Proof that derivative of Hurwitz Zeta by the first argument is not expressable in terms of Hurwitz Zeta - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:41:14Z http://mathoverflow.net/feeds/question/93406 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93406/proof-that-derivative-of-hurwitz-zeta-by-the-first-argument-is-not-expressable-in Proof that derivative of Hurwitz Zeta by the first argument is not expressable in terms of Hurwitz Zeta Anixx 2012-04-07T11:22:01Z 2013-02-09T16:55:23Z <p>The set of elementary functions is defined so that it to be closed against operation of differentiation. It is also evidently close against discrete differentiation.</p> <p>In the discrete calculus there is a similar set of functions, but with one sufficient difference. It appears not to be closed against normal (non-discrete) differentiation.</p> <p>But I need a proof.</p> <p>So I am asking for a proof for the following statement regarding Hurwitz Zeta: $$\frac{d}{dq}\zeta(q,p)$$ cannot be expressed in terms of elementary functions and Hurwitz Zeta.</p> <p><strong>UPDATE</strong></p> <p>I found the following formula which connects the two functions, but still a question remains whether one of them can be expresses explicitly.</p> <p>$\zeta '\left(z,\frac{q}{2}\right)-2^z \zeta '(z,q)+\zeta '\left(z,\frac{q+1}{2}\right)=\zeta(z,q)2^{z}\ln 2$</p>