Analytic Continuation of Riemann Zeta Function

Theorem
The analytic continuation of the Riemann zeta function to the half-plane $\left\{{z : \Re \left({s}\right)>0}\right\} \ $ is given by


 * $\displaystyle \zeta(s) = \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $

If $\Re \left({s}\right) \leq 0 \ $, the value of $\zeta(s) \ $ can be computed from the relation


 * $\displaystyle \Gamma \left({\frac{s}{2}}\right) \pi^{-s/2} \zeta(s) = \Gamma \left({ \frac{s-1}{2} }\right) \pi^{\frac{1-s}{2}} \zeta(1-s) \ $

Proof
The very first thing to show is that the zeta function, as defined, is analytic in the first place. But the zeta series converges, and since it is a Dirichlet series, it is analytic. Therefore, we need only concern ourselves with the analytic continuations presented above.

The proof must be broken into several parts:

Part 1: If $\displaystyle \Re(s)>1, \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \sum_{n=1}^\infty \frac{1}{n^s} \ $

Part 2: $\displaystyle \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $ is analytic throughout $\Re(s)>0 \ $, except $s=1 \ $.

Part 3: The symmetric equation $\displaystyle \Gamma \left({ \frac{s}{2} }\right) \pi^{-s/2} \zeta(s) = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{\tfrac{s-1}{2}} \zeta(1-s) \ $ yields an analytic function in $\Re(s) < 0 \ $

Part 4: The "seam" $\Re(s)=0 \ $ is analytic as well.

Part 1
For $\displaystyle \Re(s)>1, \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \ $.

Now in the region $\Re(s)>1, \ $ we have $\Re(1-s)<0 \ $, so setting $x=1-s \ $ we have $|2^x|<1 \ $ and so we may write:


 * $\displaystyle \frac{1}{1-2^x} = \sum_{n=1}^\infty \left({2^x}\right)^n \ $

and then:


 * $\displaystyle \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \left({  \sum_{m=0}^\infty \left({2^x}\right)^m }\right) \left({ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} }\right) \ $

Now, since $|2^x|<1 \ $, the root test guarantees the absolute convergence of the series on the left. Since the zeta series converges absolutely, the Comparison Test guarantees the series on the right is absolutely convergent. Therefore, we can determine the product as

$\displaystyle = \left({ 1 + \sum_{m=1}^\infty 2^{s-ms} }\right) \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $

$\displaystyle = \sum_{n=1}^\infty   \left({  \frac{(-1)^{n-1}}{n^s} + \sum_{ab=n} (-1)^{b-1} \frac{2^a}{ \left({2^a b }\right)^s }  }\right)    \ $

It is clear that every term of this sum will be of the form $\displaystyle \pm \frac{u}{v^s} = \pm \left({\frac{1}{v^s} + \dots + \frac{1}{v^s} }\right) $, where $u,v \in \N \ $. Let us examine how often the term $\displaystyle \frac{+1}{v^s} \ $ appears for each $v\in\N \ $.

For $v \ $ odd, there will be no $a,b \ $ such that $v^s = (2^a b)^s \ $, and so the only terms of this form will come from the $\displaystyle \frac{(-1)^{n-1}}{n^s} \ $ terms; hence for $v \ $ odd, $\displaystyle \frac{1}{v^s} \ $ will appear only once, and always be positive.

Now we turn our attention to the case where $v \ $ is even: how many $\displaystyle \frac{1}{v^s} \ $ do we have?

From the expression $\displaystyle \frac{(-1)^{n-1}}{n^s} \ $ we know we will get a single expression $\displaystyle \frac{-1}{v^s} \ $. Now, let $p \ $ be the highest power of $2 \ $ dividing $v \ $. Then $v=2^p q \ $ where $q \ $ is an odd, and $v=2^{p-r} \times (2^r q) \ $, and this will describe every single appearance of $\displaystyle \frac{1}{v^s} \ $. So we get

