Context
I was trying to solve a series: $$\sum_{k=2}^{\infty}\zeta'(k)x^{k-1}$$
Using the Fourier series \begin{equation} \begin{split} \log\Gamma(x)&=\frac{\ln(2\pi)}{2}+\sum_{k=1}^\infty \frac{\cos(2k\pi x)}{2k}+\sum_{k=1}^\infty \frac{\gamma+\ln(2k\pi)}{\pi k}\sin(2k\pi x)\\ B_{2n-1}(x)&=-2(2n-1)!\sum_{k=1}^\infty \frac{\sin(2k\pi x)}{(2k\pi)^{2n-1}}\\ \end{split} \end{equation} and using the generating function of the odd Bernoulli polynomials $$\frac{t}{2}\cdot\frac{\sinh((x-\frac{1}{2})t)}{\sinh(\frac{t}{2})}=\sum_{n=1}^\infty B_{2n-1}(x)\frac{t^{2n-1}}{(2n-1)!}$$ it can be shown that: $$\sum_{k=1}^\infty\zeta'(2k)z^{2k}=\frac{\gamma+\ln(2\pi)}{2}(1-\pi x\cot(\pi z))+\pi^2z^2\csc(\pi z)\int_0^1\sin(\pi t(2z-1))\log\Gamma(t)\mathrm{d}t$$
See this answer for more details.
I wanted to do the similar calculation in the case of $\zeta'(2k+1)$, but it comes naturally to me to use a function of this type: $$\text{Cl}_{2n}(x)=\sum_{k=1}^\infty \frac{\sin(kx)}{k^{2n}}$$ To then get to look for its generating function (beyond the multiplicative constants).
Question
Does anyone have any ideas on how to approach the following series? $$f(t,x)=\sum_{n=1}^\infty \text{Cl}_{2n}(x)\frac{t^{2n}}{(2n)!}$$ Where $\text{Cl}_{\nu}(z)$ is the Clausen function.
