I have a question regarding Riemann's first proof of the functional equation that was given in his paper on the Riemann zeta function. I am an undergraduate working on a fairly short undergraduate report on the zeta function and as such the proof of the functional equation that I give, while going into the contour integration and residue theorem, etc, also leaves out certain technicalities that would otherwise bog it down.

I am aware that the functional equation attained via this method can be used to extend the zeta function to the whole complex plane (apart from 1). However, as far as I have been able to tell, the proof of the functional equation assumes that $\operatorname{Re}(s)>1$ (this is used for instance in the proof that $\Gamma(s) \zeta(s) = \int_{0}^{\infty} \frac{u^{s-1}}{e^{u}-1} du$ for $\operatorname{Re}(s)>1$, which is later incorporated in the proof of the functional equation). As a result, the functional equation $\zeta(s)=2(2\pi)^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$ appears to my mind only to additionally define the zeta function for $\operatorname{Re}(s)<0$, which leaves the strip $0\leq\operatorname{Re}(s)\leq1$ "uncontinued".

I know that this contour proof $is$ used to extend the zeta function to the whole complex plane, so my question therefore is: where exactly during the proof can we say that the zeta function can be extended analytically to $\operatorname{Re}(s)\leq1$ rather than just $\operatorname{Re}(s)<0$, and how can I - bearing in mind I am aiming for a rather rough undergraduate-level explanation - explain that although we have started by assuming $\operatorname{Re}(s)>1$, the functional equation we obtain is actually correct for the strip $0\leq\operatorname{Re}(s)\leq1$ as well as the continued region $\operatorname{Re}(s)<0$?


The integral

$$I(s) = \int_\gamma \frac{(-z)^{s-1}}{e^z-1}\,dz,$$

where $\gamma$ is a contour enveloping the positive real axis, exists for all $s \in \mathbb{C}$ and defines an entire function. For $\operatorname{Re} s > 1$, we have

$$2 \sin \pi s\,\Gamma(s)\zeta(s) = iI(s)$$

from the known $\Gamma(s)\zeta(s) = \int_0^\infty \frac{u^{s-1}}{e^u-1}\,du$, and that defines $\zeta(s)$ via

$$\zeta(s) = \frac{i}{2\sin \pi s\,\Gamma(s)}I(s)$$

as a meromorphic function on the entire plane. The functional equation is then derived from this, after the continuation to an entire meromorphic function has been established.

  • $\begingroup$ I have an additional query: this analytic continuation of the zeta function using the contour approach doesn't require the use of the Dirichlet eta function at all? The one which allows us to extend the zeta function to the half-plane $\operatorname{Re}(s)>0$? Also, I may be incorrect but shouldn't it be $2i \sin \pi s\,\Gamma(s)\zeta(s) = I(s)$, as I get $2i \sin \pi s\,$ coming from $e^{\pi is}-e^{-\pi is}$ in the calculations? $\endgroup$ – user138908 Apr 13 '14 at 22:55
  • 1
    $\begingroup$ Riemann's continuation doesn't make use of the Dirichlet $\eta$-function. Whether you get the $i$ on the left or the right hand side depends on the orientation of $\gamma$. With $\gamma$ going from $+\infty +i\varepsilon$ to $\delta + i\varepsilon$ in the upper half-plane, then around $0$ on a circular arc, and finally from $\delta - i\varepsilon$ to $+\infty - i\varepsilon$ in the lower half-plane, you get a factor $e^{-i\pi s} - e^{i\pi s} = -2i\sin \pi s$, so the $i$ belongs on the right, where Riemann put it, or you have a $-i$ on the left. $\endgroup$ – Daniel Fischer Apr 13 '14 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.