# Serie for $\text{Arg}( \zeta (z))$ [closed]

Though I don't know if the formula I've found is useful, I decided to publish it anyway. $$\text{Arg}( \zeta (z)) = -\sum_ {k = 1}^{\infty}\sum _ {q = 1}^{\infty}\frac {1} {k P_q^{k x}}\text {Sin}( k y \text{ Log}(P_q ) ) \text{ , } |z|>1$$

where $z=x+i y$ and $p_q$ is th $q^{th}$ prime number.

Here is an example with a plot of $\text{Arg}( \zeta (1+i y))$ in blue, and a plot of the sum in red:

My question is whether the formula is known, whether someone finds it helpful.

EDIT: this post was put 'on hold'. I don't know why. I'm a graphist and foreigner to the mathematics' community, thus it's very hard for me to check the results I got, and this is the reason of this post.

EDIT2: If someone's interested, this other post shows at the end how the serie was calculated simple tools to extract Re, Im, Abs... of any complex function

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## closed as unclear what you're asking by Potato, Lord_Farin, egreg, Daniel Robert-Nicoud, azimutNov 28 '13 at 19:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

What is $P_q$? – draks ... Nov 28 '13 at 10:21
I don't know what is Pq but this is superb, even better than the previous you posted ! How many terms did you sum for the red line ? – Claude Leibovici Nov 28 '13 at 10:33
I am surprised that the sum is over the primes and not the von Mangoldt function. Wonderful result of yours. – Mats Granvik Nov 28 '13 at 13:20
@claude, only ten terms.. – Eddy Khemiri Nov 30 '13 at 3:20
@mats, I will post – Eddy Khemiri Nov 30 '13 at 3:22

I have never seen anything like this before. Here is some code for the plot, apparently with a different scale factor:

(*Mathematica*)
Clear[a, k, q, y, x]
x = 1;
Monitor[a =
N[Table[-Sum[
Sum[1/(k*Prime[q]^(k*x))*Sin[k*y*Log[Prime[q]]], {q, 1,
50}], {k, 1, 100}], {y, 1, 50, 1/4}]];, Floor[N[y]]]
g1 = ListLinePlot[a, DataRange -> {1, 50}, PlotStyle -> Red]
g2 = Plot[Arg[Zeta[1 + I*y]], {y, 1, 50}]
Show[g1, g2]

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