# Analytic continuation continued

As I promised in my previous post, here is a derivation of the analytic continuation of the Riemann zeta function to negative integer values. There are several ways of doing this but a particularly simple way is given by Graham Everest, Christian Rottger, and Tom Ward at this link. It starts with the observation that you can write

$\int_1^\infty x^{-s} dx = \frac{1}{s-1}$

if the real part of $s>0$. You can then break the integral into pieces with

$\frac{1}{s-1}=\int_1^\infty x^{-s} dx =\sum_{n=1}^\infty\int_n^{n+1} x^{-s} dx$

$=\sum_{n=1}^\infty \int_0^1(n+x)^{-s} dx=\sum_{n=1}^\infty\int_0^1 \frac{1}{n^s}\left(1+\frac{x}{n}\right)^{-s} dx$      (1)

For $x\in [0,1]$, you can expand the integrand in a binomial expansion

$\left(1+\frac{x}{n}\right)^{-s} = 1 +\frac{sx}{n}+sO\left(\frac{1}{n^2}\right)$   (2)

Now substitute (2) into (1) to obtain

$\frac{1}{s-1}=\zeta(s) -\frac{s}{2}\zeta(s+1) - sR(s)$  (3)

or

$\zeta(s) =\frac{1}{s-1}+\frac{s}{2}\zeta(s+1) +sR(s)$   (3′)

where the remainder $R$ is an analytic function when $Re s > -1$ because the resulting series is absolutely convergent. Since the zeta function is analytic for $Re s >1$, the right hand side is a new definition of $\zeta$ that is analytic for $s >0$ aside from a simple pole at $s=1$. Now multiply (3) by $s-1$ and take the limit as $s\rightarrow 1$ to obtain

$\lim_{s\rightarrow 1} (s-1)\zeta(s)=1$

which implies that

$\lim_{s\rightarrow 0} s\zeta(s+1)=1$     (4)

Taking the limit of $s$ going to zero from the right of (3′) gives

$\zeta(0^+)=-1+\frac{1}{2}=-\frac{1}{2}$

Hence, the analytic continuation of the zeta function to zero is -1/2.

The analytic domain of $\zeta$ can be pushed further into the left hand plane by extending the binomial expansion in (2) to

$\left(1+\frac{x}{n}\right)^{-s} = \sum_{r=0}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\left(\frac{x}{n}\right)^r + (s+k)O\left(\frac{1}{n^{k+2}}\right)$

Inserting into (1) yields

$\frac{1}{s-1}=\zeta(s)+\sum_{r=1}^{k+1} \left(\begin{array}{c} -s\\r\end{array}\right)\frac{1}{r+1}\zeta(r+s) + (s+k)R_{k+1}(s)$

where $R_{k+1}(s)$ is analytic for $Re s>-(k+1)$.  Now let $s\rightarrow -k^+$ and extract out the last term of the sum with (4) to obtain

$\frac{1}{-k-1}=\zeta(-k)+\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) - \frac{1}{(k+1)(k+2)}$    (5)

Rearranging (5) gives

$\zeta(-k)=-\sum_{r=1}^{k} \left(\begin{array}{c} k\\r\end{array}\right)\frac{1}{r+1}\zeta(r-k) -\frac{1}{k+2}$     (6)

where I have used

$\left( \begin{array}{c} -s\\r\end{array}\right) = (-1)^r \left(\begin{array}{c} s+r -1\\r\end{array}\right)$

The righthand side of (6) is now defined for $Re s > -k$.  Rewrite (6) as

$\zeta(-k)=-\sum_{r=1}^{k} \frac{k!}{r!(k-r)!} \frac{\zeta(r-k)(k-r+1)}{(r+1)(k-r+1)}-\frac{1}{k+2}$

$=-\sum_{r=1}^{k} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2}$

$=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \frac{\zeta(r-k)(k-r+1)}{(k+1)(k+2)}-\frac{1}{k+2} - \frac{\zeta(0)}{k+1}$

Collecting terms, substituting for $\zeta(0)$ and multiplying by $(k+1)(k+2)$  gives

$(k+1)(k+2)\zeta(-k)=-\sum_{r=1}^{k-1} \left(\begin{array}{c} k+2\\ k-r+1\end{array}\right) \zeta(r-k)(k-r+1) - \frac{k}{2}$

Reindexing gives

$(k+1)(k+2)\zeta(-k)=-\sum_{r'=2}^{k} \left(\begin{array}{c} k+2\\ r'\end{array}\right) \zeta(-r'+1)r'-\frac{k}{2}$

Now, note that the Bernoulli numbers satisfy the condition $\sum_{r=0}^{N-1} B_r = 0$.  Hence,  let $\zeta(-r'+1)=-\frac{B_r'}{r'}$

and obtain

$(k+1)(k+2)\zeta(-k)=\sum_{r'=0}^{k+1} \left(\begin{array}{c} k+2\\ r'\end{array}\right) B_{r'}-B_0-(k+2)B_1-(k+2)B_{k+1}-\frac{k}{2}$

which using $B_0=1$ and $B_1=-1/2$ gives the self-consistent condition

$\zeta(-k)=-\frac{B_{k+1}}{k+1}$,

which is the analytic continuation of the zeta function for integers $k\ge 1$.