The bounty for the best work out of Greg's answer, especially the "solving for $\pi^*(x;q,a)$ in terms of all $\Pi^*$ functions (tedious but possible)" part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...

I posted my attempt of a workout here. Comments welcome...

Original question:

I have two representations of $\pi(x)$:

  1. The Prime Counting Function $\pi(x)$ is given $$ \pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1} $$ with $ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$ and $\rho$ running over all the zeros of $\zeta$ function.

  2. This formula, while widely believed to be correct, has not yet been proved. $$ \pi(x) \approx \int\limits_2^x{\frac{dt}{\ln t}} - \frac{\sqrt x}{\ln x}\left( 1 + 2\sum_{\gamma} \ \ \frac{\sin(\gamma\ln x)}{\gamma}\right) \tag{2}, $$ with $\gamma=\text{Im}({\rho})$ being imaginary part of the roots of the $\zeta$ function.

Now I have two questions:

  1. Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $\sqrt x/\ln x$?
  2. How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $\rho$-parts fit together? How do I get $\sin$s from $\text{li}(z^{1/n})$s? Does this invoke Gram's series: $$ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n}) = 1 + \sum_{k=1}^\infty \frac{(\ln z)^k}{k! k \zeta(k+1)} ? $$ We can rewrite $\displaystyle \frac{\sin(\gamma\ln x)}{\gamma}=\frac{x^{i\gamma}-x^{-i\gamma}}{i\gamma}$ and I remember that I've seen a similar expression at Wikipedia: $$ \psi_0(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \ln 2\pi - \frac12 \ln(1-x^{-2}) , $$ But could this help, if at all? ($\psi_0(x)$ is the normalization of the Chebyshev function, see here)
  • $\begingroup$ +1: interesting question but where did you find the formula $(2)$ ? I would bet that it is only an approximation just from the lower bound $2$ of the first integral ( and... from the $\approx$ sign :-) ) ? $\endgroup$ Jan 4, 2013 at 21:42
  • $\begingroup$ It is from the linked paper here : math.stackexchange.com/q/149755/19341 . Sorry but i post from my tablet. I hate that... $\endgroup$
    – draks ...
    Jan 4, 2013 at 23:28
  • $\begingroup$ Just some preliminary numerical results : I compared $(2)$ to the actual $\pi(n)$ function in the interval $(2,100)$ !pi_100 : $(2)$ (darker and on the top) has all the features of a function growing 'by steps' and is very similar to the real $\pi(n)$ function but, as seen in $(3,1000)$ !pi_1000, it is starting too low and growing too fast (this effect is real : more than $10^4$ zeros were used). $\endgroup$ Jan 4, 2013 at 23:42
  • $\begingroup$ From your ref. the zeros seem to be the imaginary parts of the zeros of the Dirichlet beta function. I have Montgomery and Vaughan's book (I'll see tomorrow...). $\endgroup$ Jan 4, 2013 at 23:43
  • $\begingroup$ But do your plots compare to Figure 4 in Prime Number Races. Beta functions are used in the case $\pi_{4n+3}-\pi_{4n+1}$ (p.19)... $\endgroup$
    – draks ...
    Jan 5, 2013 at 9:20

3 Answers 3


Before answering to your question in the comment :

" Is it possible to write $\ \#\{\text{primes}\ 4n+3 \le x\}\,- \,\#\{\text{primes}\ 4n+1 \le x\}$ (form here) in terms like $\,\operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho})$ ? "

let's start with a sketch using von Mangoldt's derivation to obtain your equation $(1)$ that will be used for inspiration (for proofs see for example Edwards' chapter 3) :

