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

Question

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.?

Answer 1

$$f(s)=\left(\zeta(s)-\frac{1}{s-1}\right) = \int_{0}^{+\infty}\frac{x^{s-1}}{\Gamma(s)}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx $$ 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.