# Riemann's thinking on symmetrizing the zeta functional equation

In the translated version of Riemann's classic On the Number of Prime Numbers less than a Given Quantity, he quickly derives the zeta functional equation through contour integration essentially as

$$\zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin(\tfrac12\pi s)\zeta(1-s).$$

and says three lines later that it may be expressed as

$$\xi(s) = \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)=\xi(1-s).$$

After reading MO-Q7656, I started wondering whether, as his ideas evolved before he wrote the paper, he first constructed $\xi(s)$ by noticing that multiplying $\zeta(s)$ by $\Gamma(\frac{s}{2})$ introduces a simple pole at $s=0$ thereby reflecting the pole of $\zeta(s)$ at $s=1$ through the line $s=1/2$ and that the other simple poles of $\Gamma(\frac{s}{2})$ are removed by the zeros on the real line of the zeta function. The $\pi^{-s/2}$ can easily be determined as a normalization by an entire function $c^s$ where $c$ is a constant, using the complex conjugate symmetry of the gamma and zeta fct. about the real axis.

Anyone familiar with how his ideas (thinking) evolved?

(Update 5/13/2012) Riemann in his fourth equality in his paper, before he writes down the functional eqn., has

$$2sin(\pi s)(s-1)!\zeta(s)=i\int_{+\infty}^{+\infty}\frac{(-x)^{s-1}}{e^x-1}dx$$

where the contour sandwiches the positive real axis and surrounds the origin in the positive sense.

For $m=0,1,2, ...,$ this gives

$$\zeta(-m)=\frac{(-1)^{m}}{2\pi i}\oint_{|z|=1}\frac{m!}{z^{m+1}}\frac{1}{e^z-1}dz=\frac{(-1)^{m}}{m+1}\frac{1}{2\pi i}\oint_{|z|=1}\frac{(m+1)!}{z^{m+2}}\frac{z}{e^z-1}dz$$

from whence you can see, if you are familiar with the exponential generating fct. for the Bernoulli numbers, that the integral vanishes for even $m$. Riemann certainly was familiar with these numbers and states that the integral relation he gives implies the vanishing of $\zeta(s)$ for $m$ even (but gives no explicit proof). Edwards in Riemann's Zeta Function (pg. 12, Dover ed.) even speculates that ".. it may well have been this problem of deriving (2) [Euler's formula for $\zeta(2n)$ for positive $n$] anew which led Riemann to the discovery of the functional equation ...."

Riemann gives two proofs of the fct. eqn.--the first based on contour integration and the singularities of $\frac{1}{e^z-1}$, the second, on the theta function. Edwards wonders:

"Since the second proof renders the first proof wholly unnecessary, one may ask why Riemann included the first proof at all. Perhaps the first proof shows the argument by which he originally discovered the functional equation or perhaps it exhibits some properties which were important in his understanding of it."

In fact, Riemann states on page 3 of his paper, "This property of the function [$\xi(s)=\xi(1-s)$] induced me to introduce, in place of $(s-1)!$, the integral $(s/2-1)!$ into the general term of the series" for zeta. And then he proceeds to introduce the Jacobi theta function.

Edit Dec 18, 2014

In the early 1820s, both Abel and Plana separately published what is now called the Abel-Plana formula (see Wikipedia). In the title of Plana's, he mentions the Bernoulliens. I wonder how much Riemann was influenced by these papers.

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Something seems odd to me about the prospect of someone at that time noticing that $\Gamma(s/2)$'s poles absorbed $\zeta(s)$'s trivial zeros prior to having a functional equation with which to know about the trivial zeros in the first place. Ultimately, it looks to me like Riemann noticed a clever way to write $\zeta$ as an integral transform of a theta function, which already had a known functional equation. This also explains the reason for the precisely chosen normalization factor of $\pi^{-s/2}$. – anon May 10 '12 at 13:02
Riemann knew the singularities of the gamma function (and Euler's integral for it) as well as the well-known singularity of zeta at s=1 (no series convergence there). In the paper he uses these singularities to derive the functional equation. He must surely have known of the Bernoulli numbers and that given a well-behaved integrand for the MT divided by $(s-1)!$ the values of the transform at $s=0,-1,-2,...$ are the derivatives of the integrand eval at $0$, mod a sign--no need for full analytic continuation. The inventor of Riemann surfaces most likely understood these relations. – Tom Copeland May 10 '12 at 13:42
MT refers to the Mellin transform which Riemann used to define the zeta fct. Riemann would also have known that the Mellin transform of the gaussian curve $exp(-\pi x^2)$ gives $\pi^{-s/2}\Gamma(s/2)/2$, another natural reason to choose $\pi^{-s/2}$, and that $\Gamma(1/2)=\pi^{1/2}$, etc. – Tom Copeland May 10 '12 at 14:03
Erratum for my comment: Rather Riemann uses the singularities of $\frac{1}{e^z-1}$ in his first proof of the functional eqn. – Tom Copeland May 13 '12 at 0:53
– Tom Copeland May 20 '12 at 2:05

Although Riemann was the first to prove the functional equation for complex $s$, the ideas certainly predate Riemann. Euler knew something equivalent to the functional equation for integer values of $s$, via Abel summation where the series does not converge. Weil's 1975 paper "Two lectures on number theory, past and present" (in his collected works vol. 3) says that before Riemann, both Schlomilch and Malmquist published the functional equation for the Dirichlet $L$-series attached to the nontrivial character modulo $4$. This paper is a good place to begin for thinking about how Riemann's ideas evolved.