This is just a modification of the Zeta Function, if it is already present in literature, please link me to it.
$$\mathcal{Z}\left(x\right)=\sum_{n=1}^{\infty}n^{-x}\sin\left(n\right)$$ This is the Conjectured form I was able to numerically get: $$\displaystyle{\boxed{\mathcal{Z}(2k-1)=\sum_{r=1}^{k-1}\left(C_{2r}\frac{\left(-1\right)^{1+k+r}}{\left(2k-2r-1\right)!}\right)\pi^{2r}-\left(\frac{\left(-1\right)^{k}}{2\left(2k-2\right)!}\right)\pi+\left(\frac{\left(-1\right)^{k}}{2\left(2k-1\right)!}\right)}}$$ where, $$C_r=\left(\frac{1}{\pi^r}\sum_{n=1}^{\infty}\frac{1}{n^{r}}\right)$$
Some Values: $$\lim_{x\to\infty}\mathcal{Z}(x)=\sin1$$ $$\mathcal{Z}(1)=\frac{1}{2}\left(\pi\right)-\frac{1}{2}$$ $$\mathcal{Z}(3)=\frac{1}{6}\left(\pi\right)^{2}-\frac{1}{4}\left(\pi\right)+\frac{1}{12}$$ $$\mathcal{Z}(5)=\frac{1}{90}\left(\pi\right)^{4}-\frac{1}{36}\left(\pi\right)^{2}+\frac{1}{48}\left(\pi\right)-\frac{1}{240}$$ $$\mathcal{Z}(7)=\frac{1}{945}\left(\pi\right)^{6}-\frac{1}{540}\left(\pi\right)^{4}+\frac{1}{720}\left(\pi\right)^{2}-\frac{1}{1440}\left(\pi\right)+\frac{1}{10080}$$ $$\mathcal{Z}(9)=\frac{1}{9450}\left(\pi\right)^{8}-\frac{1}{5670}\left(\pi\right)^{6}+\frac{1}{10800}\left(\pi\right)^{4}-\frac{1}{30240}\left(\pi\right)^{2}+\frac{1}{80640}\left(\pi\right)-\frac{1}{725760}$$
The results seem to match numerically.
Can someone prove this?
EDIT:
I just realized the form with $\zeta(2r)$ looks more pleasing:
$$\displaystyle{\boxed{\mathcal{Z}(2k-1)=\sum_{r=1}^{k-1}\left(\zeta(2r)\frac{\left(-1\right)^{1+k+r}}{\left(2k-2r-1\right)!}\right)-\left(\frac{\left(-1\right)^{k}}{2\left(2k-2\right)!}\right)\pi+\left(\frac{\left(-1\right)^{k}}{2\left(2k-1\right)!}\right)}}$$
