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The Riemann $\zeta$-function is here with analytical extension (e.g. with her functional equation).

Definition for $|x|<1$ and $-t\in\mathbb{R}\setminus\mathbb{N}$: \begin{align*} B_t(x+1):=&-\frac{2\Gamma(1+t)}{(2\pi)^t}\cos(\frac{\pi t}{2}) \sum\limits_{k=0}^\infty (-1)^k \frac{(2\pi x)^{2k}}{(2k)!}\zeta(t-2k) \\ &-\frac{2\Gamma(1+t)}{(2\pi)^t}\sin(\frac{\pi t}{2}) \sum\limits_{k=0}^\infty (-1)^k \frac{(2\pi x)^{2k+1}}{(2k+1)!}\zeta(t-1-2k) \end{align*}

Note: It’s possible to use complex variables but it’s not necessary here.

For $t\in\mathbb{N}_0$ and $x\in\mathbb{R}$ one gets the Bernoulli polynomials.

It’s not difficult to proof, that $$\frac{\partial}{\partial x}B_t(x+1)=t B_{t-1}(x+1)\,.$$

But how can one proof $$B_t(x+1)=B_t(x)+t x^{t-1}$$ for $t\geq 1$ and $x\in(-1;0)\,$?

(The consequence is a generalization of the Bernoulli polynomials $B_t(x)$ in relation to the index with $\,(1)\, t\geq 1\,$ and $x\in\mathbb{R}$ or $\,(2)\, -t\in\mathbb{R}\setminus\mathbb{N}$ and $x>0$.)


Application example for $B_t(x)\,$:

In combination with the Polylogarithm function ( https://en.wikipedia.org/wiki/Polylogarithm ) $$Li_s(z):=\sum\limits_{k=1}^\infty \frac{z^k}{k^s}$$ and the formula $$Li_s(e^\mu)=\Gamma(1-s)(-\mu)^{s-1}+\sum\limits_{k=0}^\infty \frac{\zeta(s-k)}{k!}{\mu}^k$$ for (complex $\mu$) $|\mu|<2\pi$ and complex $s\ne 1,2,3,...$ one gets (here with real $x$ and $t$) $$\Re(\sum\limits_{k=1}^\infty \frac{e^{i2\pi kx}}{(ik)^t})=\frac{(2\pi)^t}{2\Gamma(1+t)}B_t(x)$$ for $|x|<1$ and $t>0$ which is a generalization of the Fourier expansion of the Bernoulli polynomials. For derivations with respect to $t$ is $\ln(ik)=i\frac{\pi}{2}+\ln k$ (using the main branch of the logarithm).

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Your definition seems somewhat complicated. Why don't you define your generalized Bernoulli polynomials via generaliation of the relation to the Hurwitz zeta function $B_n(x)=-n\zeta(1-n,x)?$ If you define $$B_t(x)=-t\zeta(1-t,x)$$ then you get from the well-know $\zeta$ property (http://dlmf.nist.gov/25.11.E4) $$\zeta(1-t,x)=\zeta(1-t,x+1)+\frac{1}{x^{1-t}}$$ $$\zeta(1-t,x+1)=\zeta(1-t,x)-\frac{1}{x^{1-t}}$$ $$-t\zeta(1-t,x+1)=-t\zeta(1-t,x)+t\frac{1}{x^{1-t}}$$ and finally $$B_t(x+1)=B_t(x)+tx^{t-1}$$ PS: Here is a reference to generalized Bernoulli polynomial of complex index: P.L. Butzer, M. Hauss, M. Leclerc, Bernoulli numbers and polynomials of arbitrary complex indices, Applied mathematics letters, 1992 available via ResearchGate

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  • $\begingroup$ Thank you very much for your informations. With the definition of $B_t(x)$ you use the proof is possible (thanks). But it still remains to show, that the two different types of definitions are equivalent. The literatur researchgate.net/publication/… is an interesting list of theorems, but I don't see any proof or base for a proof. Theorem 1 and 7 are useful. $\endgroup$
    – user90369
    Sep 2, 2016 at 9:05
  • $\begingroup$ Maybe 25.11.9 of dlmf.nist.gov/25.11.E4 will help. Thanks again. $\endgroup$
    – user90369
    Sep 2, 2016 at 9:23

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