TITLE: How are Zeta function values calculated from within the Critical Strip?
QUESTION [7 upvotes]: We note that for $Re(s) > 1$ 
$$ \zeta(s) = \sum_{i=1}^{\infty}\frac{1}{i^s} $$
Furthermore
$$\zeta(s) = 2^s \pi^{s-1} \sin \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s)$$
Allows us to define the zeta function for all values where
$$Re(s) < 0$$
By using the values where $$Re(s) > 1$$ But how do we define it over
$$ 0 \le Re(s) \le 1$$
Which is where most of the "action" regarding the function happens anyways...

REPLY [2 votes]: This shouldn't be taken as a complete answer....
One thing you must know is that the functional equation you mention
$\zeta(s)=\frac{(2\pi)^{s}}{\pi}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s)$
should not be used to calculate zeroes. What I think is used, and is relatively accessible, is the following reformulation
$\pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma(\frac{1-s}{2})\zeta(1-s)=\Lambda(s)$
Notice that $\Lambda(\bar{s})=\bar{\Lambda}(s)$. If you let $s=\frac{1}{2}+\imath y$, then you have $\Lambda(s)$ to be a purely real function for purely real $y$. You can use the intermediate value theorem if you can approxiamte $\Lambda$, show that for some interval it takes a positive output and a negative output, and home in on your zero within that zero (since $\Lambda$ is a continuous function). 
The following representation, previously mentioned here, is an exceptionally nice way to compute $\zeta$:
$\displaystyle \zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$
Additional Edit:
The exponential $\pi$ and $\Gamma$ factors are never zero! So if this $\Lambda$ thing is zero, the zero is coming from $\zeta$!