Riemann's Explicit Formulas

The Euler product formula gives us : $$\tag{1}\boxed{\displaystyle\zeta(s)=\prod_{p\ \text{prime}}\frac 1{1-p^{-s}}}\quad\text{for}\ \ \Re(s)>1$$ so that $$\log \zeta(s)=-\sum_{p\ \text{prime}}\log(1-p^{-s})=\sum_p\sum_{k=1}^\infty \frac{p^{-ks}}k$$ minus the derivative will be : $$\tag{2}f(s):=-\frac{\zeta'(s)}{\zeta(s)}=\sum_p\sum_{k=1}^\infty \frac{\log\,p}{p^{ks}}=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\quad\text{for}\ \ \Re(s)>1$$ with $\Lambda$ the von Mangoldt function defined by $\ \Lambda(n):=\begin{cases} \log\, p & \text{if}\ n=p^k\ \text{and}\ k>0\\ 0 & \text{else} \end{cases}$

Let's use the definition of the second Chebyshev function : $\displaystyle \psi(x)=\sum_{n\leq x}\Lambda(n)$ and Abel's sum formula applied to $a(n):=\Lambda(n)$ and $\phi(n):=n^{-s}$ : $$\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}=s\int_1^\infty \frac {\sum_{n\leq x}\Lambda(n)}{x^{s+1}}\;dx$$ to rewrite $f(s)$ as (the lower bound became $0$ since $\phi(x)=0$ for $x<1$) : $$\tag{2.1}f(s)=s\int_0^{\infty}\frac{\psi(x)}{x^{s+1}}dx$$

But this is a Mellin transform which may be reverted to get Perron's formula (let's observe that the Dirichlet series $\displaystyle f(s)=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}$ is absolutely convergent for $\Re(s)>1$ and suppose $c>1$) : $$\tag{3}\psi^*(x):=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}f(s)\frac{x^s}s\,ds=\sum_{n=1}^\infty \Lambda(n) \frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(\frac xn\right)^s\frac{ds}s$$ This last integral may be evaluated using Fourier's theorem : $$\ \displaystyle\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{y^s}s ds= \begin{cases} \ \ 0 & 0<y<1\\ 1/2 & \quad y=1\\ \ \ 1 & \quad y>1\\ \end{cases}$$ and we obtain (as wished) that : $\displaystyle \tag{4}\psi^*(x)=\sum_{n\le x}^{*}\Lambda(n)$

with $\psi^*$ the second Chebyshev function except when $x$ is an integer because in this case the last term of the sum has to be divided by $2$. This will be the meaning of the $^*$ symbols in this article : at a first order discontinuity point (i.e. a jump) the result is the mean value of the limit at the left and the right.

Now $\psi^*(x)$ may also be written as : $$\tag{5}\psi^*(x)=-\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\frac{x^s}s\,ds$$ Here the idea is simply to sum the residues of (all) the poles at the left of the vertical line of integration ($c>1$) at least if $x>1$ (for $0<x<1$ we would have to take the poles at the right and obtain $0$). These poles come from the denominator $s$, the pole of $\zeta$ at $1$ and the zeros with the correspondence : $0\mapsto -\frac{\zeta'(0)}{\zeta(0)},\ 1\mapsto x^1,\ \rho\mapsto -\frac{x^\rho}{\rho}$ for $\rho$ any zero of $\zeta\,$ (from the Weierstrass factorization of the Hadamard product) so that : $$\tag{6}\boxed{\displaystyle\psi^*(x)=x-\sum_{\rho} \frac {x^{\rho}}{\rho}-\frac{\zeta'(0)}{\zeta(0)}},\quad(x>1)$$ (in this post we won't distinguish the trivial from the non-trivial zeros $\rho$ ; for convergence the non-trivial roots should be grouped by pairs and sorted by increasing $|\Im(\rho)|$)

