# What is the analytic continuation of the Riemann Zeta Function

I am told that when computing the zeroes one does not use the normal definition of the rieman zeta function but an altogether different one that obeys the same functional relation. What is this other function that they use explicitly given?

Also if I were to take one of these non trivial zeroes and plug it into the original definition would my answer tend towards zero as I evaluate the series?

• The series that one usually uses to define the zeta function converges only when the argument $s$ has real part greater than $1$; proving this is a standard exercise. Obviously, this means that you cannot evaluate that series at a zero of the zeta function. Jul 7 '13 at 3:45

There is only one "normal" definition of the Zeta function. For $$\operatorname{Re}(s) > 1$$, the zeta function is defined as $$\displaystyle \sum_{k=1}^{\infty} \dfrac1{k^s}$$. For the rest of the $$s$$ in the complex plane, it is defined as the analytic continuation of the above function. The functional equation $$\zeta(s) = 2^s \pi^{s-1} \sin \left(\dfrac{\pi s}2\right) \Gamma(1-s) \zeta(1-s)$$ can be used to obtain the value of the $$\zeta$$ function for $$\operatorname{Re}(s) < 1$$, using the value of the zeta function for $$\operatorname{Re}(s)>1$$.

If you evaluate the series $$\displaystyle \sum_{k=1}^{\infty }\dfrac1{k^s}$$ for $$\text{Real}(s) \leq 1$$, the series will not converge. So you cannot evaluate the series at any of the zeros, let alone non-trivial zeros.

• So if I stuck one of the complex zeroes in place of s in your definition will it converge to 0? Jul 7 '13 at 3:40
• I Mean non trivial zeroes Jul 7 '13 at 3:42
• @frogeyedpeas If you evaluate the series $\displaystyle \sum_{k=1}^{\infty }\dfrac1{k^s}$ for $\text{Real}(s) \leq 1$, you will get the answer not to converge. So you cannot evaluate the series at any of the zeros, let alone non-trivial zeros.
– user17762
Jul 7 '13 at 3:42
• That will only give the value for Real$(s)\le 0$. What about the key strip? Jul 7 '13 at 5:25
• @Michael non-trivial zeros are the ones on the critical strip, so it answers your question to a degree. I'm not sure how this divergence implies it for all zeros. Jul 28 '15 at 3:21

If you want the analytic continuation of the zeta function to the zone where all the non-trivial zeros have been found so far, you can do as follows:

\begin{align*}(1)&\;\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\\ (2)&\;\sum_{n=1}^\infty \frac2{(2n)^s}=\frac1{2^{s-1}}\zeta(s)\end{align*}\;\;\;\;\left.\right\}\;\;\;\text{Re}\,(s)>1

Now, substract (2) from (1):

$$\left(1-\frac1{2^{s-1}}\right)\zeta(s)=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\ldots=\sum_{n=1}^\infty(-1)^{n-1}\frac1{n^s}=:\eta(s)\implies$$

$$\implies\;\zeta(s)=\left(1-2^{1-s}\right)^{-1}\eta(s)$$

It's a nice exercise now to prove the right hand side is analytic on $\;1\neq\;\text{Re}\,(s)>0\,$ .

Note that there are some potentially problematic points:

$$1-2^{s-1}=0\iff e^{(s-1)\log2}=1\iff (s-1)=\frac{2k\pi i}{\log2}\;,\;\;k\in\Bbb Z$$

Yet these are removable singularities, so no problem...

We have functional equation $$\zeta(s)=2^{s}\pi^{s-1}\sin\Bigl(\frac{\pi s}{2}\Bigr)\Gamma(1-s)\zeta(1-s).$$ We can prove this using contour integral.
Using this formula, we can expand Riemann Zeta Function to the whole complex plane except $s\neq1$.

To find the zeros of Riemann Zeta Function, you can use the well-known formula $$\left|Z(t)\right|=\left|\zeta(1/2+it)\right|$$ where $$Z(t)=\zeta\left(1/2+it\right)\frac{\Gamma(1/4+it/2)\pi^{-1/4-it/2}}{\left|\Gamma(1/4+it/2)\pi^{-1/4-it/2}\right|}$$ for real $t$.

• How does this functional equation work for s=0? Aug 15 '17 at 21:49

If by "normal definition of the Riemann zeta-function" you mean $$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$$ well, the thing about that series is that it doesn't converge for real part of $s$ less than or equal to $1$. In particular, it doesn't converge at any of the zeros of the zeta-function, trivial or otherwise.

To understand how the zeta-function is defined for real part less than or equal to 1, you have to be familiar with "analytic continuation." Are you?

• idk about then, but i am now :) Nov 25 '17 at 6:36