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.net2013-05-25T08:41:14Zhttp://mathoverflow.net/feeds/question/93406http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/93406/proof-that-derivative-of-hurwitz-zeta-by-the-first-argument-is-not-expressable-inProof that derivative of Hurwitz Zeta by the first argument is not expressable in terms of Hurwitz ZetaAnixx2012-04-07T11:22:01Z2013-02-09T16:55:23Z
<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>