Question follow the one answered already, zeros about Riemann Zeta function and some L-function

Let's me try my best to make it clear on what I am asking.

In his 1859 paper "On the Number of Primes Less Than a Given Magnitude" Riemann found an explicit formula for the number of primes $\pi(x)$ less than a given number x. His formula was given in terms of the related function,

$\Pi(x)=\pi(x)+1/2\pi(x^{1/2})$ +1/3$\pi(x^{1/3})$ +1/4$\pi(x^{1/4})$+1/5$\pi(x^{1/5})$+...

which counts primes where a prime power $p^n$ counts as $\frac{1}{n}$ of a prime. With help of Möbius function, Riemann gave an explicit formula for prime counting function $\pi(x)$, which relating with non-trivial zeros of $\zeta$ function.

$\Pi(x)=\mathrm{Li}(x)-\sum_{\rho}(\mathrm{Li}(x^\rho)) - log(2) +\int_x^{\infty} \frac{1}{t(t^2-1)log(t)}dt$

One can find details info from wiki,


Following Riemann's way, we can define $\Pi_q$, which counts primes where a prime power $p^n$ counts as $\frac{1}{n}$ of a prime. Here $(p,q)=1$.

$\Pi_q$ goes with following series,

$\zeta_q=\sum _{n =1}^{\infty}\frac{1}{n^{s}}, (q,n)=1$

We can see $\zeta_q$ has functional equation, too. Further more, we should be able to get similar $\Pi_q$ explicit formula for prime counting function $\pi(x)$, which exclude prime(s) (q's prime factor).

Here $\Pi_q$ should be much similar with $\Pi$, that it has also $Li(x)-\sum_\rho(Li(x^\rho)$, if $\zeta$ has exact the same non-trivial zeros with $\zeta_q$. The differences are only on other part, which relating with $q$.

If $q<x^\frac{1}{2}$, as $x$ is big enough but $x<\infty$, we can see,

<1> $\Pi(x)-\Pi_q(x) = 1+1/2+1/3+1/4+1/5+...1/n+...$

In <1>, $Li(x)-\sum_\rho(Li(x^\rho)$ is gone but remains function relation with $q$ only, here, $q$ has only 1 prime factor $p$.

so, we find another way for Euler–Mascheroni constant,

<2> $\gamma = \lim_{x\to\infty}(\Pi(x)-\Pi_q(x) - log(log_p(x))$, as $n->\infty$, $q<x^\frac{1}{2}$

If it is correct, $\gamma$ in <2> is relating with $q$ only. If so, can we go further for $\gamma$'s character?

Further more, we can see $\Pi(x)>\Pi_2(x)>\Pi_{2*3}(x)>\Pi_{2*3*5}(x)>\Pi_{2*3*5*7}(x)>...>\Pi_q(x)$, when $q$ includes prime factor ->$\infty$, $0<\Pi_q(x)\le{1}$.

Hense, we have relationship for $Li(x)-\sum_\rho(Li(x^\rho)$ and function with $q$ only, we should be able to find boundary of $\sum_\rho(Li(x^\rho)$.

My question,

(1) What the RH equivalent for Riemann prime formula $\Pi(x)$?

(2) What the formula should be for $\Pi_q(x)$?



The usual zeta function is $$ \prod_p \big( 1 - p^{-s} \big)^{-1} = \zeta(s). $$

The zeta function where you omit the single prime $q$ is $$ \prod_{p\ne q} \big( 1 - p^{-s} \big)^{-1} = \zeta(s) \big( 1-q^{-s} \big). $$

The zeta function where you omit all the primes up to and including $q$ is $$ \prod_{p>q} \big( 1 - p^{-s} \big)^{-1} = \zeta(s) \prod_{p\le q} \big( 1-p^{-s} \big). $$

Any function of the form $1-q^{-s}$ has zeros only when the real part of $s$ equals $0$. Therefore the zeros of all three zeta functions strictly inside the critical strip are identical, and so the Riemann hypothesis is exactly the same for all three functions.

  • $\begingroup$ Thanks Greg! Can you please answer my 2 questions? Note - the function $\Pi(x)$ is counting primes and prime power ($p^n$ is counting as $\frac{1}{n}$), which has explicit formula as per Riemann. $\endgroup$ – Ocean Yu Jan 13 '14 at 8:15
  • $\begingroup$ The easiest way is to just adjust the corresponding $\Pi$ function by hand. For example, if you omit simply one prime $q$: use the classical formula for $\Pi(x)$, and then subtract $\sum_{r\le (\log x)/\log q} \frac1r$ to account for the powers of $q$ that are no longer present. $\endgroup$ – Greg Martin Jan 13 '14 at 9:00
  • $\begingroup$ It is correct if doing this by the easiest way. Is this the only way? If recall Riemann's paper, he got following explicit formula with $\zeta$ function, $\Pi(x)=\mathrm{Li}(x)-\sum_{\rho}(\mathrm{Li}(x^\rho)) - log(2) +\int_x^{\infty} \frac{1}{t(t^2-1)log(t)}dt$, I am curious if we can get similar formula for $\zeta_q$ by following Riemann's way, instead of just subtracting $\sum_{r\le (\log x)/\log q} \frac1r$. If we can, we will break through RH with further discussion I believe. $\endgroup$ – Ocean Yu Jan 16 '14 at 1:45
  • $\begingroup$ It is painful that I realized that I was in a round circle, as seems we can not find a good control-able formula, even by following Riemann's way. The questions seem have no a single value, that can be closed. $\endgroup$ – Ocean Yu Jan 16 '14 at 2:32

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