Riemann zeta function at $\Re(s)=0$ and $\Re(s)=1$

Question

Riemann zeta function at $\Re(s)=0$ and $\Re(s)=1$.

What happens to the zeta function at these points? For example $\sum_{n=1}^\infty \frac1{n^s}$ is defined for $\Re(s)>1$ and for $\Re(s)>0$ you have a different formula. But none of these include 0 or 1? Or does $\Re(S)>0$ include the 1? (or maybe it was defined wrong and should be $\Re(s)>0$ excluding $s=1$ in the book)

This is like asking "what happens to the function $\sum_{n\geq 0} z^n$ along $|z| = 1$?" The explicit series formula makes no sense there, but the function can be extended analytically from $|z| < 1$ to $\mathbf C -{1}$ by rewriting it as $1/(1-z)$. Something similar, but more subtle, can be done for $\zeta(s)$ on $\mathbf C - {1}$. — KCd, Feb 06 '16 at 01:12
You don't tell us what book, or what formula, so it's hard to answer your question, but a formula for real part greater than zero should work for real part equal to one (except of course at $s=1$, where there is a pole). — Gerry Myerson, Feb 06 '16 at 03:37

score 2 · Answer 1 · answered Feb 06 '16 at 02:03

$\qquad\qquad\qquad\qquad\qquad\qquad\quad\zeta\big(1^\pm\big)=\pm\infty~$ and $~\zeta(0)=-\dfrac12$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=0.$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=1.$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=\dfrac12$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(x)$ for $x\in(-14,-1).$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(x)$ for $x\in(-20,~1).$

$\quad$ Notice how each new “hump” to the left gets exponentially bigger than the previous one.

We also have $~\lim\limits_{x\to0^\pm}\zeta(1+x)-\dfrac1x=\gamma,~$ see Euler-Mascheroni constant for more information. — Lucian, Feb 06 '16 at 02:07
See also the functional equation and the particular values at negative integers of the Riemann $\zeta$ function. — Lucian, Feb 06 '16 at 02:14

score 0 · Answer 2 · answered Feb 06 '16 at 02:10

The zeta function, written in the series form you wrote down above, is only defined for the $Re(s)>1$, but it has an "analytic continuation" that is defined everywhere on the complex plane (not just $Re(s)>0$) except at $s=1$ where it has a simple pole. Thus it will have a well defined value at $=0$, which is $\zeta(0)= -1/2$.

For more information I recommend the book by H.M. Edwards, "Riemann's Zeta Function". It might also be useful to look up the notion of analytic continuation in any complex analysis textbook.

Riemann zeta function at $\Re(s)=0$ and $\Re(s)=1$

2 Answers2