The Riemann prime-counting function is defined by : \begin{align} \tag{7}\Pi^*(x):&=\sum_{p^k\le x}^{*}\frac 1k=\sum_{n\le x}^{*}\frac {\Lambda(n)}{\log\,n}\\ &=\sum_{n\le x}^{*} \Lambda(n)\left(\int_n^x \frac{dt}{t\,\log^2 t}+\frac 1{\log\,x}\right)\\ &=\int_2^x\frac{\psi^*(t)\ dt}{t\,\log^2 t}+\frac{\psi^*(x)}{\log \,x}\\ &\tag{8}=\int_2^x\frac{\psi^{*'}(t)\ dt}{\log\,t}\\ \end{align} But $\ \displaystyle\operatorname{li}(x)=\int_2^x \frac{dt}{\log\,t}\,$ (Riemann's variant of the logarithmic integral) verifies $\ \displaystyle\operatorname{li}(x^r)'=\frac{r\,x^{r-1}}{\log\ x^r}=\frac{x^{r-1}}{\log\,x}$ so that from $(6)$ : $$\tag{9}\boxed{\displaystyle\Pi^*(x)=\operatorname{li}(x)-\sum_{\rho} \operatorname{li}(x^{\rho})},\quad(x>1)$$

The prime-counting function is defined by $\ \displaystyle\pi^*(x):=\sum_{p\le x}^{*}1\ $ while $\ \displaystyle\Pi^*(x):=\sum_{p^k\le x}^{*}\frac 1k$ will be : $$\tag{10}\Pi^*(x)=\sum_{k>0} \frac{\pi^{*}\bigl(x^{1/k}\bigr)}k$$ Applying the Möbius inversion formula $\ \displaystyle\pi^{*}(x):=\sum_{n=1}^{\infty} \frac{\mu(k)}k \Pi^*\bigl(x^{1/k}\bigr)\ $ to $(9)$ we get (with questionable convergence...) : $$\tag{11}\boxed{\displaystyle\pi^*(x)=R(x)-\sum_{\rho} R(x^{\rho})},\quad(x>1)$$ Where Riemann's $\,\displaystyle R(x):=\sum_{n=1}^{\infty} \frac{\mu(k)}k \operatorname{li}\bigl(x^{1/k}\bigr)\,$ may be written as a Gram series.

Dirichlet's L-functions

We will try to repeat the previous derivations but with $\zeta$ replaced by the Dirichlet L-function $$L(s,\chi)=\sum_{k=1}^\infty\frac{\chi(k)}{k^s}$$

With $\chi$ a Dirichlet character we get following Euler product : $$\tag{1'}\boxed{\displaystyle L(s,\chi)=\prod_{p\ \text{prime}}\frac 1{1-\chi(p)p^{-s}}}\quad\text{for}\ \ \Re(s)>1$$ so that ($\chi$ is multiplicative implying $\chi(p)^k=\chi(p^k)$) : $$\log L(s,\chi)=-\sum_{p\ \text{prime}}\log(1-\chi(p)p^{-s})=\sum_p\sum_{k=1}^\infty \frac{\chi(p^k)p^{-ks}}k$$ minus the derivative relatively to $s$ will be : $$\tag{2'}f(s,\chi):=-\frac{L'(s,\chi)}{L(s,\chi)}=\sum_p\sum_{k=1}^\infty \frac{\chi(p^k)\log\,p}{p^{ks}}=\sum_{n=1}^\infty \frac{\chi(n)\Lambda(n)}{n^s}\quad\text{for}\ \ \Re(s)>1$$

The Perron formula applied to $f(s,\chi)$ gives : $$\tag{3'}\psi^*(x,\chi):=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}f(s,\chi)\frac{x^s}s\,ds=\sum_{n=1}^\infty \chi(n)\Lambda(n) \frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(\frac xn\right)^s\frac{ds}s$$ $$\tag{4'}\psi^*(x,\chi)=\sum_{n\le x}^{*}\chi(n)\Lambda(n)$$

