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).
