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...
On
There is a well-known relation between the Riemann $\zeta$ and the Dirichlet $\eta$ functions: $$\zeta(s)~=~\frac{\eta(s)}{1-2^{1-s}}~,$$ where latter's series converges for all positive values of the argument. See Leibniz's criterion and Dirichlet's test for more information.
On
This is an outline of the steps taken in the ref. in the comments to derive an expression for $\zeta(s)$ which converges for $\sigma>0$ and which can be modified slightly to allow approximation of values of $\zeta(s).$
Letting $f(s) = 1/x^s$ and $\int_0^{\infty}\frac{dx}{x^s}=\frac{1}{s-1}$ for $\text{Re}(s)>1$ and using (proof omitted)
$$\sum_1^{\infty}f(r)-\int_1^{\infty}f(t)dt = f(1)+\int_1^{\infty}(t-[t])f'(t)dt$$
we get
$$(*)\hspace{10mm}\zeta(s)=\frac{1}{(s-1)}+1-s\int_1^{\infty}\frac{x-[x]}{x^{s+1}}dx. $$
Since $x - [x] < 1$ the integral on the right converges for all $\sigma > 0.$
The above is of no use for computation so using Euler's summation formula for partial sums,
$$ \sum_{n=2}^{\infty}f(n) = \int_1^N f(x)dx+\int_1^N(x-[x])f'(x)dx,$$
applying this to $f(x)= 1/x^s$ in which $(s \neq 1)$ we have
$$\sum_{n=1}^N\frac{1}{n^s}=1 + \frac{1}{s-1}-\frac{N^{1-s}}{s-1}-s\int_1^N\frac{x-[x]}{x^{s+1}}dx $$
in which the term $1$ is just $f(1).$
Subtracting this from (*) we get
$$\zeta(s) = \sum_{n=1}^N \frac{1}{n^s}+\frac{N^{1-s}}{s-1} +r(s)$$
in which the term $r(s)$ can be shown to vanish as $N\to\infty.$
And so we have
$$\zeta(s) = \lim_{N\to\infty}\left(\sum_{n=1}^N\frac{1}{n^s}+\frac{N^{1-s}}{s-1} \right)$$
Now if I use this to calculate $\zeta(1+it)$ along the imaginary axis from $t = 0$ to $t = 36$ as Jameson does in his text, or other values, I get good approximations for even low values of N. For $s= 3/4+i $ and $N=1000$ I get $\zeta(3/4+i)\approx 0.33-.87i$ which agrees with $Mathematica's$ figure pretty closely.
The list of results omitted to do this is not long and they are all furnished in the noted reference.
On
It's $\tt\underline{usually}$ evaluated with the alternating series $\displaystyle{\quad\color{#66f}{\large% \zeta\left(\,s\,\right) ={1 \over 1 - 2^{1 - s}} \sum_{k\ =\ 1}^{\infty}{\left(\,-1\,\right)^{k - 1} \over k^{s}}}}$.
On
Although unmentioned by the others, once you have a functional equation, you can also use a so-called "approximate functional equation". In this case, it may seem circular, because knowing the functional equation implicitly involves understanding the values in the center of the strip.
But in fact, the approximate functional equation is an extremely efficient method of approximating values in the critical strip. A reference on wikipedia the Riemann-Siegel formula. A slightly looser explanation is that for $s = \sigma + it$, we have $$\zeta(s) = \sum_{n\leq x}\frac{1}{n^s} \ + \ \gamma(1-s) \ \sum_{n\leq y}\frac{1}{n^{1-s}} \ + \ O(x^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}y^{\sigma - 1}),$$ where $\gamma(s)$ is a ratio of Gamma functions and you can choose $x,y$ to give the approximations of desired size. This works poorly on the edges of the critical strip, but very well in regions bounded away from the edges. It works especially well on the critical line, which is of special interest.
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We may start by defining the Dirichlet eta function:
$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$
We then have
$$\zeta(s)-\eta(s)=\sum_{n=1}^\infty\frac{1-(-1)^{n+1}}{n^s}=\sum_{n=1}^\infty\frac2{(2n)^s}=2^{1-s}\zeta(s)$$
You may then see that
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$
While it is true this converges for $s>0$, it is easily extended through an Euler transform:
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$
While we still have an infinite sum to deal with, this converges crazy fast, and you can easily graph approximately $\zeta(s)$ on a good graphing calculator for all $s\in\mathbb R$:
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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 $\zeta$!
An extension of the area of convergence can be obtained by rearranging the original series. The series
converges for $\Re s > 0$. See here.