But $\psi^*(x,\chi)$ may also be written as : $$\tag{5'}\psi^*(x,\chi)=-\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{L'(s,\chi)}{L(s,\chi)}\frac{x^s}s\,ds$$ Again we compute the sum of the residues at (all) the poles to the left of the vertical line of integration (for $c>1$). The contributions from the different poles will be : $0\mapsto -\frac{L'(0,\chi)}{L(0,\chi)},\ \rho\mapsto -\frac{x^\rho}{\rho}$ for $\rho$ any zero of $L(s,\chi)\,$ (from Weierstrass factorization) so that : $$\tag{6'}\boxed{\displaystyle\psi^*(x,\chi)=-\sum_{\rho} \frac {x^{\rho}}{\rho}-\frac{L'(0,\chi)}{L(0,\chi)}},\quad(x>1)$$ (the sum over $\rho$ combines the trivial and the non-trivial zeros and we will suppose that $\chi$ is a non principal character so that $\,L(s,\chi)\,$ is regular everywhere, for a principal character we should add a $x$ contribution from the pole at $1$)

An analogue to the Riemann prime-counting function will be : \begin{align} \tag{7'}\Pi^*(x,\chi):&=\sum_{p^k\le x}^{*}\frac {\chi(p^k)}k=\sum_{n\le x}^{*}\frac {\chi(n)\Lambda(n)}{\log\,n}\\ &=\sum_{n\le x}^{*} \chi(n)\Lambda(n)\left(\int_n^x \frac{dt}{t\,\log^2 t}+\frac 1{\log\,x}\right)\\ &=\int_2^x\frac{\psi^*(t,\chi)\ dt}{t\,\log^2 t}+\frac{\psi^*(x,\chi)}{\log \,x}\\ &\tag{8'}=\int_2^x\frac{\psi^{*'}(t,\chi)\ dt}{\log\,t}\\ \end{align} From $\ \displaystyle\operatorname{li}(x^r)'=\frac{x^{r-1}}{\log\,x}$ and $(6')$ we get : $$\tag{9'}\boxed{\displaystyle\Pi^*(x,\chi)=-\sum_{\rho} \operatorname{li}(x^{\rho})},\quad(x>1)$$ (for $\chi$ a principal character there is an additional $\,\operatorname{li}(x)$ term)

The $\chi$-prime-counting function is $\ \displaystyle\pi^*(x,\chi):=\sum_{p\le x}^{*}\chi(p) $ while $\ \displaystyle\Pi^*(x,\chi):=\sum_{p^k\le x}^{*}\frac {\chi(p^k)}k$ so that : $$\tag{10'}\Pi^*(x,\chi)=\sum_{k>0} \frac{\pi^{*}\bigl(x^{1/k},\chi^k\bigr)}k,\quad(x>1)$$

Broken analogy...

The problem here is that we have $\,\pi^{*}\bigl(x^{1/k},\chi^k\bigr)$ in the sum and not $\,\pi^{*}\bigl(x^{1/k},\chi\bigr)$.
For principal characters (when $\,\chi(p)=0$ or $1$ so that $\,\chi^k(p)=\chi(p)\,$ for $k$ positive) we may apply the Möbius inversion formula to $(9')$ and get : $$\tag{11'}\displaystyle\pi^*(x,\chi)=[R(x)]-\sum_{\rho} R(x^{\rho})$$ (since $\chi$ is a principal character we must add the $R(x)$ term from the pole at $1$)

but this doesn't work in general and the equality becomes questionable after the first prime $p$ such that $\,\chi^k(p)\not =\chi(p)\,$ as we will see by considering your specific question.

