# Proof of $\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1)$

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.?

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

$$f(s)=\left(\zeta(s)-\frac{1}{s-1}\right) = \int_{0}^{+\infty}\frac{x^{s-1}}{\Gamma(s)}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx$$ 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.
• Still not getting !! How from dominated convergence theorem the integral becomes $O(s-1)$ ?
• If $f(x)$ is holomorphic in a neigbourhood of the origin, $$f(x) = f(0)+f'(0)x+o(x)=f(0)+O(x).$$ Commented Oct 28, 2018 at 10:21