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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?

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    $\begingroup$ Have you ever read Emil Artin's monograph about the gamma function? $\endgroup$ – Harry Gindi Dec 3 '09 at 14:20
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    $\begingroup$ They are both conceptually related to sums of powers. The $\zeta$ function itself is defined as a non-alternating sum of powers for $\Re(z)>1$, and as an alternating sum of powers (times a certain factor) for $\Re(x)\in(0,1)$ On the other hand, geometric shapes of the form $x^n+y^m=1$, called superellipses or Lame curves, are also bounded sums of powers. But by integrating $y=\sqrt[m]{1-x^n}$ or $x=\sqrt[n]{1-y^m}$ on $(0,1)$ we get the multiplicative inverse of the binomial coefficient ${m+n\choose n}={m+n\choose m}$, which is obviously expressible in terms of the $\Gamma$ function. $\endgroup$ – Lucian Jun 1 '14 at 16:52
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    $\begingroup$ For any even function $f$ belonging to the Schwartz space, we have $\widetilde f (s) \zeta(s) = \widetilde{\hat f}(1-s) \zeta(1-s)$, where $\widetilde g$ is the Mellin transform of $g$. Taking $f(y) = e^{-\pi y^2}$ yields the result. $\endgroup$ – Watson Apr 2 '17 at 16:33
  • $\begingroup$ Note that this gamma formula does not complete the whole thing: in particular, you cannot derive the hallowed critical strip from it and the base sum-of-powers definition - which should make sense, as it suggests there is, in a sense, "too much complexity" and "I'm not gonna make it quiiite so eassie on youue" for the gamma function, alone, to capture. In that regard, perhaps, the gamma formula may not be as surprising as one may at first think. $\endgroup$ – The_Sympathizer Apr 18 at 2:15

10 Answers 10

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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.

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    $\begingroup$ That is the content of John Tate's PhD thesis. It is in Cassels & Froehlich - Algebraic Number Theory (Last chapter). In my opinion, this is still the best reference for this matter. $\endgroup$ – Marc Palm Feb 9 '11 at 10:45
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    $\begingroup$ The factor $\pi ^{-s/2}\Gamma (s/2)$ is the Mellin transform of $e^{-\pi x^2}$ which is its own Fourier transform. This yields a functional equation for the theta function whose Mellin transform is the Zeta function (together with the Gamma factor); therefore, this zeta function gets a functional equation. $\endgroup$ – Venkataramana Oct 25 '13 at 14:38
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    $\begingroup$ The expression $\pi^{-s/2} \Gamma(s/2)$ looks similar to the volume of the $s$-dimensional unit ball (say $s$ is an integer). Is this a coincidence? $\endgroup$ – mlbaker Jan 23 '15 at 7:02
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    $\begingroup$ @mlbaker: According to one of Quillen's manuscript in the Clay's math site, up to a factor of 2, the reciprocal of the expression $\pi^{-s/2}\Gamma(s/2)$ is the area of an $(n-1)$-dimensional unit sphere in ${\mathbb R}^n$. That is the archimedean factor of Riemann's completed zeta function. And, using the Euler product expansion of the zeta function, the local Euler factors $(1-p^{-s})^{-1}$ are the $p$-adic areas of the unit sphere in ${\mathbb Q}_p^n$ : these are the non-archimedean factors of the complete zeta function. Of course all of these are for positive integer values of $s$... $\endgroup$ – F Zaldivar Apr 18 at 1:55
  • $\begingroup$ ... There are some extra comments by Quillen in his notes for non integer positive real values of $s$. All of these may be, some way or another, in Tate's thesis, but I don't have it at hand now. $\endgroup$ – F Zaldivar Apr 18 at 1:55
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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.