We want $\chi$ to be the Dirichlet character modulo $4$ ($\chi_2$ in the link) defined by $\ \displaystyle\chi(0)=0,\ \chi(1)=1,\ \chi(2)=0,\ \chi(3)=-1\ $ and get : $$\tag{12'}L(s,\chi)=\sum_{n=0}^\infty\frac {(-1)^n}{(2n+1)^s}=\beta(s)$$ with $\beta$ the Dirichlet beta function we could hope that : $$\pi^*(x,\chi)=\displaystyle\pi_{4,1}^*(x)-\pi_{4,3}^*(x)\approx -\sum_{\rho} R(x^{\rho})$$ with $\pi_{q,a}(x)=\#\{p:p\ \text{is prime and}\ \,p\le x\,\ \text{and}\ \,p\equiv a\pmod q\}$

But, from $(10')$ and $(9')$, we have in fact : \begin{align} \Pi^*(x,\chi)&=\sum_{k>0} \frac{\pi_{4,1}^{*}\bigl(x^{1/k}\bigr)+(-1)^k\,\pi_{4,3}^{*}\bigl(x^{1/k}\bigr)}k=-\sum_{\rho} \operatorname{li}(x^{\rho})\\ &=\sum_{k>0} \frac{\pi_{4,1}^{*}\bigl(x^{1/k}\bigr)-\,\pi_{4,3}^{*}\bigl(x^{1/k}\bigr)}k+2\sum_{j>0} \frac{\pi_{4,3}^{*}\left(\sqrt{x}^{1/j}\right)}{2j}\\ \end{align} so that the Möbius transformation will return the exact (with $\rho$ any zero of $\beta$) : $$\tag{13'}\boxed{\displaystyle\pi_{4,1}^{*}(x)-\pi_{4,3}^{*}(x)+\pi_{4,3}^{*}\left(\sqrt{x}\right)=-\sum_{\rho} R(x^{\rho})}$$

To show the effect of the $\,\pi_{4,3}^{*}\left(\sqrt{x}\right)\,$ term let's exhibit an approximation of the expression at the right (taking the $40$ first terms of the sum $R$ and the $300$ first non-trivial zeros of the Dirichlet-beta function) that we compare to the exact (dark) result for $\,\pi^*(x,\chi)=\displaystyle\pi_{4,1}^*(x)-\pi_{4,3}^*(x)\,$ in the interval $\,[5,80]$ :

Interval 5-80

The initial behavior seems right : $-1$ at $3$ followed by $+1$ at $5$ and $-1$ at $7$ but just after that there is a first 'irregularity' at $9=3^2$ because $\chi(3^2)=1$ while $\chi(3)=-1$. After that the behavior is correct (shifted $+1$ of course) until another irregularity at $49=7^2$ again because $\chi(7^2)=1$. The next one will be at $121=11^2$ and at all the following squares of primes of gender $4n+3$...

ADDITION: Draks telescoping

The situation was blocked at this point until draks, using a suggestion of Greg Martin, proposed in february his powerful telescoping method to obtain (the $\rho_\zeta$ and $\rho_\beta$ are respectively all the zeros of the $\zeta$ and $\beta$ function) : $$\tag{14'}\pi_{4,3}(x):=\sum_{k=0}^\infty\;2^{-k-1}\left( R\left(x^{1/2^{k}}\right)-\sum_{\rho_\zeta} R\left(x^{\rho_\zeta/2^k}\right) +\sum_{\rho_\beta} R\left(x^{\rho_\beta/2^k}\right) \right)$$

