Question

Defining $$\xi(s) := \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)$$ yields $\xi(s) = \xi(1 - s)$ (where $\zeta$ is the Riemann Zeta function).

Is there any conceptual explanation - or intuition, even if it cannot be made into a proof - for this? Why of all functions does one have to put the Gamma-function there?

Whoever did this first probably had some reason to try out the Gamma-function. What was it?

(Best case scenario) Is there some uniform way of producing a factor out of a norm on the rationals which yields the other factors for the p-adic norms and the Gamma factor for the absolute value?

Answer 1

To the best of my understanding, the answer is yes, and this uniform way consists of some integration over the local field. This is explained in John Tate's dissertation. One starts with a certain smooth rapidly decreasing function, for which one takes the characteristic function of the p-adic integers in the nonarchimedean case and the function $e^{-|x|^2}$ for an archimedean field. This is being multiplied with $|x|^s$ (approximately) and integrated over the Haar measure of the additive group of the field. This produces the $\Gamma$-factor for an archimedean field and $(1-p^{-s})^{-1}$ for a p-adic field.

Answer 2

One way to get started is to look at the integral for the gamma function: $$\Gamma(s)=\int_0^\infty t^{s-1} e^{-t}\,dt$$ Subsitute $t=nx$ in the integral to arrive at $$\frac{\Gamma(s)}{n^s}=\int_0^\infty e^{-nx}x^{s-1}\,dx$$ which we then sum up to get $$\Gamma(s)\zeta(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx$$ which already shows that there is some connection between the gamma and zeta functions, and it does in fact allow us to extend the definition of the zeta function into the critical strip.

What comes next is far less obvious, but the idea is to introduce a branch cut for $x^{s-1}$ along the positive real axis, and to replace the above integral by one running from $+\infty$ along the bottom of the positive real axis, around the origin, and back to $+\infty$ along the top of the real axis. This introduces an extra factor $1-e^{2\pi i s}$. Now start expanding the circle around the origin, taking account of the poles of the integrand along the imaginary axis as we go, and end up with $$\Gamma(s)\zeta(s)=(2\pi)^{s-1}\Gamma(1-s)\sin(\tfrac12\pi s)\zeta(1-s).$$ From there, some cleanup still remains. As I said, this is not terribly intuitive, so it doesn't answer your question, but the first paragraph should at least give you a notion how the gamma and zeta functions are interrelated.

Answer 3

As has been explained above, the zeta function has a factor for each completion of $\mathbb{Q}$. The factor at $\mathbb{R}$ has to do with integrating $e^{- \pi x^2}$ and the factor at $\mathbb{Q}_p$ has to do with integrating the characteristic function of $\mathbb{Z}_p$.

Some people might wonder why these two functions were chosen. The answer is simple: they are both their own Fourier transforms.

Also, I don't think anyone has recommended Terry Tao's expository post on this material yet. It is quite good.

Answer 4

Whoever did this first probably had some reason to try out the Gamma-function. What was it?

The first one to do this was, precisely, Riemann in his famous (and 150 years old) paper: Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse. There he proved the functional equation as well, with the method that Harald explained above.

Answer 5

"Why of all functions does one have to put the Gamma-function there?"

zeta(s) has trivial zeroes at -2, -4, -6, etc. zeta(1-s) thus has trivial zeroes at s=3, 5, 7, etc - a completely different set of zeroes.

To make a reflection formula where zeta(s) is somehow equal zeta(1-s), you have to get rid of the two differing sets of trivial zeroes. Multiplying by the gamma is perfect for this since its poles will cancel out those zeroes. For example, gamma(s/2) has poles at 0, 2, 4, 6, etc. and should go with zeta(s). gamma((1-s)/2) has poles at s=1, 3, 5, etc. and should go with zeta(1-s).

It's possible to prove that gamma is the right choice, but Euler no doubt discovered that gamma is the right function through numerical experimentation - when he discovered the zeta reflection formula like 250 years ago.

Answer 6

I'm not sure of the history of the gamma factor, though I would suggest that no one "tried it out", but rather it simply arose in trying to prove of the functional equation. Riemann was the first to prove the functional equation, and his proof essentially follows that in Harald Hanche-Olsen's answer, which makes my explanation plausible. Alternatively, the functional equation of the zeta function comes out of the functional equation of a theta series, and the Mellin transform of a theta series gives rise to a Gamma function. This latter explanation arises more naturally for modular forms: the L-function of a modular form is also completed by a gamma factor to obtain a functional equation; in this case, the completed L-function is simply the Mellin transform of the modular form itself.

Furthermore, as Leonid Positselski answers, it is indeed true that Tate's thesis provides a uniform way of obtain the gamma factors at infinity in the same manner as one obtains the local L-factors at finite places.

More generally, there is a recipe given an arbitrary motive for the expected gamma factors that should give a functional equation for the motivic L-functions. These are due to Deligne and Serre (I believe) and are determined by the Hodge structure of the motive (see Deligne's corvallis article "Valeurs de fonctions L..."). This shows that there's a uniform way of obtaining the gamma factors as one varies the L-function one is studying, an orthogonal question to the one Leonid Positselski answered.