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    $\begingroup$ So far, this answer has received four upvotes and two downvotes. I am curious about the reason for the downvotes: I thought they were intended for off topic or wrong answers, especially the sort of answers you want to discourage, and I can't see that this answer is either. Perhaps I should have left the second half out of it, since it does not contribute much to the why question, but to me, that doesn't seem sufficient reason for a downvote. $\endgroup$ – Harald Hanche-Olsen Dec 3 '09 at 15:24
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    $\begingroup$ I guess Riemann's opinions about the Riemann zeta function aren't good enough for some people? $\endgroup$ – Ryan Budney Dec 3 '09 at 22:18
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    $\begingroup$ I was one of the downvotes. I'll say, I don't think your answer is wrong or problematic, I just think there's a much better answer, which hasn't been written properly: "Q has a real prime." I'm not familiar enough with the subject to write a good answer like that, but I'm familiar enough with the subject to say I don't think an answer which leaves it out should be at the top. $\endgroup$ – Ben Webster Dec 3 '09 at 23:14
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    $\begingroup$ @Ben Webster: though it is true that it is the real prime of Q that allows for the appearance of the gamma factor, the question was "Why the Gamma function?", not "Why is there another factor?". The real prime could be considered as a reason to have another factor, not a reason for that factor to be the gamma function. Harald's answer illustrates how the gamma function arises in the proof. $\endgroup$ – Rob Harron Dec 3 '09 at 23:40
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    $\begingroup$ @Harald: What Ben means about a real prime is that if you consider the set of equivalence classes of absolute values on Q there is one for each prime p, and the usual absolute value. This leads number theorists to consider the usual absolute value as an "infinite" prime. It is called real as it comes from the embedding of Q into R (whereas finite extensions of Q might embed into C, but not R, and hence have complex primes). The Riemann zeta function can be viewed as an Euler product of factors 1/(1-p^-s) and the gamma factor can be viewed as the factor coming from the infinite prime. $\endgroup$ – Rob Harron Dec 4 '09 at 1:07
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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.

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    $\begingroup$ Ah, Terry Tao's post is clearly the answer I was looking for. $\endgroup$ – Ben Webster Dec 4 '09 at 22:32
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    $\begingroup$ But there is a lot of functions which are their own Fourier transform, isn't it ? why this one in particular ? what would happen with other functions with the same property ? $\endgroup$ – Simon Henry Oct 26 '15 at 9:59
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There may also be some interest in the point of the "local functional equation", namely, that in fact the Gamma function (with the power of $\pi$) is just one (optimized) possibility, and somehow making a suboptimal choice doesn't really matter:

For a Schwartz function $f$ on $\mathbb R$, let $\Gamma(f,s)=\int_{\mathbb R^\times} |x|^s\,f(x)\;{dx\over |x|}$. The usual Gamma factor is obtained by taking a Gaussian. The local functional equation (proven by changing variables in the defining integrals, in the range $0<{\rm Re}(s)<1$, is $$\Gamma(f,s)\cdot \Gamma(\hat{g},1-s)\;=\; \Gamma(\hat{f},1-s)\cdot \Gamma(g,s)$$ for any two Schwartz functions $f,g$. And Riemann's argument proves $$ \Gamma(f,s)\cdot \zeta(s) \;=\; \Gamma(\hat{f},1-s)\cdot \zeta(1-s) $$ for any Schwartz $f$.

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    $\begingroup$ So one can use any fixed point of the Fourier transform for $f$. Any idea what do we get for the appropriate Hermite functions? $\endgroup$ – Vít Tuček Sep 10 '14 at 16:41
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    $\begingroup$ If I'm thinking correctly, $\zeta(s)$ is formally $\Gamma(g,1-s)$ where $g$ is a sum of $\delta$ functions at the positive integers. Is there some sense in which $g = \hat{g}$ here? $\endgroup$ – David E Speyer Sep 10 '14 at 16:47
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    $\begingroup$ I guess that would be the Dirac comb. See mathoverflow.net/a/39187/6818 $\endgroup$ – Vít Tuček Sep 10 '14 at 16:53
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    $\begingroup$ For the record, the Hermite functions yield just a polynomial multiple of the standard Gamma factor. For example $H_{16}$ leads to $2027025 + 16 (-1 + s) s (274455 + 2 (-1 + s) s (27051 + 8 (-1 + s) s (127 + (-1 + s) s)))$ $\endgroup$ – Vít Tuček Sep 10 '14 at 22:46
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    $\begingroup$ Starting from $\Gamma(f,s)\zeta(s) = \Gamma(\widehat{f},1-s)\zeta(1-s)$ and multiplying both sides by $\Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{1-s}{2}\right)$ I obtain, after cancellation of the classical functional equation for $\xi(s)$, the following equation $$ \Gamma\left(\frac{1-s}{2}\right)\Gamma\left(\frac{1+s}{2}\right) = -\imath \frac{\pi}{\sin(\pi s)}, $$ which can't be true since for real $s$ the left hand side is real whereas the right hand side is imaginary. Where did I make mistakes? $\endgroup$ – Vít Tuček Sep 11 '14 at 22:00
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"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.