We may obtain this result too (with a minor correction) by rewriting $(11)$ as (for $x>2$) : $$\pi^*(x)=1+\pi_{4,1}^{*}(x)+\pi_{4,3}^{*}(x)=R(x)-\sum_{\rho_\zeta} R(x^{\rho_\zeta})$$ ($1$ was added since all primes except $2$ are equal to $1$ or $3\bmod{4}$) and using $(13')$ : $$\pi_{4,1}^{*}(x)-\pi_{4,3}^{*}(x)+\pi_{4,3}^{*}\left(\sqrt{x}\right)=-\sum_{\rho_\beta} R(x^{\rho_\beta})$$ Subtracting these equations gives : $$2\pi_{4,3}^{*}(x)-\pi_{4,3}^{*}\left(\sqrt{x}\right)=R(x)-1-\sum_{\rho_\zeta} R(x^{\rho_\zeta})+\sum_{\rho_\beta} R(x^{\rho_\beta})$$ while draks' telescoping sum $\ \displaystyle\frac 12\sum_{k=0}^\infty \frac{f\left(x^{1/2^{k}}\right)}{2^k}$ is : $$\sum_{k=0}^\infty \frac {2\,\pi_{4,3}^{*}(x^{1/2^{k}})-\pi_{4,3}^{*}\bigl(x^{1/2^{k+1}}\bigr)}{2^{k+1}}=\sum_{k=0}^\infty \frac {R\left(x^{1/2^{k}}\right)-1-\sum_{\rho_\zeta} R\left(x^{\rho_\zeta/2^k}\right)+\sum_{\rho_\beta} R\left(x^{\rho_\beta/2^k}\right)}{2^{k+1}}$$ producing draks' formula : $$\tag{15'}\boxed{\displaystyle\pi_{4,3}^{*}(x)=\sum_{k=0}^\infty \frac {R\left(x^{1/2^{k}}\right)-\sum_{\rho_\zeta} R\left(x^{\rho_\zeta/2^k}\right)+\sum_{\rho_\beta} R\left(x^{\rho_\beta/2^k}\right)}{2^{k+1}}-1}$$ A similar formula may be written for $\pi_{4,1}$ and this answers positively the initial question !

Let's finish with a plot of $\pi_{4,3}^{*}(x)$ from $4$ to $100$ (more details here with the pari/gp scripts used)

plot of pi_4,3

  • 1
    $\begingroup$ Raymond, first of all: Wow! What an answer. "+1". Thank you so much for everything including the link to Edward's book. My copy is buried somewhere in a box because I moved. It'll take me a while to verify all that, but like you, I think that the reasoning by analogy is correct. Specially the last equation thrills me. You might have noticed my attempts with "Summing over General Functions of Primes" ... $\endgroup$
    – draks ...
    Jan 20, 2013 at 22:44
  • $\begingroup$ ... or the follow-on question: "An Expression for $\log\zeta(ns)$ derived from the Limit of the truncated Prime $\zeta$ Function". So I think it's possible to combine these result to get an analytic expression for the truncated Prime $\zeta$ function with a restriction to primes $4n\pm 1$. This is ever so cool and I can't leave this unawarded... $\endgroup$
    – draks ...
    Jan 20, 2013 at 22:51
  • $\begingroup$ Thanks @draks... I think that most things are right here except tiny things like the - from the citation, the starting value $n=1$ of the Dirichlet beta function, and perhaps very important things like the transitions from $(9')$ to $(11')$ : these should be (more than!) double checked (because of things like $\chi(3^2)=1\not =-1$ and of warnings from different authors like Ingham that Riemann's method to obtain $\pi$ may not be as easily generalized as $\psi\cdots$ but works only for some specific Dirichlet characters ($\chi(p)=0,1$ only?)). But this clear path may at least be checked! $\endgroup$ Jan 21, 2013 at 0:05
  • $\begingroup$ @draks... I corrected the end of my answer. $\endgroup$ Jan 26, 2013 at 23:04
  • $\begingroup$ aha, looks like we've gottta add $(\mu^2(n)-1)$ to get rid of the squares...interesting, what do you think? And thanks for your thoughts. At least my ${\rm li}(x)$ works... $\endgroup$
    – draks ...
    Jan 27, 2013 at 16:03

Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :

"The precise formula he (Riemann) proposed is a bit too technical for this article, but we can get a good sense of it from the following approximation when x is large. This formula, while widely believed to be correct, has not yet been proved."

$$ \tag{1}\boxed{\displaystyle\frac{\int\limits_2^x{\frac{dt}{\ln t}}-\pi(x)}{\sqrt x/\ln x} \approx 1 + 2\sum_{\gamma} \ \ \frac{\sin(\gamma\ln x)}{\gamma}} $$

I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.

Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $\displaystyle \frac{\sin(\gamma\ln x)}{\gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' : $$\tag{2}\frac{\cos(\gamma\ln x-\arg \rho)}{|\rho|},\quad\text{with}\ \ \rho=\frac 12+i\gamma$$ since $\ \gamma\gg 1 $ we have indeed $\ |\rho|\approx \gamma\ $ and $\ \arg\,\rho\approx \frac{\pi}2$
(note that $\gamma > 14.13\ $ so that $\ 1<\frac {|\rho|}{\gamma}<1.00063$)

Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) : $$\tag{3}\operatorname{R}(x^{\rho}) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n}\operatorname{li}(x^{\rho/n})$$ with the first term : $$\tag{4}\operatorname{R}(x^{\rho}) \approx\left[\operatorname{li}(x^{\rho})=\operatorname{li}(e^{\rho\;\ln x})=\operatorname{Ei}(\rho\;\ln x)\right]$$ and to use the asymptotic expansion for $\operatorname{Ei}$ : $$\tag{5}\operatorname{Ei}(z)\sim \frac{e^z}{z}\left(1+\frac {1!}z+\frac{2!}{z^2}+\cdots\right)$$ to get : $$\operatorname{Ei}\left(\left(\frac 12+i\gamma\right)\;\ln x\right)\approx \frac{x^{1/2+i\gamma}}{(1/2+i\gamma)\;\ln x}$$ so that combining the two 'mirror' zeros : $$\operatorname{Ei}\left(\left(\frac 12+i\gamma\right)+\;\ln x\right)+\operatorname{Ei}\left(\left(\frac 12-i\gamma\right)+\;\ln x\right)\approx \frac{\sqrt{x}}{\ln x}\left(\frac{e^{i\gamma\ln x}}{1/2+i\gamma}+\frac{e^{-i\gamma\ln x}}{1/2-i\gamma}\right)$$ $$\approx\frac{\sqrt{x}}{\ln x}\frac{e^{i\gamma\ln x-i\arg{\rho}}+e^{-i\gamma\ln x+i\arg{\rho}}}{|\rho|}\quad\text{since}\ \ \rho=\frac 12+i\gamma$$ we get Riesel's approximation : $$\tag{6}\operatorname{R}(x^{\rho})+\operatorname{R}(x^{\overline{\rho}})\approx\frac{2\;\sqrt{x}\;\cos(\gamma\ln x-\arg{\rho})}{|\rho|\;\ln x}$$ while $\operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) : $$\tag{7}\operatorname{R}(x)\approx \operatorname{li}(x)-\frac 12\operatorname{li}\bigl(x^{1/2}\bigr)\quad \text{with}\ \operatorname{li}\bigl(x^{1/2}\bigr)\approx \frac{2\sqrt{x}}{\ln x}$$ this allows too to get better approximations if wished ($\operatorname{R}(x)\approx \operatorname{li}(x)-\frac 12\operatorname{li}\bigl(x^{1/2}\bigr)-\frac 13\operatorname{li}\bigl(x^{1/3}\bigr) $ for example, adding the trivial zeros contribution and so on...)

Now to answer your questions :

  1. We have to suppose the R.H. to be able to write $\rho=\frac 12+i\gamma\ $ for any root $\rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)
  2. The implication from $(1)\to (2)$ was provided without using the Gram series

