left|\mu(n)\right|\,,\qquad \frac{\zeta(s)}{\zeta(2\,s)}=\sum\limits_{n=1}^\infty\frac{\left|\mu(n)\right|}{n^s}$

(2) $\quad M(x)=\sum\limits_{n=1}^x \mu(n)\,,\qquad \frac{1}{\zeta(s)}=\sum\limits_{n=1}^\infty\frac{\mu(n)}{n^s}$

(3) $\quad\Phi(x)=\sum\limits_{n=1}^x \phi(n)\,,\qquad \frac{\zeta(s-1)}{\zeta(s)}=\sum\limits_{n=1}^\infty\frac{\phi(n)}{n^s}$

(4) $\quad L(x)=\sum\limits_{n=1}^x \lambda(n)\,,\qquad \frac{\zeta(2\,s)}{\zeta(s)}=\sum\limits_{n=1}^\infty\frac{\lambda(n)}{n^s}$

I've read the explicit formulas for the four functions defined above are as follows.

(5) $\quad Q_o(x)=\frac{6\,x}{\pi^2}+\sum\limits_\rho\frac{x^{\frac{\rho}{2}}\,\zeta\left(\frac{\rho}{2}\right)}{\rho\,\zeta'\rho)}+1+\sum\limits_{n=1}^N\frac{x^{-n}\,\zeta(-n)}{(-2\,n)\,\zeta'(-2\,n)}\,,\quad N\to\infty$

(6) $\quad M_o(x)=\sum\limits_\rho\frac{x^{\rho}}{\rho\,\zeta'(\rho)}-2+\sum\limits_{n=1}^N\frac{x^{-2\,n}}{(-2\,n)\,\zeta'(-2\,n)}\,,\quad N\to\infty$

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

(8) $\quad L_o(x)=\frac{\sqrt{x}}{\zeta\left(\frac{1}{2}\right)}+\sum\limits_\rho\frac{x^\rho\,\zeta(2\,\rho)}{\rho\,\zeta'(\rho)}+1$

The four explicit formulas defined in (5) to (8) above are illustrated in the following four plots in orange and the corresponding reference functions defined in (1) to (4) above are illustrated in blue. All plots are evaluated over the first 200 pairs of zeta zeros and the sum over $n$ is also evaluated with the upper limit $N=200$. The red discrete portions of the plots illustrate the evaluations of the explicit formulas at integer values of $x$.

$\text{Figure (1): Illustration of$Q_o(x)$}$

$\text{Figure (2): Illustration of$M_o(x)$}$

$\text{Figure (3): Illustration of$\Phi_o(x)$}$

$\text{Figure (4): Illustration of$L_o(x)$}$

I initially thought perhaps the formula for $\Phi_o(x)$ was wrong as it seemed to exhibit a different convergence than the the formulas for $Q_o(x)$ and $M_o(x)$ which was the motivation for my earlier question, but I subsequently noticed the formulas for $Q_o(x)$ and $M_o(x)$ also seem to exhibit slightly different convergences. Note $Q_o(x)$, $M_o(x)$, and $\Phi_o(x)$ illustrated in figures (1), (2), and (3) above seem to converge for $x>b$, $x>c$, and $x>1$ respectively where $0<b<c<1$. I'm now trying to understand why explicit formulas such as $Q_o(x)$, $M_o(x)$, and $\Phi_o(x)$ seem to exhibit different lower convergence bounds.

Question (1): Is there a simple explanation as to what determines the lower convergence bound with respect to $x$ of explicit formulas such as $Q_o(x)$, $M_o(x)$, and $\Phi_o(x)$?

Question (2): Is there an explicit formula analogous to those above that actually converges for $x>0$?

Note the explicit formula $L_o(x)$ illustrated in Figure (4) above doesn't seem to converge.

Question (3): Is the explicit formula $L_o(x)$ defined in (8) above incorrect and if so, what is the correct explicit formula for $L(x)$?