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  • $\begingroup$ Nice, thank you! I can imagine that this was maybe the first hint, then leading to the connection pointed out by Harald Hanche-Olsen... $\endgroup$ – Peter Arndt Aug 2 '10 at 10:34
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    $\begingroup$ Peter: that was not the first hint. That the zeta-function even makes sense for negative numbers (in a rigorous sense) was first worked out by Riemann through his proof of analytic continuation. Before that there was not a known relation between zeta(s) and zeta(1-s) to motivate using the Gamma-function. Although Euler, long before Riemann, had derived a non-rigorous formula that is equivalent to the functional equation of the zeta-function just at integers, I don't think it was something that influenced Riemann's work which brought in the Gamma-function explicitly. $\endgroup$ – KConrad Mar 10 '11 at 5:45
  • $\begingroup$ I posted an MSE question asking about Riemann's thinking on symmetrizing the functional equation, math.stackexchange.com/questions/143449/…. $\endgroup$ – Tom Copeland May 10 '12 at 23:03
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Multiple answers and comments have already pointed out that the conceptual role of $\pi^{-s/2}\Gamma(s/2)$ comes from the viewpoint of Iwasawa and Tate, which for $\text{Re}(s) > 1$ creates this function as $\int_{\mathbf R^\times} e^{-\pi x^2}|x|^s\,dx/|x|$, an integral over the multiplicative group $\mathbf R^\times$ of the function $e^{-\pi x^2}$ that is self-dual for the Fourier transform on the additive group $\mathbf R$ relative to the self-duality $\langle x,y\rangle = e^{2\pi ixy}$ or $\langle x,y\rangle = e^{-2\pi ixy}$ on $\mathbf R$. (If we use another self-duality of $\mathbf R$ then $e^{-ax^2}$ would be self-dual for some $a \not= \pi$ instead.)

It's also been said elsewhere on this page that there are many self-dual Schwartz functions on $\mathbf R$, or more specifically many even self-dual Schwartz functions on $\mathbf R$: for Schwartz $f$ on $\mathbf R$ and $\text{Re}(s) > 0$, we have $\int_{\mathbf R^\times} f(x)|x|^s\,dx/|x| = \int_{0}^\infty (f(x) + f(-x))x^s\,dx/x$ and this is $0$ when $f$ is odd, so we may as well assume $f$ is even since $f(x) + f(-x)$ is even anyway and we want to avoid the silly equation $0=0$ even if it is a valid equation.

For arbitrary Schwartz $f$ on $\mathbf R$, set $\Gamma_f(s) = \int_{0}^\infty f(x)x^s\,dx/x$, which is a mild modification of the function $\Gamma(f,s)$ in Paul Garrett's answer (his $\Gamma(f,s)$ is my $\Gamma_{f(x)+f(-x)}(s)$ by a formula I wrote in the previous paragraph). This function converges absolutely and is analytic for $\text{Re}(s) > 0$, and it extends meromorphically to $\mathbf C$ by repeated integration by parts (the same way the $\Gamma$-function can be extended to $\mathbf C$ from its integral definition for $\text{Re}(s) > 0$), and Tate's thesis shows there is a general functional equation $\Gamma_f(s)\zeta(s) = \Gamma_{\hat{f}}(1-s)\zeta(1-s)$ where $\hat{f}$ is the Fourier transform of $f$ (for the self-duality on $\mathbf R$ given by $\langle x,y\rangle = e^{-2\pi ixy}$), so if $f$ is self-dual then we get $$\Gamma_f(s)\zeta(s) = \Gamma_{f}(1-s)\zeta(1-s),$$ a very nice functional equation indeed, especially if we use even $f$ to avoid $0 = 0$.

