As the title says, I'm interested to know how $\zeta(i)$ is calculated.
I know the functional equation for the zeta function, but if I put it in that in there, I must know $\zeta(1-i)$. Is it a good idea to use the Riemann-Siegel-formula? If so, can you give a hint how to begin with it?
Riemann-Siegel-formula is for large values of the imaginary part $(*)$.
For $\zeta(i)$ Euler Maclaurin's expansion should be more useful (and simpler to implement!).
Set $z=i$ here :
$$\zeta(z) \sim \sum_{k=1}^N \frac 1{k^z} \color{#bb0000}{+\frac 1{(z-1)\;N^{z-1}}}\\\color{#006600}{-\frac 1{2\;N^z}+\frac z{12\;N^{z+1}}-\frac{z(z+1)(z+2)}{720\;N^{z+3}}+\frac{z(z+1)(z+2)(z+3)(z+4)}{30240\;N^{z+5}}}$$
Table of values for some values of $N$ : \begin{array}{cc|c} N&S(N)&|S(N)-\zeta(I)|\\ \hline 1&0.00284391534392 - 0.418386243386\,i\quad &0.000511354364339\\ 2&0.00329397373604 - 0.418154520290\,i\quad &6.31859409486 E-6\\ 3&0.00329985801481 - 0.418155237552\,i\quad &4.22474873780 E-7\\ 4&0.00330018215444 - 0.418155406437\,i\quad &5.95693232453 E-8\\ 5&0.00330021691875 - 0.418155438230\,i\quad &1.28391196012 E-8\\ 10&0.00330022369863 - 0.418155449038\,i\quad &1.042857637 E-10\\ 20&0.00330022368592 - 0.418155449141\,i\quad &8.23113727 E-13\\ \end{array} pari/gp script for Euler Maclaurin's expansion :
zeta(n,z)=sum(k=1,n,1/k^z)+1/((z-1)*n^(z-1))-1/(2*n^z)+z/(12*n^(z+1))-z*(z+1)*(z+2)/(720*n^(z+3))+z*(z+1)*(z+2)*(z+3)*(z+4)/(30240*n^(z+5))
$(*)$ Further details concerning Riemann-Siegel formula.
We can use the identity $$\zeta\left(s\right)=\frac{1}{s-1}\sum_{n\geq1}\frac{n\left(n+1\right)}{2}\left(\frac{2n+3-s}{\left(n+1\right)^{s+2}}-\frac{2n-1-s}{n^{s+2}}\right) $$ which holds for $\textrm{Re}\left(s\right)>-1 $.
