Can Bernoulli polynomials be extended to fractional orders without losing elementarity?

Can Bernoulli polynomials $B_s(x)$ be extended to fractional $s$ in such a way so that for any given $s$ the function $B_s(x)$ still could be expressed in elementary functions of $x$?

edited Aug 25 at 20:20

MTS
5,306●1●9●27

asked Jun 19 at 8:02

Anixx
2,072●1●9●19

You asked this also at math.stackexchange.com/questions/160233/… Please in the extreme case where you must ask in both sites, add mutual links to the posts. – Mariano Suárez-Alvarez Jun 19 at 8:14

I've done this some years ago for the integrals of the Bernoulli-polynomials (which are then the "Faulhaber"-polynomials); the polynomials at fractional s extend to infinitely many terms (according to their connection with binomial-coefficient at fractional orders) and thus become series instead of polynomials. For the general case I'd call then "ZETA-polynomials" as one can express the most natural extension as polynomials/series of binomial coeffients and zeta at noninteger arguments. (This does not exactly answer your question but I think it's sufficiently near) – Gottfried Helms Jun 19 at 9:24

Well, the Bernoulli polynomials can be extended by meant of Hurwitz zeta function, but in that case the non-integer orders would not produce an elementary function. I really doubt that an extension that always gives elementary functions is possible but who knows? – Anixx Jun 19 at 10:15

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