All of what I wrote so far has appeared explicitly or implicitly in some of the other comments or answers. Since there are many self-dual even Schwartz functions $f$ on $\mathbf R$, what is it about the choice $f(x) = e^{-\pi x^2}$, leading to $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$ (an extra $1/2$ on both sides of the functional equation can be cancelled) that is so nice? I have not seen the following property pointed out yet: with this choice of $f$ and familiarity with the $\Gamma$-function we know $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (in fact for $\text{Re}(s) > 0$), so therefore $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) > 1$ from $\zeta(s)$ being nonvanishing there, and then by the functional equation $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) < 0$, which means all zeros of $\Gamma_f(s)\zeta(s)$ have $0 \leq \text{Re}(s) \leq 1$. If you want to use a totally random even Schwartz function for $f$ in order to define a factor $\Gamma_f(s)$ that completes the Riemann zeta-function, you will get the nice-looking nontrivial functional equation displayed above but how are you going to use $\Gamma_f(s)\zeta(s)$ to analyze the location of zeros of $\zeta(s)$ (including discovering its trivial zeros, whether or not you consider those important) if you do not know where $\Gamma_f(s)$ has its zeros and poles?

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually know the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)? The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders at most $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$. Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the zeros of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are all simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $G(s) := \Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2))$ is an entire function. Thus $\Gamma_f(s) = G(s)\pi^{-s/2}\Gamma(s/2)$ with $G$ entire, so $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. The exponential factor $\pi^{-s/2}$ was kind of irrelevant to drag through the calculation since it has no zeros or poles, but it's traditionally seen alongside $\Gamma(s/2)$ so I used it. This addresses comments below by Will Sawin and Venkataramana.)

Example: the function $f(x) = 1/(e^{\pi x} + e^{-\pi x})$ is an even self-dual Schwartz function on $\mathbf R$. Can someone determine in a self-contained way (i.e., not using $\zeta(s)$) where $\Gamma_f(s)$ has its zeros on $\mathbf C$, or determine if it has no zeros?

Edit: Ignoring the wacky example just above, in some comments below I work out an example with $f(x)$ being a 4th degree Hermite polynomial times a Gaussian and find that $\Gamma_f(s)$ has two zeros with positive real part, at $s = (1\pm \sqrt{-2})/2$.

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    $\begingroup$ (+1) for emphasizing that self-duality is not special to the Gaussian. One could also add the following complete characterization of self-duality. Hermite functions (Gaussian times Hermite polynomials) form a Schauder basis of Schwartz space which is also an eigenbasis for the Fourier transform with four eigenvalues corresponding to the four roots of unity. I wonder if anyone computed $\Gamma_f$ for one of these more general $f$'s in the $\lambda=1$ eigenspace. $\endgroup$ – Abdelmalek Abdesselam Apr 17 at 16:30
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    $\begingroup$ @KConrad We should be able to show that given any even self-dual Schwartz function, its Mellin transform has the form $ g( s (1-s)) \Gamma (s)$ for some holomorphic $g$. If $g$ has no zeroes, it is the exp of some function of $s (1-s)$, which if it is nonconstant has growth order 2, thus grows faster than the gamma function. If $g$ is constant, we can do an inverse Mellin transform. So the Gaussian/ Gamma pair is optimal in some precise sense... $\endgroup$ – Will Sawin Apr 18 at 0:20
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    $\begingroup$ The issue of finding a function whose integral is the gcd of integrals of all possible functions in some class reoccurs in the theory of (local) $L$-functions of (higher rank) automorphic forms. For the standard $L$-function, at non-archimedean places , newforms fill this role. $\endgroup$ – Will Sawin Apr 18 at 12:37
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    $\begingroup$ @paulgarrett I found Lachaud's paper (MR2022610) and it was in 2003, not 2009. He says on the top of p. 182 that for all Schwartz $f$, the Mellin transform $\Gamma_{f(x)+f(-x)}(s)$ is nonvanishing for $\text{Re}(s) = 1/2$. Do you know such a general nonvanishing result? Determining where $\Gamma_f(s)$ is $0$ seems subtle. Moreover, by calculations I mentioned in earlier comments, if $F(x) = (8b^2x^4−24bx^2+6)e^{-(b/2)x^2}$ for either self-duality $\langle x,y\rangle = e^{\pm ibxy}$, then $\Gamma_{F(x) + F(-x)}(s) = \Gamma_{2F}(s)$ has zeros $1/2 \pm \sqrt{2}i/2$, contradicting that claim. $\endgroup$ – KConrad Apr 18 at 20:47
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    $\begingroup$ @paulgarrett since you write that you have not seen Muntz's 1922 paper cited much, another place is Albeverio and Cebulla, "Müntz formula and zero free regions for the Riemann zeta function" (MR2285583). $\endgroup$ – KConrad Apr 18 at 20:54
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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.

