Is there a closed form of $\sum_{k=1}^{\infty}\frac{\sin (2k-1)\pi x}{(2k-1)^5\pi^5}$

Question

Is there a closed form of $$\sum_{k=1}^{\infty}\frac{\sin (2k-1)\pi x}{(2k-1)^5\pi^5}$$

The above sum if a solution of some problem so I need to simplify it to get a clear equation if possible .......thanks

Answer 1

It is pretty well-known (and not difficult to prove) that $$ \sum_{n\geq 0}\frac{\sin((2n+1)\pi x)}{(2n+1)\pi} $$ is the Fourier (sine) series of the rectangle wave which equals $\frac{1}{4}$ over $(0,1)$ and $-\frac{1}{4}$ over $(1,2)$.
By applying termwise integration four times, it follows that $$ \sum_{n\geq 0}\frac{\sin((2n+1)\pi x)}{(2n+1)^5 \pi^5} $$ is a piecewise-quartic polynomial which equals $\color{red}{\frac{x(1-x)(1+x-x^2)}{96}}$ over $[0,1]$ and fulfills $f(x+1)=-f(x)$.

Answer 2

Mathematica gives:

$$-\frac{i e^{-i \text{$\pi $x}} \left(e^{2 i \text{$\pi $x}} \Phi \left(e^{2 i \text{$\pi $x}},5,\frac{1}{2}\right)-\Phi \left(e^{-2 i \text{$\pi $x}},5,\frac{1}{2}\right)\right)}{64 \pi ^5}$$