For $\sigma>1$, summation over the intervals $[n,n+1)$ gives
$$
\zeta(s)=s\int_1^\infty\frac{\lfloor u\rfloor}{u^{s+1}}\,du.
$$
Since $\lfloor u\rfloor=u-\{u\}$,
$$
\boxed{\zeta(s)=\frac{s}{s-1}
-s\int_1^\infty\frac{\{u\}}{u^{s+1}}\,du.}
$$
The integral converges absolutely and defines a <holomorphic function> for $\sigma>0$. This formula therefore gives the <Meromorphic continuation of the Riemann zeta function to the right half-plane>, with only a simple pole at $s=1$ and residue one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 3 / a / Solution

{c}

For $\Re s>0$,
$$
\zeta(s)=\frac{s}{s-1}
-s\int_1^\infty\frac{\{u\}}{u^{s+1}}\,du.
$$
The integral defines a <holomorphic function> in that half-plane, so this continues $\zeta$ meromorphically across $\Re s=1$, with a simple pole of residue one at $s=1$.

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