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  • $\begingroup$ Related to mathoverflow.net/questions/58004/…. $\endgroup$ – Tom Copeland May 19 '12 at 15:33
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    $\begingroup$ Actually, not quite. Define $\eta(z)=(1-2^{1-z})\zeta(z)=\sum_{n\ge 1}\frac{(-1)^{n+1}}{n^z}$. This sum converges (conditionally) when Re$\,z>0$, thus $\eta$ is defined in the same half-plane (modulo considerations for Re$\,z=1$. The functional equation for $\zeta$ leads to a functional equation for $\eta$. The latter makes sense without complex analysis since $\eta(s)$ and $\eta(1-s)$ are both defined if $0<s<1$. This functional equation was already published by Euler! See: 1) E. Landau: Euler und die Funktionalgleichung der Riemannschen Zetafunktion. 2) A. Weil: Prehistory of the zeta-func. $\endgroup$ – M Mueger Nov 11 '15 at 22:01
  • $\begingroup$ See also math.stackexchange.com/questions/143449/… $\endgroup$ – Tom Copeland Jan 5 '17 at 21:21
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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.

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Gamma function arises when we consecutively differentiate an Appell sequence. An example of Appell polynomials are Bernoulli polynomials. When we differentiate it, the factors combine with themselves:

$$B_n'(x)=nB_{n-1}(x)$$

$$B_n''(x)=n(n-1)B_{n-2}(x)$$

$$B_n'''(x)=n(n-1)(n-2)B_{n-3}(x)$$

They are just another name for Hurwitz Zeta function:

$$B_n(x) = -n \zeta(1-n,x)$$

Thus, for $f(s,q)=\zeta(s,-q)$

$$\frac\partial{\partial q}f(s,q)= s f(s+1,q)$$

$$\frac{\partial^2}{\partial q^2}f(s,q)= s(s+1) f(s+2,q)$$

$$\frac{\partial^3}{\partial q^3}f(s,q)= s(s+1)(s+2) f(s+3,q)$$

Since Reihmann zeta is Hurwitz zeta evaluated at $q=1$, the expression you give is apparently consecutive derivative of Hurwitz Zeta, with factor $\pi^{-s}$ appearing if we normalize Hurwitz Zeta by stretching it horizontally by factor of pi.

Consecutive derivatives of Hurwitz Zeta in turn are nothing more than just polygamma function.


For instance, here is the function $-1/x$:

enter image description here

If we add infinitely many similar functions with a shift of pi/2 each in both directions, we get $\tan x$. But if we do the same only in one direction, we get "incomplete tangent":

http://storage7.static.itmages.ru/i/14/0910/h_1410326921_7988832_91f3fd7d7d.png

The yellow one is $\operatorname{pg}(x)=\frac 1\pi \psi (\frac x\pi)$, the blue one is $\operatorname{cpg}(x)=-\frac 1\pi \psi (1-\frac x\pi)$. They obey $\operatorname{cpg}(x)+\operatorname{pg}(x)=-\cot(x)$.

Now if we differentiate cpg(x) we get:

$$(\operatorname{cpg}(x))^{(s-1)}=\pi^{-s}\Gamma(s)\zeta(s,1-\frac x\pi)$$

Compare it with yours formula:

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

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Although there is already an answer of mine, I want to add another answer.

This is TL;DR.

Short answer. This is because logarithmic function lacks factorial in its Taylor expansion.

Medium answer. Riemann's functional equation links exponential and trigonometric functions with logarithms and inverse trigonometric. It contains everything what you need to make an exponent from a logarithm.

Long answer.

