# Detailed proof of $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$

I'm trying to understand the proof of PNT by Don Zagier. But his proof is too simplified so I can't understand it. I got stumped at step II: $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$.

The proof is available here: http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf

Can anyone post a detailed proof that $\zeta(s)-1/(s-1)$ extends holomorphically to $\Re(s)>0$？

Or please tell me which book has a detailed proof so I can look it up?

edit:
Zagier's proof:
$\displaystyle \zeta(s)-\frac{1}{s-1} = \sum_{n=1}^{\infty} \int_{n}^{n+1} (\frac{1}{n^s}-\frac{1}{x^s}) dx$
$\displaystyle \left|\int_{n}^{n+1} (\frac{1}{n^s}-\frac{1}{x^s}) dx\right| \leq \frac{|s|}{(n^{\Re(s)+1})}$
therefore $\displaystyle \zeta(s)-\frac{1}{s-1}$ extends holomorphically to $\Re(s)>0$

My question: We have a function
(i) $f(s)=\sum_{1}^{\infty} g_n(s) \quad \forall \Re(s)>0$
(ii) $g_n(s) \leq |s|/(n^{\Re(s)+1}) \quad \forall \Re(s)>0$

Is (i) and (ii) the necessary and sufficient condition that $f(s)$ has holomorphic continuation?
Does there exist functions $f$ and $g_n$ that satisfy (i) and (ii), but does not have holomorphic continuation?
Do we need to prove other conditions, for example $g_n$ must be continuous/holomorphic?

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What part exactly you don't understand in that proof? It looks to me pretty straightforward... – DonAntonio Aug 25 '12 at 18:35
@alek: You might notice that I merged the account you used to write the post with the account you wrote the answer/edit. This will let you comment and edit your own question. Best of luck. – mixedmath Aug 26 '12 at 4:45

Start with the defining series, convergent for $\Re(s)>1$: $$\sum_{n=1}^\infty \frac{1}{n^s} = \sum_{n=1}^\infty \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \mathrm{e}^{-n t} \mathrm{d} t = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \frac{\mathrm{e}^{-t}}{1- \mathrm{e}^{-t}} \mathrm{d} t$$ Also notice that, for $\Re(s)>1$, $$\frac{1}{s-1} = \frac{\Gamma(s-1)}{\Gamma(s)} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-2} \mathrm{e}^{-t} \mathrm{d} t$$ Now subtracting we have: $$\zeta(s) - \frac{1}{s-1} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} \mathrm{e}^{-t} \left( \frac{1}{1- \mathrm{e}^{-t}} - \frac{1}{t} \right) \mathrm{d} t$$ The integral on the right-hand side converges now for $\Re(s)>0$, since at the lower integration bound $\frac{1}{1-\exp(-t)} - \frac{1}{t} = \frac{1}{2} + \mathcal{o}(1)$.

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Trying to address what I think was troubling the OP.

Fix constants $a,b$ such that $0<a<b$. Consider the set $$L(a,b)=\{x+yi\in\mathbb{C}\mid a<x, |x+yi|<b\}.$$ The functions $g_n(z)$ are holomorphic in the set $L(a,b)$. Item $(ii)$ means that for all $z\in L(a,b)$ and all positive integers $n$ we have $$|g_n(z)|\le\frac{b}{n^{1+a}}.$$ Therefore, by Weierstrass' M-test (note that here the lower bound on the exponent, $1+a$, does not depend on $z$, only on the set $L(a,b)$), the series $f(z)=\sum_{n=1}^\infty g_n(z)$ converges uniformly in the set $L(a,b)$ and therefore gives a holomorphic function on $L(a,b)$.

The right half-plane is the union of the sets $L(a,b)$, so $f(z)$ is holomorphic there.

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In other words, use the techniques familiar from proving that a converging power series gives a holomorphic function: Shrink the set in a suitable way such that your estimates prove the convergence to be uniform in that smaller set. – Jyrki Lahtonen Aug 26 '12 at 9:50