Consider : $\displaystyle f(x)= \sum_{n=1}^{\infty} \frac{\sin nx }{n^4}$
Find : $\displaystyle \int_0^{x} f(t)\ \mathrm{d}t$.
Consider : $\displaystyle f(x)= \sum_{n=1}^{\infty} \frac{\sin nx }{n^4}$
Find : $\displaystyle \int_0^{x} f(t)\ \mathrm{d}t$.
Note that, $$ \int_{0}^{x}f(t)\ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\int_{0}^{x}\sin(nt) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\left( \frac{1}{n}-\frac{\cos(nx)}{n} \right) $$
$$ \implies \int_{0}^{x}f(t) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^5}-\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^5} $$
$$\implies f(x)=\zeta(5)-\frac{1}{2}\sum_{n=1}^{\infty} \frac{e^{inx} }{n^5} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{e^{-inx} }{n^5} $$
$$ \implies f(x)=\zeta(5)-\frac{1}{2}( \operatorname{Li}_{5}(e^{ix})+ \operatorname{Li}_{5}(e^{-ix}) ),$$
where $\operatorname{Li}_{s}(z)$ is the Polylogarithm function.