$\displaystyle \frac{-1}{v^s} + \frac{2^p}{v^s} - \left({\frac{2^{p-1}}{v^s} + \frac{2^{p-2}}{v^s} + \dots + \frac{2}{v^s} }\right) = \frac{1}{v^s} \left({ 2^p - \sum_{n=0}^{p-1} 2^n }\right) \ $

Examine this expression on the right, $\displaystyle 2^p - \sum_{n=0}^{p-1} 2^n \ $. Clearly, for $p=1 \ $, this is $1 \ $. Now suppose this expression is equal to $1 \ $ for some $p \ $; then

$\displaystyle 2^{p+1} - \sum_{n=0}^p 2^n = 2 \cdot 2^p - \left({ 2^p + \sum_{n=0}^{p-1} 2^n }\right) = 2^p + 2^p - 2^p - \sum_{n=0}^{p-1} 2^n =1 $

and the induction is complete; this expression is $1 \ $ for all $p \ $. Therefore, the term $\displaystyle \frac{1}{v^s} \ $ appears once and is positive for $v \ $ even; combining this with our previous results means that

$\displaystyle \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \sum_{n=1}^\infty \frac{1}{n^s} \ $

for $\Re(s)>1 \ $.

Part 2
$\displaystyle \frac{1}{1-2^{1-s}} \ $ is analytic whenever $s \neq 1 \ $, since it has complex derivative $\displaystyle \frac{2^{1-s}}{2^{2-s}-2^{2-2s}-1}\log(2) \ $. Hence we must show that the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $ converges in the region in question, except $s=1 \ $. It will then follow that the sum is analytic, and then it will follow that the product of the two, the continuation itself, is analytic.

Let $s=x+iy \ $; then

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x n^{iy}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x} e^{-iy \log(n)} \ $

For all $j, k \geq N, j \geq k \ $, the difference between the $k^{th} \ $ term and the $j^{th} \ $ partial sums will be

$\displaystyle \left|{ \sum_{n=1}^j (-1)^{n-1} n^{-x} e^{-iy\log(n)} - \sum_{n=1}^k (-1)^{n-1} n^{-x} e^{-iy\log(n)} }\right| = \left|{ \sum_{n=k}^j (-1)^{n-1} n^{-x} e^{-iy\log(n)} }\right| \ $

but observe that since



$\displaystyle \frac{d}{dx} \log(x) = \frac{1}{x}, \log(x+1) \sim \log(x) + \frac{1}{x} \ $.

$\displaystyle \left|{  (-1)^{k-1} k^{-x} e^{-iy\log(k)} + (-1)^{k} (k+1)^{-x} e^{-iy \log(k+1)} +  \dots + (-1)^{j-1} j^{-x} e^{-iy\log(j)}  }\right| \ $

$\displaystyle \sim \left|{  (-1)^{k-1} k^{-x} e^{-iy\log(k)} + (-1)^{k} (k+1)^{-x} e^{-iy \log(k) } e^{-\tfrac{iy}{k}} +  \dots + (-1)^{j-1} j^{-x} e^{-iy\log(j)}  }\right| \ $

approximately. Now, for very large $k \ $, $e^{-\tfrac{iy}{k}} \sim 1 \ $, and so this sum will be approximately

$\displaystyle \left|{ \pm \left({ k^{-x} - (k+1)^{-x} }\right) e^{-iy \log(k) }  +  \dots + (-1)^{j-1} j^{-x} e^{-iy\log(j)}  }\right| \ $

$\displaystyle = \left|{ \pm O \left({ \frac{1}{k^2} }\right)^{-x} e^{-iy \log(k) }  +  \dots + (-1)^{j-1} j^{-x} e^{-iy\log(j)}  }\right| \ $

by the Binomial Theorem. Now if $j-i \ $ is even, all the terms of the sum can be paired up in this way; if it is odd, there will be a single term not paired. Considering the odd case as an even case, plus another term, we examine only the odd case to find the sum is then

