B How to write this in terms of ##\zeta (x)##?

[MevsEinstein]

TL;DR Summary

I wrote $$\zeta (x+yi)$$ as $\zeta(x)\zeta(yi) - \displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$. I want to simplify the second term in terms of the zeta function so I can solve for $\zeta (x)$.

How do I re-write $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$$ in terms of $\zeta (x)$ ? I want to solve for $\zeta (x)$ and simplifying the above expression in terms of $\zeta (x)$ would avoid repetition.

 

[Office_Shredder]

You have to fix the latex in your post. You can use dollar signs instead of hashes to make the whole line display style which might make it look better.

 

[MevsEinstein]

You have to fix the latex in your post.

It took me forever to find the error. Turns out it was just a parentheses >D.

 

[MevsEinstein]

Oh wait instead of \ it gives me $\S=$

 

[Mark44]

How do I re-write $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$$

Before you can simplify the 2nd term, you need to say what is being summed.

 

[MevsEinstein]

Before you can simplify the 2nd term, you need to say what is being summed.

OOPS! $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x}*(\displaystyle\sum_{k \in S, \mathbb{Z} \S =n} \frac{1}{k^{yi}})$$ There you go. Note that $\mathbb{Z} \S =n$ is actually $\mathbb{Z}$\S = n, I couldn't fix the bug.