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Wikipedia's article for the Selberg class tell us that the Riemann's Zeta function belongs to the class, thus satisfies the Definition.

I would like to know the details using this terminology of the axiom $3$ taking as specialization our Riemann's Zeta function $\zeta(s)$. Note also that seems that such axiom presumes the unicity of such representation (and of course I am agree with the axiom , but if you can clarify the meaning of this axiom with respect the unicity feel free to add it, if make sense this discussion).

Question. Can you prove rigurously that using the terminology explained in previous Wikipedia's article, the Riemann's Zeta function satisfies that has a functional equation? That is, can you evidence rigurously that the Riemann's Zeta function satisfies a functional equation in the spirit of the Selberg class? If you prefer provide us only hints to get such comparison between the functional equation and this description in the language of the Selberg class, free feel to do it as an answer. Thanks in advance.

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  • $\begingroup$ I'm not really sure what you are asking. Are you asking if RZ has a functional equation in that form? Because it does, not to mention it would be a very pointless class of functions otherwise... $\endgroup$ – Ali Caglayan Mar 14 '17 at 11:42
  • $\begingroup$ From my ignorance: the $\zeta(s)$ satisfies the axiom $3$, thus has a functional equation. What am I asking is that you explain it, in the spirit of this Wikipedia's article. I hope that my words has meaning. Many thanks for your help @AliCaglayan I don't understand this theory thus I don't know if my question is right. I am interested in learn more about it. $\endgroup$ – user243301 Mar 14 '17 at 11:45
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See the proofs of the functional equation (you need to understand the Fourier series and the Fourier transform).

For example thisone (Démonstration) $$(1-2^s)\zeta(s) = -s\int_0^\infty (\{x\}-\{2x\}) x^{-s-1}dx \quad \text{and} \quad \{x\}-\{2x\} = \sum_{n=1}^\infty (-1)^n\frac{\sin( \pi n x)}{\pi n} $$ from which $$(1-2^s)\zeta(s) = -s \sum_{n=1}^\infty (-1)^n\int_0^\infty \sin(2 \pi n x) x^{-s-1} dx \\ = -s \sum_{n=1}^\infty (-1)^n n^{s-1}\int_0^\infty \sin(2 \pi x) x^{-s-1} dx = -s (1-2^s)\zeta(1-s)\int_0^\infty \sin(2 \pi x) x^{-s-1} dx$$

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  • $\begingroup$ And Wolfram Alpha tell me that the last integral is $$-(2\pi)^s\sin\left(\frac{\pi s}{2}\right)\Gamma(-s)$$ when $-1<\Re s<1$, then one simplifies the factor $(1-2^s)$ and I believe that one get the functional equation for the critical band. But what am I asking if about the comparison of this functional equation and the description as an axiom of the Selberg class. Any case many thanks you are a great professor. $\endgroup$ – user243301 Mar 14 '17 at 16:29
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    $\begingroup$ @user243301 That is to say $\int_0^\infty \sin(x)x^{-s-1}dx$ is just a $\Gamma$ factor, and all the $\Gamma$ factors in the Selberg class are of this form. What's important is to understand the functional equation comes from a self-similarity of $\{x\}$ under the Fourier series / Fourier transform, and the same for any Dirichlet series in the (extended) Selberg class. $\endgroup$ – reuns Mar 14 '17 at 16:41
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    $\begingroup$ Hence this isn't related to the Euler product, and many Dirichlet series have a functional equation but no Euler product nor Riemann hypothesis (for example $\sum_{n=1}^\infty e^{i a n} n^{-s}$) $\endgroup$ – reuns Mar 14 '17 at 16:43
  • $\begingroup$ PLease what's means then that the ...from a self-similarity of the fractional part under the Fourier transform? $\endgroup$ – user243301 Mar 14 '17 at 16:44
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    $\begingroup$ @user243301 the coefficients of $\zeta(s+1) = \sum_{n=1}^\infty a_n n^{-s}$ are $a_n = \frac{1}{n}$ the same as $\{x\} =\frac{1}{2}- \sum_{n=1}^\infty a_n \frac{\sin( 2 \pi n x)}{\pi}$. This is the necessary and sufficient condition for having such a functional equation $\endgroup$ – reuns Mar 14 '17 at 16:52

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