Extending $\zeta$ to a meromorphic function on $\mathbb{C} - \{1\}$

Question

I know that we can extend $\zeta (s)$, originally defined on $\Re(s)>1$ by the sum $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$, to the domain $\mathbb{C} - \mathbb{Z}$ by this definition: $$\zeta(s) = \frac{i}{2\Gamma(s)\sin(\pi s)}\displaystyle\int_C \frac{(-z)^{s-1}}{e^z -1}dz,$$ where $C$ is a small circle around the origin.

But it seems that we can extend the function even further, and include all integers except $1$, by this formula: $$\zeta(s) = \frac{s}{s-1} -s\displaystyle\int_{1}^{\infty}x^{-s-1}\{x\}dx,$$ where $\{x\}$ is the fractional part of the real number $x$. This will also show us that $\zeta$ can be extended to a meromorphic function on $\mathbb{C}$ with a simple pole at $s=1$. I really don't know how this formula can be derived. Any help would be great.

Answer 1

For every positive integer $n$ and every complex number $s$, consider the integral $$I_n(s)=\int_n^{n+1}sx^{-s-1}\{x\}dx$$ Using the change of variable $x\to x+n$, then the fact that $x=(x+n)-n$, one gets $$I_n(s)=\int_0^1s(n+x)^{-s-1}xdx=\int_0^1s(n+x)^{-s}dx-n\int_0^1s(n+x)^{-s-1}dx$$ Primitives are direct for both integrals on the RHS hence, for every $s\ne1$, $$I_n(s)=\frac s{s-1}\left(n^{1-s}-(n+1)^{1-s}\right)-a_n(s)$$ where one defines $$a_n(s)=n(n^{-s}-(n+1)^{-s})$$ If $\Re s>1$, $n^{1-s}\to0$ when $n\to\infty$, hence, summing the formulas for $I_n(s)$ and using the obvious cancellations, one gets, for every $s$ such that $\Re s>1$, $$\int_1^\infty sx^{-s-1}\{x\}dx=\sum_{n=1}^\infty I_n(s)=\frac s{s-1}-\sum_{n=1}^\infty a_n(s)$$ Now, $$a_n(s)=n^{1-s}-(n+1)^{1-s}+(n+1)^{-s}$$ hence, using again some obvious cancellations, one gets, for every $s$ such that $\Re s>1$, $$\sum_{n=1}^\infty a_n(s)=1+\sum_{n=1}^\infty(n+1)^{-s}=\zeta(s)$$ Reordering yields the desired result.