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Prove that $\displaystyle \zeta(s)=\frac{1}{s-1}+\gamma+O(s-1)$,near $s=1$, where $\gamma$ is Euler's Constant.

I've proved $\displaystyle \zeta(s)=s\int_1^{\infty}\frac{[x]-x+1/2}{x^{s+1}}\,dx+\frac{1}{s-1}+\frac 12$. Also I've $$ \lim_{s\to 1}\left\{\zeta(s)-\frac{1}{s-1}\right\}=\gamma.$$I've stuck from where $O(s-1)$ comes ?

Any hint.?

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    $\begingroup$ $F(s) =\zeta(s)-\frac{s}{s-1}=\zeta(s)-\frac{1}{s-1}-1 = s \int_1^\infty (\lfloor x \rfloor -x) x^{-s-1}dx$ converges absolutely so is analytic for $\Re(s) > 0$, whence for $|s-1| < 1$, $F(s) = \sum_{k=0}^\infty \frac{F^{(k)}(1)}{k!} (s-1)^k= F(1)+O(s-1)$ $\endgroup$
    – reuns
    Commented Oct 27, 2018 at 21:55
  • $\begingroup$ @reuns Thanks ! Got it $\endgroup$
    – Topo
    Commented Oct 28, 2018 at 4:35

1 Answer 1

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$$f(s)=\left(\zeta(s)-\frac{1}{s-1}\right) = \int_{0}^{+\infty}\frac{x^{s-1}}{\Gamma(s)}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx $$ is a holomorphic function in a neighbourhood of $s=1$.
Once $\lim_{s\to 1}f(s)=\gamma$ has been proved through the dominated convergence theorem, $$ f(s)-\gamma = O(s-1) $$ as $s\to 1$ is automatic.

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  • $\begingroup$ Still not getting !! How from dominated convergence theorem the integral becomes $O(s-1)$ ? $\endgroup$
    – Topo
    Commented Oct 28, 2018 at 3:58
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    $\begingroup$ If $f(x)$ is holomorphic in a neigbourhood of the origin, $$ f(x) = f(0)+f'(0)x+o(x)=f(0)+O(x).$$ $\endgroup$ Commented Oct 28, 2018 at 10:21

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