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I'm just curious, has anyone ever proved that the Riemann Zeta function is not an elementary function?

Here I am using the term "elementary" in the sense of Liouville or as defined in this paper. Roughly speaking, "elementary" means "can be built up from the rational functions using a finite amount of logarithms or exponentials".

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Hilbert and followers proved that the Riemann zeta function could not be solution of an algebraic differential equation (as opposed to elementary functions).

The idea is shortly that the meromorphic Riemann zeta function satisfies its famous functional equation involving the Gamma function :$$\tag{1}\zeta(z) = 2^z\pi^{z-1}\sin\frac{\pi z}{2} \;\Gamma(1-z)\;\zeta(1-z)$$ Supposing that $\zeta$ is differentially algebraic would clearly imply that the $\Gamma$ function is also differentially algebraic (by isolating $\Gamma(1-z)$ in $(1)\,$) but this was proved false by Hölder (see the detailed proofs in the references). $$-$$ Gamma and zeta are not solution of algebraic differential equations (and so "hypertranscendental functions") while elementary functions and their integrals (recursively) define the "Liouvillian functions" which are all solutions of algebraic differential equations (this includes many special functions like the exponential and logarithmic integral, the error function and so on).

There are also solutions of algebraic differential equations which are not Liouvillian (if we exclude the special cases) : Bessel functions, hypergeometric functions and their generalizations like Meijer G-functions.

Ref:

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  • $\begingroup$ Well done, thanks $\endgroup$ – ASKASK Sep 7 '15 at 1:13
  • $\begingroup$ Glad it helped @ASKASK! $\endgroup$ – Raymond Manzoni Sep 7 '15 at 6:44

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