$\displaystyle = \left|{ \pm O \left({ \frac{1}{k^2} }\right)^{-x} e^{-iy \log(k) }  \pm   O \left({ \frac{1}{(k+2)^2} }\right)^{-x} e^{-iy \log(k+2) } \pm \dots + (-1)^{j-1} j^{-x} e^{-iy\log(j)}  }\right| \ $

which, by the Triangle Inequality, is

$\displaystyle \leq \left|{ O \left({ \frac{1}{k^2} }\right)^{-x} }\right| + \left|{ O \left({ \frac{x}{(k+2)^2} }\right)^{-x}}\right| + \dots + \left|{j^{-x}  }\right| \ $

Hence

$\displaystyle \lim_{N\to\infty} \left|{ \sum_{n=k}^j (-1)^{n-1} n^{-x} e^{-iy\log(n)} }\right| \leq \lim_{N\to\infty} \left({ \left|{ O \left({ \frac{1}{k^2} }\right)^{-x} }\right| + \left|{ O \left({ \frac{x}{(k+2)^2} }\right)^{-x}}\right| + \dots + \left|{j^{-x}  }\right| }\right) = 0 \ $

But this is precisely what it means for a series to be Cauchy; hence, the series converges.

Part 3
If $\Re \left({s}\right) \leq 0 \ $, the value of $\zeta(s) \ $ is precisely

$\displaystyle \zeta(s)=\frac{\Gamma \left({ \frac{1-s}{2} }\right) \pi^{\frac{s-1}{2}} \zeta(1-s) }{\Gamma(\frac{s}{2}) \pi^{-s/2}} = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{s-\frac{1}{2}} \zeta(1-s)   \frac{1}{\Gamma(\frac{s}{2})} $

But the Gamma function is analytic for complex terms with positive real part, and terms like $\displaystyle \pi^{\frac{1-s}{2}} \ $ and $\pi^{-s/2} \ $ are analytic everywhere, since they have easy to compute complex derivatives well-defined everywhere, and we know already that $\zeta(1-s) \ $ is analytic for $\Re(s)<0 \ $, so we only need apply Sums, Products, and Quotients of Analytic Functions repeatedly:

$\displaystyle \zeta(s) = \left({ \text{Some analytic function} }\right) \times \frac{1}{\Gamma(\frac{s}{2})} \ $

So it suffices to show that the inverse gamma function is analytic for the left half-plane. But since the gamma function is analytic except for its poles, and it is never zero, the inverse gamma function is analytic for the entire complex plane.

Hence $\zeta(s) \ $ is analytic for the whole left half-plane.

Part 4
Since $\zeta(s) \ $ is analytic for $\Re(s)=1 \ $, by part 2, we need only show that the symmetric equation holds in the critical strip to demonstrate that $\zeta(0+iy) \ $ is an analytic function of the complex variable.

The symmetric equation is

$\displaystyle \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{s-\frac{1}{2}} \left({ \frac{1}{1-2^{s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{1-s}} }\right)   \frac{1}{\Gamma(\frac{s}{2})} \ $

where $\Gamma(z) \ $ is the Gamma function. Re-arranging terms and inserting definition, we have

$\displaystyle \frac{1-2^s}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \left({ \frac{2}{1-s} \prod_{n=1}^\infty \left({ \left({ 1+\frac{1}{n} }\right)^\frac{1-s}{2} \left({1+\frac{1-s}{2n} }\right)^{-1} }\right) }\right) \pi^{s-\frac{1}{2}} \left({  \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{1-s}}  }\right)   \left({ \frac{s}{2} e^{\gamma  s/2} \prod_{n=1}^\infty \left({ \left({ 1+ \frac{s}{2n} }\right) e^{-\frac{s}{2n}} }\right) }\right) \ $

Which becomes

$\displaystyle \frac{1-2^s}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}= \frac{s}{1-s} \pi^{s-\frac{1}{2}} e^{\gamma  s/2} \left({  \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{1-s}}  }\right) \prod_{n=1}^\infty \left({ \left({ 1+\frac{1}{n} }\right)^\frac{1-s}{2} \left({1+\frac{1-s}{2n} }\right)^{-1}   \left({ 1+ \frac{s}{2n} }\right) e^{-\frac{s}{2n}}  }\right) \ $

which is obviously true, right?