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

0" title="Analytic continuation continued" />0" class="latex" title="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)

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

(3′)

where the remainder

is an analytic function when

-1" title="Analytic continuation continued" /> -1" class="latex" title="Re s > -1" /> because the resulting series is absolutely convergent. Since the zeta function is analytic for

1" title="Analytic continuation continued" />1" class="latex" title="Re s >1" />, the right hand side is a new definition of

$\zeta$

that is analytic for

0" title="Analytic continuation continued" />0" class="latex" title="s >0" /> aside from a simple pole at

. Now multiply (3) by

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

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

-(k+1)" title="Analytic continuation continued" />-(k+1)" class="latex" title="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

-k" title="Analytic continuation continued" /> -k" class="latex" title="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

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

and

gives the self-consistent condition

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

which is the analytic continuation of the zeta function for integers

$k\ge 1$

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