# Continuation of the Riemann Zeta Function

I am actually aware of the argument showing $\zeta$ has a meromorphic extension to $\mathbb{C}$ with a single pole at $z = 1$. On a recent number theory exam, however, one of the questions asked to show $\zeta$ has a meromorphic extension to $\mathrm{Re} \; z > -1$ (and to show the same for the Hurwitz zeta function). Since this was a timed exam with multiple problems I doubt the professor was looking for the argument mentioned earlier. Instead I think there must be a comparatively elementary way of showing this, which leads to my question:

Is there a relatively simple way to prove that $\zeta$ has a meromorphic extension to $\mathrm{Re} \; z > -1$ with only a simple pole at $z = 1$?

If it helps, the first part of this problem on the exam asked us to establish the following: For a continuously differentiable function $f$ on $\{x \in \mathbb{R} : x > 0\}$ there is an equality

$$\sum_{k=1}^n f(k) = \int_1^n f(x) dx + \frac{1}{2}(f(1)+f(n)) + \int_1^n \left(x - [x] -\frac{1}{2}\right)f'(x) dx$$

Not sure if this last piece of information helps or is simply a red herring, but I figured it would be best to include it.

## 1 Answer

Yes, your instinct was right to include the first part of the problem! The left-hand side of the equality becomes $\zeta(s)$ when you set $f(k)=k^{-s}$ and let $n$ go to infinity; so let's do the same to the right-hand side.

But first: let $B_2(y) = \int_1^y (x-\lfloor x\rfloor-\frac12)\,dx$. Note that $B_2$ is a continuous periodic function with period $1$ (taking the value $0$ at every positive integer) and in particular is bounded. Note that by integration by parts, $$\int_1^n \bigg(x - [x] -\frac{1}{2}\bigg)x^{-s-1} \,dx = (s+1) \int_1^n B_2(x) x^{-s-2}\,dx.$$

So now: when Re $s>1$, we have \begin{align*} \zeta(s) &= \lim_{n\to\infty} \sum_{k=1}^n k^{-s} \\ &= \lim_{n\to\infty} \bigg( \int_1^n x^{-s} \,dx + \frac{1}{2}(1^{-s}+n^{-s}) - s \int_1^n \bigg(x - [x] -\frac{1}{2}\bigg)x^{-s-1} \,dx \bigg) \\ &= \int_1^\infty x^{-s} \,dx + \frac{1}{2} - s \lim_{n\to\infty} (s+1) \int_1^n B_2(x) x^{-s-2}\,dx \\ &= \frac1{s-1} + \frac{1}{2} - s(s+1) \int_1^\infty B_2(x) x^{-s-2}\,dx, \tag{$*$} \end{align*} provided the integral converges. But in fact it converges absolutely, since $|B_2(x)|$ is bounded by some $C$: $$\int_1^\infty |B_2(x) x^{-s-2}|\,dx \le C \int_1^\infty x^{-s-2}\,dx = \frac C{s+1}.$$

This gives the representation ($*$) for $\zeta(s)$, which we derived assuming that Re $s>1$. However, the right-hand side of ($*$) is convergent and well-defined for Re $s>-1$ (by the argument we just did). Therefore ($*$) gives an analytic continuation of $\zeta(s)$ to the region Re $s>-1$.

If this seems a little magical (including my choice to call that function $B_2$), you can read about Euler-Maclaurin summation to get some context.

• Thanks! So yes it definitely appears my professor was aiming for this elementary proof. – James Miller Mar 10 '14 at 0:19