• Didn't you post a piece of this question recently? – Gerry Myerson Apr 15 '18 at 1:41
• Ah, yes, here it is: math.stackexchange.com/questions/2731806/… – Gerry Myerson Apr 15 '18 at 1:42
• @GerryMyerson Yes, I originally thought perhaps the explicit formula for $\Phi(x)$ was wrong as it seemed to exhibit a different convergence than the others. Since an answer to my earlier question indicates the formula is correct, I'm now trying to understand what determines the differences in convergence between different explicit formulas. – Steven Clark Apr 15 '18 at 1:47
• Fine, but since the two questions are so closely related, I think you really ought to have linked them. – Gerry Myerson Apr 15 '18 at 5:38
• The explicit formula for $\sum_{n \leq x} \varphi(x)$ is wrong; when you shift the contour, the shifted contour integral is not small. One can use this to show that the error term for this sum is at least as large as a constant multiple of $x\sqrt{\log \log x}$ infinitely often. – Peter Humphries Apr 17 '18 at 10:08

you would need to introduce a test function $f(x)$ to make it convergent for example

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\gamma}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x}$$

Also for the Liouville function we have

$$\sum_{n=1}^{\infty} \frac{\lambda(n)}{\sqrt{n}}g(\log n) = \sum_{\gamma}\frac{h( \gamma)\zeta(2 \rho )}{\zeta '( \rho)}+ \frac{1}{\zeta (1/2)}\int_{-\infty}^{\infty}dx g(x)$$

For the Euler-Phi function the explicit formula reads

$$\sum_{n=1}^{\infty} \frac{\varphi(n)}{\sqrt{n}}g(\log n)= \frac{6}{\pi ^2} \int_{-\infty}^{\infty}dx g(x)e^{3x/2}+ \sum_{\gamma}\frac{h( \gamma)\zeta(\rho/2 )}{\zeta '( \rho)}+\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}\frac{\zeta (-2n-1)}{\zeta ' (-2n)}dx g(x) e^{-x(2n+1/2}$$ for the square-free function

$$\sum_{n=1}^{\infty} \frac{|\mu(n)|}{n^{1/4}}g(\log n)= \frac{6}{\pi ^2} \int_{-\infty}^{\infty}dx g(x)e^{3x/4}+ \sum_{\gamma}\frac{h( \gamma)\zeta(\rho -1 )}{\zeta '( \rho)}+ \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta (-n)}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dx g(x)e^{-x(n+1/4)}$$

here $g(x)$ and $H(x)$ form a fourier transform pair , these formulae are generalizatons of possion summation formula

$\rho = 1/2+i\gamma$

• I've noticed several discrepancies between your answer above and a couple of your papers. For the Liouville function, should $\frac{1}{\zeta(1/2)}$ be $\frac{1}{2\,\zeta(1/2)}$ preceding the integral? For the Euler-Phi function, should $\zeta(\rho/2)$ be $\zeta(\rho-1)$? Also, the closing parenthesis is missing on $e^{-x\,(2\,n+1/2)}$. For the square-free function, should $\sum\limits_\gamma\frac{h(\gamma)\,\zeta(\rho-1)}{\zeta'(\rho)}$ be $\sum\limits_\gamma\frac{h(\gamma/2)\,\zeta(\rho/2)}{2\,\zeta'(\rho)}$? – Steven Clark Apr 22 '18 at 16:07
• I also noticed the same discrepancies in your answer at math.stackexchange.com/q/2217448. Also, I believe there are some typo errors in formulas (3.7) and (3.8) in your paper at vixra.org/abs/1310.0048 where in the sums over $n$ I believe the four references to $k$ should be $n$. – Steven Clark Apr 22 '18 at 16:09
• Also, in the last sentence of your answer above I believe $H(x)$ should be $h(x)$. In the last sentence of your answer at math.stackexchange.com/q/2217448 I believe f and g should be g and h. – Steven Clark Apr 22 '18 at 16:24
• I also noticed there are similar discrepancies in the Wikipedia article at en.wikipedia.org/wiki/Explicit_formulae_(L-function). – Steven Clark Apr 22 '18 at 22:16