This is Taylor series for logarithm:

$$\ln(z+1)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+\frac{z^7}{7}-\frac{z^8}{8}+\frac{z^9}{9}-\frac{z^{10}}{10}+O\left(z^{11}\right)$$

This is Taylor series for exponent:

$$\exp (z)-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+\frac{z^6}{6!}+\frac{z^7}{7!}+\frac{z^8}{8!}+\frac{z^9}{9!}+\frac{z^{10}}{10!}+O\left(z^{11}\right)$$

What should we add to the former to get the later? Well, we have to add the factorial and remove the counter from the denominator.

Consider such algebraic element $\omega_+$ (not a real number) on which a function "standard part" is implemented in such a way, that $\operatorname{st} \omega_+^n=B_n^*$ where $B_n^*$ are Bernoulli numbers (with $B_1^*=1/2$), or more generally, $\operatorname{st}\omega_+^x=-x\zeta(1-x)$.

Now consider the function

$$\frac{z}{2\pi} \log \left(\frac{\omega _+-\frac{z}{2 \pi }}{\omega _++\frac{z}{2 \pi }}\right)$$ Its Taylor series is

$$-\frac{z^2}{2 \left(\pi ^2 \omega _+\right)}-\frac{z^4}{24 \left(\pi ^4 \omega _+^3\right)}-\frac{z^6}{160 \left(\pi ^6 \omega _+^5\right)}-\frac{z^8}{896 \left(\pi ^8 \omega _+^7\right)}-\frac{z^{10}}{4608 \left(\pi ^{10} \omega _+^9\right)}+O\left(z^{11}\right)$$

Following Riemann's functional equation and our definition, we have:

$$\operatorname{st}\omega_+^{-x}=\operatorname{st}\frac{-\omega_+^{x+1} 2^x\pi^{x+1}}{\sin(\pi x/2)\Gamma(x)(x+1)}$$

So we can substitute the negative powers of $\omega_+$ with positive powers without changing the standard part of the whole expression.

The non-zero terms are

$$\frac{2 \left(-\frac{1}{2 \pi }\right)^n \left(-\omega _+\right){}^{1-n}}{n-1}$$

and after substitution we have

$$\frac{\omega _+^n \sec \left(\frac{\pi n}{2}\right)}{\Gamma (n+1)}$$

The resulting series is

$$\frac{1}{2} \omega _+^2 z^2+\frac{1}{24} \omega _+^4 z^4-\frac{1}{720} \omega _+^6 z^6+\frac{\omega _+^8 z^8}{40320}-\frac{\omega _+^{10} z^{10}}{3628800}+O\left(z^{11}\right)$$

oh, wait... is not it similar to

$$\cos \left(\omega _+ z\right)=1-\frac{1}{2} \omega _+^2 z^2+\frac{1}{24} \omega _+^4 z^4-\frac{1}{720} \omega _+^6 z^6+\frac{\omega _+^8 z^8}{40320}-\frac{\omega _+^{10} z^{10}}{3628800}+O\left(z^{11}\right)$$

Well, we got:

$$\operatorname{st}\frac{z}{2 \pi } \log \left(\frac{\omega _+-\frac{z}{2 \pi }}{\omega _++\frac{z}{2 \pi }}\right)=\operatorname{st}(\cos \left(\omega _+ z\right)-1)$$

In a similar way one can establish other impressive relations:

$$\operatorname{st}(\exp \left(\omega _+ z\right)-\omega _+ z-1)=\operatorname{st}\frac{i z}{2 \pi } \log \left(\frac{\omega _+-\frac{i z}{2 \pi }}{\omega _++\frac{i z}{2 \pi }}\right)$$

$$\operatorname{st}\cos \left(\omega _+ z\right)=\operatorname{st}\frac{ z}{2 \pi } \log \left(\frac{\omega _+-\frac{ z}{2 \pi }}{\omega _-+\frac{ z}{2 \pi }}\right)$$

$$\operatorname{st}\cosh \left(\omega _+ z\right)=\operatorname{st}\frac{i z}{2 \pi } \log \left(\frac{\omega _+-\frac{i z}{2 \pi }}{\omega _-+\frac{i z}{2 \pi }}\right)$$

(where $\omega_-=\omega_+-1$).

In other words, Riemann's functional equation is a direct bridge that connects exponential function to logarithm, trigonometric functions to inverse trigonometric, transforming each term of the series separately.

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