Hoping this clarified things,

  • $\begingroup$ thank you very much. Again :-) $\endgroup$
    – draks ...
    Jan 6, 2013 at 22:14
  • $\begingroup$ You are welcome @draks : this gave me the occasion to compute some zeros (say 331) of Dirichlet beta, see a fine approximation for $\pi(n)$ and ways to make it better. Thanks for the journey ! $\endgroup$ Jan 6, 2013 at 23:43
  • $\begingroup$ Wow 331 sounds great. I was just out to find a table where these zeros are listed. Would you mind sharing the zeros of beta with me? And to be honest I still have one more question: Is it possible to write $\#\{\text{primes}\ 4n+3 \le x\} - \#\{\text{primes}\ 4n+1 \le x\}$(form [here](math.stackexchange.com/q/149755/19341)) in terms like $\operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho})$ ? I hope I don't bother you too much... $\endgroup$
    – draks ...
    Jan 6, 2013 at 23:54
  • $\begingroup$ Thanks a lot. Concerning the last question, I thought about using a simple analogy $(\pi_{4n+3}-\pi_{4n+1})-\text{Li}(x)=\text{R}(x)-\sum_\rho \text{R}(x^\rho)$ with $\rho$ being the values you've kindly provided...? $\endgroup$
    – draks ...
    Jan 7, 2013 at 0:21
  • $\begingroup$ @draks: I gave the zeros in your other thread and will erase the values here... Concerning your other question I don't know but will look at it in the coming days (it's late...). I must add that from your formula $(4')$ there is probably a similar (or identical) formula as you suppose ! $\endgroup$ Jan 7, 2013 at 0:30

You probably know that $$ \Pi^*(x;q,a) = \sum^*_{\substack{n\le x \\ n\equiv a\pmod q}} \frac{\Lambda(n)}{\log n} = \frac1{\phi(q)} \sum_{\chi\pmod q} \overline{\chi(a)}\Pi^*(x,\chi). $$ I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that \begin{align*} \Pi^*(x;q,a) &= \sum^*_{\substack{p\le x \\ p\equiv a\pmod q}} 1 + \sum^*_{\substack{p^2\le x \\ p^2\equiv a\pmod q}} \tfrac12 + \sum^*_{\substack{p^3\le x \\ p^3\equiv a\pmod q}} \tfrac13 + \cdots \\ &= \pi^*(x;q,a) + \tfrac12 \sum_{\substack{b\pmod q \\ b^2\equiv a\pmod q}} \pi^*(x^{1/2};q,b) + \tfrac13 \sum_{\substack{c\pmod q \\ c^3\equiv a\pmod q}} \pi^*(x^{1/3};q,c) + \cdots \end{align*} and solving for $\pi^*(x;q,a)$ in terms of all $\Pi^*$ functions (tedious but possible).

  • $\begingroup$ +1 thanks, doesn't it boil down a little/lot when $a=1$? Could you give me a reference hint or something else to read on that topic? $\endgroup$
    – draks ...
    Jan 31, 2013 at 20:43
  • $\begingroup$ I'm not sure it gets much better when $a=1$. Except that all the values $\chi(a)$ (which I forgot to put in in the first place - they're there now) equal $1$ and thus disappear. But $a=1$ will yield the largest possible number of terms in the sums over $b$, $c$, etc. in the last displayed equation line. $\endgroup$ Feb 1, 2013 at 1:25
  • $\begingroup$ hmm, when I look at the second term in the last line and insert $a=1$, the sum's running over $b=1,3,5,7,\dots$. But $b>3$ doesn't make sense since it's included already before, right? If this is correct and I continue, I get $x^m\equiv 1 \bmod 4\to \cases{x=1,3 \text{ if $m$ even }\\x=1\;\;\,\, \text{ if $m$ odd}} $. But doesn't summing $\pi(x^{1/m};4,\pm 1)$ equal $\pi(x^{1/m})$? How to continue from here? Could you kindly sketch the way...? $\endgroup$
    – draks ...
    Feb 8, 2013 at 23:01
  • $\begingroup$ Right, when I write a sum over $b\pmod q$, I mean a sum over residue classes - so don't separately include both $b=1$ and $b=5$, since they're the same residue class. And in this particular case, when $m$ is even you do in fact count all odd primes in the sum, yes. $\endgroup$ Feb 9, 2013 at 6:13
  • $\begingroup$ you are awarded the bounty since it was your prime race paper which gave me new hope. I'm currently studying your arXiv stuff. Could you point me to something that would help me once more? $\endgroup$
    – draks ...
    Feb 16, 2013 at 13:35

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