How are Zeta function values calculated from within the Critical Strip?

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...

• have you checked in literature? – Mhenni Benghorbal Dec 27 '14 at 6:28
• I explored wikipedia, didn't find anything that explicitly remedied the issue. Unless it is one of those integration relations example : $\zeta(s) = \frac{1}{\Gamma(s)} \int_0^{\infty}\left[ \frac{x^{s-1}}{e^x-1} \ dx \right]$ – frogeyedpeas Dec 27 '14 at 6:31
• But it was not explained if that or any other relation is defined for $0 \le Re(s) \le 1$ – frogeyedpeas Dec 27 '14 at 6:32

An extension of the area of convergence can be obtained by rearranging the original series. The series

$$\zeta(s)=\frac{1}{s-1}\sum_{n=1}^\infty \left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)$$

converges for $\Re s > 0$. See here.

• Wiki doesn't explain the derivation of it, where is the best place to go for that? – frogeyedpeas Dec 27 '14 at 6:47
• @frogeyedpeas: Try to consult one of the books on zeta function. Try to manipulate the expression $(1-s)\zeta(s)$ using the series $\zeta(s)=\sum_{k=1}^{\infty} \frac{1}{k^s}$. – Mhenni Benghorbal Dec 27 '14 at 6:52
• @frogeyedpeas: Chapter 3 of The Prime Number Theorem by G.J.O. Jameson has a good explanation of extending the definition of the $\zeta$ function essentially as the accepted answer suggests. – daniel Dec 27 '14 at 7:04
• @daniel: Thank you for giving a reference. I really appreciate it. – Mhenni Benghorbal Dec 27 '14 at 7:06
• Good to have you back, Mhenni ! :-$)$ – Lucian Dec 27 '14 at 8:14

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.

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.

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.

It's $\tt\underline{usually}$ evaluated with the alternating series $\displaystyle{\quad\color{#66f}{\large% \zeta\left(\,s\,\right) ={1 \over 1 - 2^{1 - s}} \sum_{k\ =\ 1}^{\infty}{\left(\,-1\,\right)^{k - 1} \over k^{s}}}}$.

For example, $\displaystyle{\zeta\left(\,1 \over 2\,\right) ={1 \over 1 - \,\sqrt{\,2\,}\,} \sum_{k\ =\ 1}^{\infty}{\left(\,-1\,\right)^{k - 1} \over \,\sqrt{\,k\,}\,} \approx {\tt -1.4603545088095868129}}$

• Or through this extension: $$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n}{(n+1)^{s}}} + \frac{2\,s-1}{n^s} - {\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad 0<\Re(s)<1$$ we get an even simpler (I believe the simplest) series-expression for: $$\displaystyle \zeta\left(\frac12\right) = \sum _{n=1}^{\infty } \left(\sqrt{n-1} -{\frac {n}{\sqrt{n+1}}}\right)$$ – Agno Jan 2 '15 at 13:57
• @Agno These are very nice expressions. It's curious that textbooks seldom mention these ones. Thanks. – Felix Marin Jan 2 '15 at 21:59
• I prefer applying the Euler transform to get a globally convergent series that converges very rapidly. See my answer @Agno – Simply Beautiful Art Apr 14 '17 at 20:32
• @SimplyBeautifulArt I agree with you. Euler-T... is better whenever we want to perform a numerical evaluation. Indeed, Euler invented that to apply to the Basel-Problem because he had to calculate manually. Thanks for your remark. – Felix Marin Apr 14 '17 at 20:51
• :-) Not to mention how easy it is. – Simply Beautiful Art Apr 14 '17 at 20:58

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 $s\in\mathbb R$:

https://www.desmos.com/calculator/my05uiirse

• Your last representation is not just crazy fast, but it's "smooth" when you plot the partial sums for complex $s$ (no wiggles). – Dave huff Apr 14 '17 at 23:33
• @Davehuff Well, it is one of the fastest converging series representations of the Riemann zeta function is what I meant – Simply Beautiful Art Apr 14 '17 at 23:55

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 $\zeta$:

$\displaystyle \zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$

The exponential $\pi$ and $\Gamma$ factors are never zero! So if this $\Lambda$ thing is zero, the zero is coming from $\zeta$!